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4
MOTION CHAPTER IN MORE THAN ONE DIMENSION
Problem
3. An object is moving in the x direction at 1.3 m/s when it is subjected to an acceleration given by a = 0.52 m/s2 . What is its velocity vector after 4.4 s of acceleration?
ActivPhysics can help with these problems: All Activities in Section 3, Projectile Motion Section 4-1: Velocity and Acceleration Problem
1. A skater is gliding along the ice at 2.4 m/s, when she undergoes an acceleration of magnitude 1.1 m/s2 for 3.0 s. At the end of that time she is moving at 5.7 m/s. What must be the angle between the acceleration vector and the initial velocity vector?
Solution
From Equation 4-3, v = v0 + at = (1.30 m/s) + 2 (0.52 m/s )(4.4s) = (1.30+ 2.29 m/s. )
Problem
4. An airliner is ying at a velocity of 260 m/s, when a wind gust gives it an acceleration of 0.38+ 2 0.72 m/s for a period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been deected from its original course?
Solution
For constant acceleration, Equation 4-3 shows that the vectors v0 , at and v form a triangle as shown. The 2 law of cosines gives v 2 = v0 + (at)2 2v0 at cos(180 0 ). When the given magnitudes are substituted, one can solve for 0 : (5.7 m/s)2 = (2.4 m/s)2 + (1.1 m/s2 )2 (3.0 s)2 + 2(2.4 m/s) 2 (1.1 m/s )(3.0 s) cos 0 , or cos 0 = 1.00 (exactly), and 0 = 0 . Since v0 and a are colinear, the change in speed is maximal.
Solution
(a) Equation 4-3 gives v = 260 m/s + (0.38+ 2 0.72 )(m/s )(24 s) = (269 + 17.3 m/s. (b) Since v0 ) is along the x-axis, the angular deection is just tan1 (vy /vx ) = tan1 (17.3/269) = 3.67 .
Section 4-2: Constant Acceleration Problem
5. The position of an object as a function of time is given by r = (3.2t + 1.8t2 ) + (1.7t 2.4t2 ) m, where t is the time in seconds. What are the magnitude and direction of the acceleration?
Problem 1 Solution.
Solution Problem
2. In the preceding problem, what would have been the magnitude of the skaters nal velocity if the acceleration had been perpendicular to her initial velocity? One can always nd the acceleration by taking the second derivative of the position, a(t) = d2 r(t)/dt2 . However, collecting terms with the same power of t, one can write the position in meters as r(t) = (3.2 + 1.7 + (1.8 2.4 2 . Comparison with )t )t Equation 4-4 shows that this represents motion with constant acceleration equal to twice the coecient of 2 the t2 term, or a = (3.6 4.8 m/s . The magnitude ) 2 + (4.8)2 m/s2 = and direction of a are (3.6) 2 4.49 m/s , and tan1 (4.8/3.6) = 307 (CCW from the x-axis, in the fourth quadrant) or 53.1 (CW from the x-axis).
Solution
When v0 and at are perpendicular, Equation 4-3 and 2 the Phythagorean theorem imply v = v0 + (at)2 = (2.4 m/s)2 + (1.1 m/s )(3.0 s)2 = 4.08 m/s.
2
22
CHAPTER 4 displacement during the rocket ring, not the pathlength of its trajectory.)
Problem
6. An airplane heads northeastward down a runway, accelerating from rest at the rate of 2.1 m/s2 . Express the planes velocity and position at t = 30 s in unit vector notation, using a coordinate system with x-axis eastward and y-axis northward, and with origin at the start of the planes takeo roll.
Problem
8. An object is moving initially in the x direction at 4.5 m/s, when an acceleration is applied in the y direction for a period of 18 s. If it moves equal distances in the x and y directions during this time, what is the magnitude of its acceleration?
Solution
Since the acceleration is constant and the airplane starts from rest (v0 = 0) at the origin (r0 = 0), Equations 4-3 and 4-4 give v(t) = at and r(t) = 1 at2 . 2 Both vectors are in the NE direction, parallel to 2 2 a = (2.1 m/s )( cos 45 + sin 45 ) = (1.48 m/s ) 2 ( + Thus, v(30 s) = (1.48 m/s )(30 s)( + = ). ) 1 (44.5 m/s)( + and r(30 s) = 2 (1.48 m/s)(30 s)2 ), ( + = (668 m) ( + (Note: ( + ) ). )/ 2 is a unit vector in the NE direction.)
Solution
The x component of the displacement is due only to the initial velocity, x = v0x t. The y component is 1 just due to the acceleration, y = 2 at2 = x. Thus, 2 a = 2v0x /t = 2(4.5 m/s)/18 s = 0.5 m/s .
Problem
9. A hockey puck is moving at 14.5 m/s when a stick 2 imparts a constant acceleration of 78.2 m/s at a 90.0 angle to the original direction of motion. If the acceleration lasts 0.120 s, what is the magnitude of the pucks displacement during this time?
Problem
7. An asteroid is heading toward Earth at a steady 21 km/s. To save their planet, astronauts strap a giant rocket to the asteroid, giving it an acceleration of 0.035 km/s2 at right angles to its original motion. If the rocket ring lasts 250 s, (a) by what angle does the direction of the asteroids motion change? (b) How far does it move during the ring?
Solution
Take the x-axis in the direction of the initial velocity, v0 = 14.5 m/s, and the y-axis in the direction of the 2 acceleration, a = 78.2 m/s . The displacement during the 0.120 s interval of constant acceleration is 1 (Equation 4-4) r = r r0 = v0 t + 2 at2 = 2 1 (14.5 m/s)(0.120 s) + 2 (78.2 m/s )(0.120 s)2 = (1.74 + 0.563 m. This has magnitude ) (1.74)2 + (0.563)2 m = 1.83 m (and makes an angle of tan1 (0.563/1.74) = 17.9 with the direction of the initial velocity).
Solution
Problem
10. Repeat the preceding problem, except that now the acceleration makes a 65.0 angle with the original direction of motion.
figure 4-2a Problem 7 Solution. (a) The change in velocity, v = at, is at right angles to the initial velocity, v0 , so they form the legs of a right triangle with hypotenuse v (see Fig. 4-2a). The angle between v and v0 is = tan1 (|v| |v0 |) = tan1 (0.035 km/s)(250 s)/(21 km/s) = 22.6 . (b) For constant acceleration, the displacement is (see 1 Equation 4-4) r = r r0 = v0 t + 2 at2 . Again, the 1 2 two vectors v0 t and 2 at form the legs of a right triangle, whose hypotenuse, r, has magnitude
1 (21 km/s)2 (250 s)2 + ( 2 )2 (0.035 km/s )2 (250 s)4 = 3 5.3610 km. (Note: this is the asteroids 2
Solution
Now the acceleration is (78.2 m/s ) cos 65.0 + sin 65.0 ) and r = (14.5 m/s)(0.120 s) + 2 1 (33.0 + 70.9 )(m/s ) (0.120 s)2 = (1.98+ 0.510 m. ) 2 This has magnitude 2.04 m (and makes an angle of 14.5 with the initial velocity).
2
Problem
11. A particle leaves the origin with initial velocity v0 = 11+ 14 m/s. It undergoes a constant 2 acceleration given by a = 1.2 + 0.26 m/s . (a) When does the particle cross the y-axis?
CHAPTER 4 (b) What is its y-coordinate at the time? (c) How fast is it moving, and in what direction, at that time? acceleration is assumed constant, the electron trajectory is parabolic in the deecting region.
y
23
Solution
(a) Since the particle leaves from the origin (r0 = 0), 1 its position is r(t) = v0 t + 2 at2 . It crosses the y-axis 1 2 when x(t) = v0x t + 2 ax t = 0, or t = 2v0x /ax = 2 2(11 m/s)/(1.2 m/s ) = 18.3 s. (b) y(t) = v0y t + 2 1 1 2 2 ay t = [14 m/s + 2 (0.26 m/s )(18.3 s)](18.3 s) = 300 m. (c) v(t) = v0 + at = (11 + 14 m/s + ) (1.2+ 0.26 )(18.3)m/s = (11 + 18.8 m/s. Then ) |v(t)| = (11)2 + (18.8)2 m/s = 21.8 m/s and x = tan1 (18.8/(11)) = 120.
Electron gun
Deflecting region 4.2 cm 15
x
figure 4-27 Problem 13 Solution.
Problem
12. A particle starts from the origin with initial velocity v0 = v0 and constant acceleration a = a . Show that the particles distance from the origin and its direction relative to the x-axis are given by d=t
2 v0 + 1 a2 t2 and = tan1 (at/2v0 ). 4
Section 4-3: Projectile Motion Problem
14. You toss an apple horizontally at 8.7 m/s from a height of 2.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground?
Solution
From Equation 4-4, the particles displacement from its original position is r = r r0 = v0 t + 1 at2 = 2 (v0 + 2 at This has magnitude t 1 )t.
2 v0 + ( 1 at)2 and 2
Solution
The time of ight for either projectile can be determined from the vertical component of the motion, which is the same for both, since v0y = 0. Thus, Equation 4-8 gives t = 2(y y0 )/g = 2(2.6 m)/(9.8 m/s ) = 0.728 s.
2
direction (CCW from the x-axis) = tan1 (at/2v0 ).
Problem
13. Figure 4-27 shows a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientic instruments. Electrons are accelerated by the electron gun, then move down the center of the tube at 2.0109 cm/s. In the 4.2-cm-long deecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration steers them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is needed to deect the electrons through 15 , as shown in the gure? (b) What is the shape of an electrons path in the deecting region?
Problem
15. A carpenter tosses a shingle o a 8.8-m-high roof, giving it an initially horizontal velocity of 11 m/s. (a) How long does it take to reach the ground? (b) How far does it move horizontally in this time?
y v0 (11ms)
y0 8.8 m
Solution
(a) With x-y axes as drawn on Fig. 4-27, the electrons emerge from the deecting region with velocity v = v0 + at after a time t = x/v0 , where x = 4.2 cm , and v0 = 2109 cm/s. The angle of deection (direction of v) is 15 , so tan 15 = vy /vx = at/v0 = 2 2 2 ax/v0 . Thus, a = v0 tan 15 /x = 2.551017 cm/s (when values are substituted). (b) Since the
x x
Problem 15 Solution.
24
CHAPTER 4 land on another roof 1.9 m lower. If the gap between the buildings is wide, how fast must he run?
Solution
(a) The shingle reaches the ground when y(t) = 0 = 1 y0 2 gt2 , or t= 2y0 = g 2(8.8 m) (9.8 m/s )
2
= 1.34 s.
Solution
The horizontal and vertical distances covered by 1 the stuntman are x x0 = v0 t and y0 y = 2 gt2 (since v0x = v0 , and v0y = 0). Eliminating t, one nds v0 = (x x0 ) g/2(y0 y) = (4.5 m) (9.8 m/s )/2(1.9 m) = 7.23 m/s. (Note that Equation 4-9 with 0 = 0 and y0 = 0 provides an equivalent solution.)
2
(b) The horizontal displacement is x = v0 t = (11 m/s)(1.34 s) = 14.7 m.
Problem
16. An arrow red horizontally at 41 m/s travels 23 m horizontally before it hits the ground. From what height was it red?
Problem
19. Ink droplets in an ink-jet printer are ejected horizontally at 12 m/s, and travel a horizontal distance of 1.0 mm to the paper. How far do they fall in this interval?
Solution
From x x0 = v0x t, one nds the time of ight 1 t = 23 m/(41 m/s) = 0.561 s, and from y0 y = 2 gt2 (recall that v0y = 0) one nds the height y0 y = 2 1 2 2 (9.8 m/s )(0.561 s) = 1.54 m.
Solution
From x x0 = v0x t, the time of ight can be found. 1 Substitution into y0 y = 2 gt2 (recall that 1 2 v0y = 0) yields y0 y = 2 g(x x0 )2 /v0 = 2 1 3 m)2 /(12 m/s)2 = 3.40108 m = 2 (9.8 m/s )(10 34 nm for the distance fallen, practically negligible. Note that this analysis is equivalent to using Equation 4-9 with 0 = 0.
Problem
17. A kid res water horizontally from a squirt gun held 1.6 m above the ground. It hits another kid 2.1 m away square in the back, at a point 0.93 m above the ground (see Fig. 4-28). What was the initial speed of the water?
Problem
20. Protons in a particle accelerator drop 1.2 m over the 1.7-km length of the accelerator. What is their approximate average speed?
0.93 m
1.6 m
Solution
The horizontal and vertical distances in projectile motion (range and drop) are related by the trajectory equation (Equation 4-9). With 0 = 0, 2 y = gx2 /2v0 , or v0 = x (g)/2y =
2
2.1 m
figure 4-28 Problem 17.
Solution
Since the water was red horizontally (v0y = 0), the time it takes to fall from y0 = 1.6 m to y = 0.93 m is given by Equation 4-8, t = 2(y0 y)/g = 2(1.6 0.93) m/(9.8 m/s ) = 0.370 s. Its initial speed, v0 = v0x , can be found from Equation 4-7, v0 = (x x0 )/t = 2.1 m/0.370 s = 5.68 m/s.
2
(1.7103 m) (9.8 m/s )/2(1.2106 m) = 3.44106 m/s. Since vy = gt = g(x x0 )/v0 = 4.85103 m/s is negligible compared to this, v0 is hardly dierent from the average speed.
Problem
21. Youre standing on the ground 3.0 m from the wall of a building, and you want to throw a package from your 1.5-4.2m shoulder level to someone in a second-oor window above the ground. At what speed and angle should you throw it so it just barely reaches the window?
Problem
18. In a chase scene, a movie stuntman is supposed to run right o the at roof of one city building and
CHAPTER 4
25
Solution
We suppose that just barely means that the maximum height of the package equals the height of the window sill. When the package reaches the sill (in 2 2 the coordinate system shown), vy = 0 = v0y 2gy,
impact, was the pedestrian in the crosswalk?
Solution
What is an issue here is the horizontal range of a piece of signal light lens in projectile motion, starting from a height of y0 y = 0.63 m o the ground, with an initial horizontal velocity of v0 = v0x = (40 m/3.6 s), and v0y = 0. Eliminating the time of ight from Equations 4-7 and 4-8 (see the solution to Problem 19), one obtains x x0 = v0 2(y0 y)/g = (11.1 m/s) 2(0.63 m)/(9.8 m/s ) = 3.98 m. If the pieces of lens did not bounce very far from the point where they hit the ground, this places the point of impact of the accident just 4.00 m 3.98 m = 2 cm from the center of the crosswalk. (Forensic physics is crucial to the prosecutions case.)
2
so v0y = 2(9.8 m/s2 )(2.7 m) = 7.27 m/s. Since vy = 0 = v0y gt, the time of ight is t = v0y /g. 2 Therefore v0x = x/t = (3.0 m)(9.8 m/s )/(7.27 m/s) = 4.04 m/s. From these components, we nd: v0 =
2 2 v0x + v0y = 8.32 m/s, and 1
0 = tan
(v0y /v0x ) = 60.9 .
Problem
24. Repeat Problem 15 for the case when the shingle is thrown with a speed of 11 m/s at 14 below the horizontal.
Solution
(a) If the initial velocity of the shingle has a downward component, v0y = v0 sin 0 = (11 m/s) sin(14 ) = 2.66 m/s, the time t to reach the ground is now given by the positive solution (t > 0) of the 1 equation y(t) = 0 = y0 + v0y t 2 gt2 . From the quadratic formula, this is t = (v0y +
2 v0y + 2gy0 )/g = [2.66 m/s + 2 2
Problem 21 Solution.
Problem
22. Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed v0 from a height h.
(2.66 m/s)2 + 2(9.8 m/s )(8.8 m)]/(9.8 m/s ) = 1.10 s (b) The horizontal range is x x0 = v0x t = (11 m/s)(cos(14 ))(1.10 s) = 11.7 m.
Solution
This problem is solved by reasoning identical to that in Problem 15, where it was shown that x = v0 2h/g. (By referring the student to a similar problem, the wide applicability of the laws of projectile motion is better appreciated.)
Problem
25. In part (b) of the exercise following Example 4-5, what is the vertical component of the velocity with which the cyclist strikes the ground?
Solution
One rst needs to work out part (a) of the exercise, which is similar to the rst part of Example 4-5. At the minimum speed at takeo, the cyclist covers a horizontal range x = 48 m, and a vertical drop y = 5.9 m (recall that x0 = y0 = 0 in Equation 4-9). Then (9.8 m/s )(48 m)2 2 15 )[(48 m) tan 15 (5.9 m)] 2(cos
2 1/2
Problem
23. A car moving at 40 km/h strikes a pedestrian a glancing blow, breaking both the cars front signal light lens and the pedestrians hip. Pieces of the lens are found 4.0 m down the road from the center of a 1.2-m-wide crosswalk, and a lawsuit hinges on whether or not the pedestrian was in the crosswalk at the time of the accident. Assuming that the lens was initially 63 cm o the ground, and that the lens pieces continued moving horizontally with the cars speed at the time of the
v0,min =
= 25.4 m/s.
26
CHAPTER 4 Equation 4-8 when y(t) y0 = 0 = v0y t 1 gt2 . Thus 2 2 t = 2v0y /g = 2(6.64 km/s) sin 45 /(9.8 m/s ) = 958 s = 16.0 min. (c) At a 20 launch angle, v0 = (4500 km)(0.0098 km/s2 )/ sin 40 = 8.28 km/s.
The cyclists actual initial speed is v0 = (1 + 50%)(25.4 m/s) = 38.1 m/s. Equation 2-11 gives the vertical component of the velocity (negative downward) for the given drop, vy =
2 2 v0y 2gy =
(38.1 m/s)2 sin2 15 2(9.8 m/s )(5.9 m) = 14.6 m/s
Problem
28. A rescue airplane is ying horizontally at speed v0 at an altitude h above the ocean, attempting to drop a package of medical supplies to a shipwreck victim in a lifeboat. At what line-of-sight angle (Fig. 4-29) should the pilot release the package?
Problem
26. Compare the travel times for the projectiles launched at 30 and 60 in Fig. 4-15, both of which have the same starting and ending points.
Solution
The package, when dropped, is a horizontally launched projectile, like that in Problem 15 (the expressions for t and x are the same). The line-of-sight angle is = tan1 (h/x), where x = v0 2h/g is the horizontal 2 distance at drop time. Thus, = tan1 gh/2v0 .
v0 v0
A
figure 4-15 Problem 26.
Solution
The time of ight (t > 0) for a projectile trajectory beginning and ending at the same height (y(t) = y0 ) can be found from Equation 4-8, y(t) y0 = 0 = v0y t 1 2 2 gt or t = 2v0y /g = 2v0 sin 0 /g. In Fig. 4-15, v0 = 2 50 m/s, so t30 = 2(50 m/s) sin 30 /(9.8 m/s ) = 5.10 s, and t60 = 3 t30 = 8.84 s.
y0
h
x
Problem
27. A submarine-launched missile has a range of 4500 km. (a) What launch speed is needed for this range when the launch angle is 45 ? (b) What is the total ight time? (c) What would be the minimum launch speed at a 20 launch angle, used to depress the trajectory so as to foil a space-based antimissile defense?
figure 4-29 Problem 28 Solution.
Problem
29. At a circus, a human cannonball is shot from a cannon at 35 km/h at an angle of 40 . If he leaves the cannon 1.0 m o the ground, and lands in a net 2.0 m o the ground, how long is he in the air?
Solution
(a) Assuming Equation 4-10 applies (i.e., the trajectory begins and ends at the same height, or y(t) = y0 ), one nds v0 = xg/ sin 2 = (4500 km)(0.0098 km/s )/ sin 90 = 6.64 km/s. (b) The time of ight is the positive solution of
2
CHAPTER 4
27
Solution
The time of ight can be calculated from Equation 4-8, where the trajectory begins at y0 = 1.0 m and t = 0, ends at y(t) = 2.0 m, and v0y = v0 sin 0 = (35 m/3.6 s) sin 40 = 6.25 m/s. The 1 equation is a quadratic, 2 gt2 v0y t + (y y0 ) = 0,
2 v0y 2g(y y0 )]/g = 2
Appendix E, we nd xmoon = (gEarth /gmoon )xEarth = (9.81/1.62)(180 m) = 1090 m.
Problem
32. Prove that a projectile launched on level ground reaches its maximum height midway along its trajectory.
with solutions t = [v0y [6.25 m/s
2
(9.8 m/s ) = 0.188 s or 1.09 s. The trajectory crosses the height 2.0 m twice, once going up, at the smaller time of ight, and once going down into the net, at the larger time of ight. The latter is the answer to the question asked here.
(6.25 m/s)2 2(9.8 m/s )(2 m 1 m)]
Solution
From the trajectory (Equation 4-9), the maximum 2 height occurs when dy/dx = tan gx/v0 cos2 0 = 0, 2 or x = v0 sin 0 cos /g, which is midway, or half of the horizontal range (Equation 4-10). This result can be derived in other ways; for example, the maximum height is reached at time v0y /g (when vy = 0), whereas the total time of ight is twice this (the solution of 1 0 = v0y t 2 gt2 ).
Problem
30. Preparing to give an injection, a physician ejects a drop of medicine to ensure theres no air in the syringe. The syringe is pointing upward, at 20 to the vertical, with the tip of the needle 45.0 cm above a tabletop. The drop leaves the needle at 1.30 m/s. (a) What maximum height does it reach? (b) How long is it before the drop hits the tabletop? (c) How far does the drop travel horizontally?
Problem
33. A projectile launched at an angle 0 to the horizontal reaches a maximum height h. Show that its horizontal range is 4h/ tan 0 .
Solution
The intermediate expression for the horizontal range (when the initial and nal heights are equal) is 2 x = 2v0 sin 0 cos 0 /g = 2v0x v0y /g (see the equation before Equation 4-10). The components of the initial velocity are related by v0y /v0x = tan 0 . The maximum height, h = ymax y0 , can be found from 2 Equation 2-11 (when vy = 0) or v0y = 2gh. Then x = 2v0y v0x /g = 2v0y (v0y / tan 0 )/g = 2(2gh)/g tan 0 = 4h/ tan 0 . (This result reects a classical geometrical property of the parabola, namely, that the latus rectum is four times the distance from vertex to focus.)
Solution
(a) Take the tip of the syringe to be at y0 = 45.0 cm above the tabletop at y = 0. At its maximum height, just the vertical component of the drops velocity is 2 2 instantaneously zero, so vy = 0 = v0y 2g(ymax y0 ), 2 or ymax y0 = v0y /2g. The vertical component of the initial velocity is v0y = (1.30 m/s) cos 20 , which gives ymax y0 = 7.61 cm. This is the maximum height above the tip, corresponding to 52.6 cm above the tabletop. (b) The time of ight is the positive solution 1 of the equation y(t) = 0 = y0 + v0y t 2 gt2 , or t = (v0y + part (a), we obtain t = 0.452 s. (c) The horizontal distance traveled is x x0 = v0x t = (1.30 m/s)(sin 20 )(0.452 s) = 20.1 cm.
Problem
34. Youre 5.0 m from the left-hand wall of the house shown in Fig. 4-30, and you want to throw a ball to a friend 5.0 m from the right-hand wall. (a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle should it be thrown? Assume the throw and catch both occur 1.0 m above the ground.
2 v0y + 2gy0 )/g. Substituting the numbers from
Problem
31. If you can hit a golf ball 180 m on Earth, how far can you hit it on the moon? (Your answer is an underestimate, because the distance on Earth is restricted by air resistance as well as by a larger g.)
Solution
Since the trajectory is symmetrical (begins and ends at the same height), one can use the result of the previous problem with h = ymax y0 = 6 m 1 m = 5 m and horizontal range x = 5 m + 6 m + 5 m = 16 m. Then (b) 0 = tan1 (4h/x) = 51.3 , and (a) v0 = v0y / sin 0 = 2gh/ sin 0 = 12.7 m/s.
Solution
For given v0 , the horizontal range is inversely proportional to g. With surface gravities from
28
CHAPTER 4 jumped at sea level, where g = 9.81 m/s , at the same angle and initial speed as in Mexico City, how far would Beamon have gone? Neglect air resistance in both cases (although its eect is actually more signicant than the change in g).
1.0 m 6.0 m 5.0 m
2
45 3.0 m 1.0 m 5.0 m
Solution
The length of the jump is inversely proportional to g (just as in Problem 31), so x = (g/g )x = (9.786/9.810)(8.90 m) = 8.88 m.
figure 4-30 Problem 34.
Problem Problem
35. A circular fountain has jets of water directed from the circumference inward at an angle of 45 . Each jet reaches a maximum height of 2.2 m. (a) If all the jets converge in the center of the circle and at their initial height, what is the radius of the fountain? (b) If one of the jets is aimed at 10 too low, how far short of the center does it fall? 37. In 1991 Mike Powell shattered Bob Beamons 1968 world long jump record with a leap of 8.95 m (see Fig. 4-31). Studies show that Powell jumps at 22 to the vertical. Treating him as a projectile, at what speed did Powell begin his jump?
Solution
The horizontal range formula (Equation 4-10) gives v0 = xg/ sin 20 = (8.95 m)(9.8 m/s )/ sin 2(90 22 ) = 11.2 m/s (Air resistance and body control are important factors in the long jump also.)
2
Solution
(a) The radius is the horizontal range, r = (Equation 4-10 with = 45 ). The maximum height is 2 2 h = v0y /2g = v0 /4g (Equation 2-11 with vy = 0 and v0y = v0 cos 45 = v0 / 2). Therefore, r = (4gh)/g = 4h = 4(2.2 m) = 8.8 m. (b) If one jet is directed at 35 2 with the same initial speed (v0 = rg), it would fall short by r x, where x is given by Equation 4-10. 2 Therefore, r x = r (v0 /g) sin(2 35 ) = (8.8 m)(1 sin 70 ) = 0.531 m.
2 v0 /g
Problem
38. A motorcyclist driving in a 60-km/h zone hits a stopped car on a level road. The cyclist is thrown from his bike and lands 39 m down the road. Was the cyclist speeding? To answer, nd the minimum speed he could have been going just before the accident.
Solution
If the motorcyclist was deected upward from the road at an angle of 45 , the horizontal range formula (Equation 4-10) implies a minimum initial speed of 2 v0 = xg = (39 m)(9.8 m/s ) = 19.5 m/s = 70.4 km/h. In fact, some speed would be lost during impact with the car, so the cyclist probably was speeding.
Problem 35 Solution.
Problem
36. When the Olympics were held in Mexico City in 1968, many sports fans feared that the high altitude would result in poor performances due to reduced oxygen. To their surprise, new records were set in track and eld events, probably as a result of lowered air resistance and a decrease in g to 9.786 m/s2 , both ultimately associated with the high altitude. In particular, Robert Beamon set a new world record of 8.90 m in the long jump. Photographs suggest that Beamon started his jump at a 25 angle to the horizontal. If he had
Problem
39. Show that, for a given initial speed, the horizontal range of a projectile is the same for launch angles 45 + and 45 , where is between 0 and 45 .
Solution
The trigonometric identity in Appendix A for the sine of the sum of two angles shows that sin 2(45 ) = sin(90 2) = sin 90 cos 2 cos 90 sin 2 = cos 2,
CHAPTER 4 so the horizontal range formula (Equation 4-10) gives the same range for either launch angle, at the same initial speed.
29
pass through the basket, although in this case they are not symmetrically placed about 45 .
Problem
40. One model of the Scud missile used in the 1991 Persian Gulf war has a range of 630 km. (a) What is its launch speed, given a 40 launch angle? (b) What is the missile ight time?
Solution
(a) From Equation 4-10, v0 =
2
xg/ sin 20 =
(630 km)(.0098 km/s = 2.50 km/s. (b) From Equation 4-7, the time of ight is t = x/v0 cos 40 = (630 km)/(2.50 km/s) cos 40 = 328 s = 5.47 min . (We assumed the same initial and nal trajectory elevations and neglected air resistance.)
)/ sin 80
Problem
41. A basketball player is 15 ft horizontally from the center of the basket, which is 10 ft o the ground. At what angle should the player aim the ball if it is thrown from a height of 8.2 ft with a speed of 26 ft/s?
Solution
With origin at the point from which the ball is thrown, the equation of the trajectory (Equation 4-9), evaluated at the basket, becomes 2 (32 ft/s )(15 ft)2 , y = (10 8.2)ft = (15 ft) tan 0 2(26 ft/s)2 cos2 0 2 or 1.8 = 15 tan 0 5.33/ cos 0 . Using the trigonometric identity 1 + tan2 0 = 1/ cos2 0 , we can convert this equation into a quadratic in tan 0 : 7.13 15 tan 0 + 5.33 tan2 0 = 0, so 0 = tan1 15 152 4(5.33)(7.13) 2(5.33)
= 31.2 or 65.7 Like the horizontal range formula (see Fig. 4-13), for given v0 there are two launch angles whose trajectories
Problem 41 Solution.
30
CHAPTER 4 in radians before dividing.) Most calculus texts contain a proof that (sin x/x) 1 for x 0.
Section 4-4: Circular Motion Problem
42. How fast would a car have to round a turn 75 m in radius in order for its acceleration to be numerically equal to that of gravity?
Solution
For circular motion with constant speed, ar = v 2 /r or v = ar r = (9.8 m/s2 )(75 m) = 27.1 m/s = 97.6 km/h = 60.7 mi/h.
Problem 44 Solution.
Problem
43. Estimate the acceleration of the moon, which completes a nearly circular orbit of 385,000 km radius in 27 days.
Problem
45. When Apollo astronauts landed on the moon, they left one astronaut behind in a circular orbit around the moon. For the half of the orbit spent over the far side of the moon, that individual was completely cut o from communication with the rest of humanity. How long did this lonely state last? Assume a suciently low orbit that you can use the moons surface gravitational acceleration (see Appendix E) for the spacecraft.
Solution
The centripetal acceleration is given in terms of the period for uniform circular motion by Equation 4-12 in Example 4-8. In the case of the moon, a = 4 2 r/T 2 = 2 4 2 (3.85108 m)/(27.386, 400 s)2 = 2.73103 m/s , where we used more accurate data from Appendix E. (Note: centripetal is a purely kinematic adjective descriptive of circular motion. In this case, the origin of the moons centripetal acceleration is the gravitational attraction of the Earth.)
Solution
Consider a circular orbit around the moon with radius slightly larger than the lunar radius, r = 1.74106 m, and centripetal acceleration approximately equal to the lunar surface gravity, ac = 1.62 m/s2 (see Appendix E). The orbital period is related to r and ac by Equation 4-12 as in Example 4-8, T = 2 r/ac , and radio communications with Earth were 1 blocked for half of this period, or 2 T = (1.74106 m)/(1.62 m/s ) = 3.26103 s or about 54.3 min.
2
Problem
44. An object is in uniform circular motion. Make a graph of its average acceleration, measured in units of v 2 /r, versus angular separation for points on the circular path spaced 40 , 30 , 20 , and 10 apart. Your graph should show the average acceleration approaching the instantaneous value v 2 /r as the angular separation decreases.
Solution
In uniform circular motion, the average acceleration is a multiple of the base of an isosceles triangle with apex angle equal to the angular separation of the positions at the intervals endpoints (see Fig. 4-21). Using some trigonometry, we nd |v| = 2v 2 (1 cos ) = 2v sin( 1 ). Since, for constant 2 speed, v = (arc length)/(time) = r/t,
1 1 2v sin( 2 ) |v| v 2 sin( 2 ) |aav | = = = 1 t (r/v) r ( 2 )
Problem
46. A 12-in-diameter circular saw blade rotates at 3500 revolutions per minute. What is the acceleration of one of the saw teeth? Compare with the acceleration of gravity.
Solution
If the saw blade rotates at 3500 rpm, a point on its circumference has a linear speed of v = (3500)(12 in)/60 s = 183 ft/s. The (radial) acceleration is ar = v 2 /r = (183 ft/s)2 /(6 ft/12) = 2 2 6.72 104 ft/s 2103 g, where g = 32.2 ft/s . Alternatively, one could use Equation 4-12 with period equal to (3500)1 min .
The graph of |aav |, in units of v 2 /r, versus is the 1 same as the graph of f () = sin( 1 )/( 2 ) versus . 2 For = 40 , 30 , 20 , and 10 , f () = 0.980, 0.989, 0.995, and 0.999 respectively. (Dont forget to express
CHAPTER 4
31
Problem
47. A jet is diving vertically downward at 1200 km/h (see Fig. 4-32). If the pilot can withstand a maximum acceleration of 5g (i.e., 5 times Earths gravitational acceleration) before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive? Assume the speed remains constant.
r = 4.30 cm 55
Solution
The height at the start of the 90 -turn must be greater than the radius of the turn, in order to avoid hitting the ground. The radius of the turn must be great enough that the centripetal acceleration not exceed 5g, i.e., ac = v 2 /r 5g or r v 2 /5 g = 2 (1200 m/3.6 s)2 = 5(9.8 m/s ) = 2.27 km.
figure 4-33 Problem 48.
Problem
49. How long would a day last if Earth were rotating so fast that the acceleration of an object on the equator were equal to g?
v
Solution
The Earths equitorial radius is about 6378 km. If the centripetal acceleration at the equator were ac = 9.8 m/s2 , the Earths period of rotation would have to be T = 2 RE /ac =
r
2 (6378 km)/(0.0098 km/s ) = 5.07103 s = 1 h 24.5 min. (See Equation 4-12.)
2
Problem
50. A runner rounds the semicircular end of a track whose curvature radius is 16 m. The runner moves at constant speed, with an acceleration of 2 0.94 m/s . How long does it take to complete the turn?
figure 4-32 Problem 47 Solution.
Problem
48. Electrons in a TV tube are deected through a 55 angle, as shown in Fig. 4-33. During the deection they move at constant speed in a circular path of radius 4.30 cm. If they experience 2 an acceleration of 3.351017 m/s , how long does the deection take?
Solution
Since the runner has a constant speed along a circular arc, the acceleration must be purely a centripetal acceleration, namely ac = 0.94 m/s2 = v 2 /r, and the speed is v = (0.94 m/s2 )(16 m) = 3.88 m/s. At this speed, it takes time t = r/v = (16 m)/(3.88 m/s) = 13.0 s to complete a semicircle. (Of course, Equation 4-12 would give the same result, 1 T = r/ac .) 2
Solution
The speed of the electrons is v =
ar r =
(3.35 1017 m/s2 )(0.043 m) = 1.20108 m/s. The length of a circular arc of 55 and radius 0.043 m is 2(0.043 m)(55/360) = 0.0413 m. Therefore, the time for the deection is t = 0.0431 m/1.2108 m/s = 0.344 ns. (This problem will appear more transparent after studying Section 12-1.)
Section 4-5: Nonuniform Circular Motion Problem
51. A space station 120 m in diameter is set rotating in order to give its occupants articial gravity. Over a period of 5.0 min, small rockets bring the station steadily to its nal rotation rate of 1 revolution every 20 s. What are the radial and
32
CHAPTER 4 tangential accelerations of a point on the rim of the station 2.0 min after the rockets start ring?
Solution
The tangential acceleration (in the direction of motion) is perpendicular to the radial acceleration, so the resultant total acceleration (their vector sum) is at 45 between them at the instant when at = ar = v 2 /r. The linear speed along the circle depends on at only (since ac is perpendicular to the velocity), v so = at t for constant at (provided the object is set into motion with v0 = 0 at t = 0). Thus, at = (at t)2 /r or t = r/at .
Solution
If the rotation rate increases steadily from 0 to 1 revolution in 20 s, over a 5-minute interval, the rotation rate after 2 minutes is (2/5)(1 rev/20 s) = 1 rev/50 s, or the (instantaneous) period after 2 minutes is 50 s. Thus, by Equation 4-12, the centripetal (radial) acceleration is ac = 4 2 r/T 2 = 4 2 (60 m)/(50 s)2 = 0.947 m/s2 . The tangential speed increases steadily from 0 to 2(60 m)/20 s = 6 m/s, in 5 minutes, so the tangential acceleration is at = (6 m/s)/5 min = 6.28102 m/s2 . (The solution to this problem may appear more straightforward after the discussion of angular motion in Chapter 12.)
Problem
54. A car moving at 65 km/h enters a curve that describes a quarter turn of radius 120 m. The driver gently applies the brakes, giving a constant 2 tangential deceleration of magnitude 0.65 m/s . Just before emerging from the turn, what are (a) the magnitude of the cars acceleration and (b) the angle between the acceleration vector and the direction of motion?
Problem
52. A plane is heading northward when it begins to turn eastward on a circular path of radius 9.10 km. At the instant it begins to turn, its acceleration vector points 22.0 north of east and 2 has magnitude 2.60 m/s . (a) What is the planes speed? (b) At what rate is its speed increasing?
Solution
The cars tangential acceleration is constant, 2 a = 0.65 m/s along the direction of motion, so Equation 2-11 can be used to determine the speed at 2 the end of the turn, v 2 = v0 + 2at s. Here, s = 2r/4 is the linear distance around the quarter turn. Then the 2 centripetal acceleration is ac = v 2 /r = (v0 + rat )/r = 2 2 2 (65 m/3.6 s) /120 m + (0.65 m/s ) = 0.675 m/s . (a) The magnitude of the total acceleration is 2 a2 + a2 = 0.937 m/s . (b) The angle of the total c t acceleration with respect to the tangent to the curve is = tan1 (ac /at ) = 134 . (This angle is in the second quadrant because the tangential acceleration is opposite to the direction of motion.)
Solution
(a) The east-component of a is the radial acceleration 2 ar = (2.60 m/s ) cos 22 = v 2 /r. Therefore, the speed is v = (2.60 m/s ) cos 22(9.1 km) = 148 m/s. (b) The north-component of a is the tangential acceleration (the rate of increase in speed), so 2 2 at = (2.60 m/s ) sin 22 = 0.974 m/s .
2
Paired Problems Problem
55. An alpine rescue team is using a slingshot to send an emergency medical packet to climbers stranded on a ledge, as shown in Fig. 4-34. What should be the launch speed from the slingshot? Problem 52 Solution.
Solution Problem
53. An object is set into motion on a circular path of radius r by giving it a constant tangential acceleration at . Derive an expression for the time t when the acceleration vector points at 45 to the direction of motion. If we take the origin of coordinates at the slingshot and the stranded climbers at x = 390 m and y = 270 m, we can use Equation 4-9 for the trajectory to solve for v0 : v0 = x cos 0 g 2(x tan 0 y)
CHAPTER 4 390 m cos 70 9.8 m/s = 89.2 m/s. 2(390 m tan 70 270 m)
2
33
=
Solution
The solution to the previous problem yields 1 h = 2 x = 90 km. (The approximations implicit in this result are less valid for a rocket than for a stone.)
Problem
56. A cat leaps onto a counter 90 cm o the oor, starting 65 cm from the edge of the counter. It leaps at an initial angle of 79 to the horizontal and lands on the counter 22 cm from the edge. What was its initial speed?
Problem
59. I can kick a soccer ball 28 m on level ground, giving it an initial velocity at 40 to the horizontal. At the same initial speed and angle to the horizontal, what horizontal distance can I kick the ball on a 15 upward slope?
Solution
If the cat jumps from the origin, it lands at x = 65 cm + 22 cm = 0.87 m and y = 0.9 m. Equation 4-9 solved for v0 gives v0 = 0.87 m cos 79 9.8 m/s = 5.34 m/s. 2(0.87 m tan 79 0.9 m)
2
Solution
We need to nd the intersection of the trajectory of the ball (Equation 4-9) with a 15 slope through the same origin, y = x tan 15 . The appearance of the trajectory equation can be simplied by use of the fact that y = 0 when x = 28 m and 0 = 40 . Thus, 2 y = 0 = x tan 0 (g/2v0x )x2 = (28 m)[tan 40 2 2 (g/2v0x )(28 m)], or the coecient (g/2v0x ) equals tan 40 /28 m. The trajectory equation simplies to y = x tan 40 x2 (tan 40 /28 m) = x(1 x/28 m) tan 40 . The intersection of this with the slope occurs when y also equals x tan 15 , or x tan 15 = x(1 x/28 m) tan 40 . The x-coordinates of the two points of intersection are x = 0 (the origin) and x = (28 m)(1 tan 15 / tan 40 ) = 19.1 m (the horizontal distance queried in this problem).
Problem
57. If you can throw a stone straight up to a height of 16 m, how far could you throw it horizontally over level ground? Assume the same throwing speed and optimum launch angle.
Solution
To throw an object vertically to a maximum height of h = 16 m = ymax 0 requires an initial speed of v0 = y 2g(ymax y0 ) = 2gh. With this value of v0 and the optimum launch angle 0 = 45 , Equation 4-10 gives a maximum horizontal range on level ground of 2 x = v0 /g = 2h = 32 m. (The maximum horizontal range on level ground is twice the maximum height for vertical motion with the same initial speed. This result holds in the approximation of constant g and no air resistance.)
Problem
58. In a conversion from military to peacetime use, a missile with a maximum horizontal range of 180 km is being adapted for studying the upper atmosphere. What is the maximum altitude it can achieve, if launched vertically?
Problem 59 Solution.
Problem
60. A model rocket has a horizontal range of 280 m on level ground, when given a 45 launch angle. What horizontal distance will the rocket cover when launched at 45 to the horizontal from the top of a hill whose sides slope down at 21 ?
270 m
Solution
70 390 m
figure 4-34 Problem 55.
We are given two points on the trajectory (Equation 4-9) of the rocket at the altitude of the hilltop, the origin x = y = 0, and the horizontal range at level x = 280 m and y = 0. As in the previous problem, we seek the x-coordinate of the intersection of the
34
CHAPTER 4 drops range from 10 below the horizontal to 25 above. If the nozzle is 1.7 m above the ground, how wide is the region (marked w in Fig. 4-35) that gets wet?
25 10
trajectory with the slope y = x tan 21 through the 2 origin. Since 0 = (280 m) tan 45 (g/2v0x )(280 m)2 , 2 we can eliminate the constant (g/2v0x ) = 1/280 m from the trajectory equation and nd the intersection of it with the slope as before: x tan 21 = x x2 (1/280 m), or x = (280 m)(1 + tan 21 ) = 387 m. (Recall that tan 45 = 1.)
w
figure 4-35 Problem 62. Problem 60 Solution.
Solution
We can nd where the drops hit the ground from the trajectory equation (Equation 4-9), as in the previous problem. Here, the origin is at the nozzle, and the ground has y = 1.7 m. The positive root of the 2 quadratic 1.7 m = x tan 0 (g/2v0x )x2 is 2 2 2 x = [tan 0 + tan 0 + 4(1.7 m)(g/2v0x )](v0x /g). When data for the upper and lower extremes of spray (0 = 25 or 10 , v0 = 4.6 m/s) are substituted, one nds a spread in x of w = 3.42 m 2.32 m = 1.09 m.
Problem
61. A reworks rocket is 73 m above the ground when it explodes. Immediately after the explosion, one piece is moving at 51 m/s at 23 to the upward vertical direction. A second piece is moving at 38 m/s at 11 below the horizontal direction. At what horizontal distance from the explosion site does each piece land?
Problem
63. You toss a chocolate bar to your hiking companion located 8.6 m up a 39 slope, as shown in Fig. 4-36. Determine the initial velocity vector so that the chocolate bar will reach your friend moving horizontally.
Solution
In the trajectory equation (Equation 4-9) with origin at the position of the explosion, the coecients are known for each piece. One can solve this quadratic equation for x, when y = 73 m (ground level), and select the positive root (since the trajectories start at x = 0 and end on the ground in the direction of v0x , which is chosen positive). 2 For the rst piece, tan 23 = 0.424 and g/2v0x = 2 2 3 1 (9.8 m/s )/2(51 cos 23 m/s) = 2.2210 m , so the quadratic is 73 m = 0.424x (2.22103 m1 )x2 . This has positive root x = [0.424 + (0.424)2 + 4(73 m)(2.22103 m1 )] (4.44103 m1 ) = 300 m. Similarly, for the second piece, tan(11 ) = 2 2 0.194 and g/2v0x = (9.8 m/s ) 2(38 cos(11 ) m/s)2 = 3.52103 m1 , so x = [0.194 + (0.194)2 + 4(73 m)(3.52103 m1 )] (7.04103 m1 ) = 119 m.
8.6 m
39
figure 4-36 Problem 63.
Problem
62. A hose nozzle sprays water drops in a fan-shaped pattern as suggested in Fig. 4-35. The drops leave the nozzle moving at 4.6 m/s. With the hose aimed as shown, the directions of the emerging
Solution
The candy bar moves horizontally only at the apex of its trajectory (where vx = v0x and vy = 0). Thus,
CHAPTER 4 ymax y0 = (8.6 m) sin 39 = 5.41 m, and v0y = 2g(ymax y0 ) = 2(9.8 m/s )(5.41 m) = 10.3 m/s (see Equation 2-11). The time to reach the apex is t = v0y /g, so v0x = (x x0 )/t = (x x0 )g/v0y (see Equations 4-6 and 4-7). The horizontal distance from apex to origin is x x0 = (8.6 m) cos 39 = 6.68 m, so 2 v0x = (6.68 m)(9.8 m/s )/(10.3 m/s) = 6.36 m/s. v0 can be expressed in unit vector notation as 2 2 (6.36+ 10.3 m/s, or by its magnitude v0x + v0y = ) 12.1 m/s and direction = tan1 (v0y /v0x ) = 58.3 (CCW from the x-axis).
2
35
ac = 1 [(390/3.6)2 + (740/3.6)2](m2 /s2 )/(7.1 km) = 2 2 3.80 m/s . Finally, = tan1 (3.80/0.456) = 83.2 . (Instead of working out each component of acceleration numerically, we could have written the nal result symbolically as follows: ac = at
2 2 vi + vf 2r
2(3/4)(2r) 3 = 2 2 (vf vi ) 2
2 2 vf + vi 2 v2 vf i
,
and = tan1 [3(7402 + 3902 )/2(7402 3902 )] as before.)
Problem
66. On its landing approach a plane makes a semicircular turn of radius 6.7 km, remaining at constant altitude while its speed drops steadily from 840 km/h to 290 km/h. At the midpoint of the turn, what is the angle between the planes velocity and acceleration vectors?
Problem
64. A circus lion prepares to leap through a aming hoop. A line from the lion to the hoop is 2.2 m long and makes a 26 angle with the oor. With what initial velocity should the lion leap so as to pass through the hoop moving horizontally?
Solution
The apex of the lions trajectory (where its moving horizontally with vx = v0x and vy = 0) is displaced from the origin from which it leaps by x x0 = (2.2 m) cos 26 = 1.98 m, and y y0 = (2.2 m) sin 26 = 0.964 m. As in the previous problem, v0y = 2g(y y0 ) = 4.35 m/s, and v0x = (x x0 )g/v0y = 4.46 m/s. The magnitude and direction of v0 are 6.23 m/s and 44.3 (CCW from x-axis).
Solution
The symbolic result of the previous problem can be 3 1 used if we change the length of turn from 4 to 2 of a 1 2 2 2 2 circle. Then = tan [(vf + vi )/(vf vi )] = tan1 [(2902 + 8402 )/(2902 8402 )] = 104 . Since the plane decelerates, the argument of the arctan is negative, and the angle is in the second quadrant (i.e., at is opposite to the velocity).
Problem
65. After takeo, a plane makes a three-quarter circle turn of radius 7.1 km, maintaining constant altitude but steadily increasing its speed from 390 to 740 km/h. Midway through the turn, what is the angle between the planes velocity and acceleration vectors?
Supplementary Problems Problem
67. Verify the maximum altitude and ight time for the 15 launch angle trajectory of the missile described at the end of the Application: Ballistic Missile Defense (page 79).
Solution
The missile has a range (on level ground) of R = xmax x0 = 1000 km at a launch angle of 0 = 15 , so Equation 4-10 gives the launch speed as 2 v0 = gR/ sin 20 . (Numerically, this is (0.0098 km/s )(1000 km)/ sin 2(15 ) = 4.43 km/s, as stated in the text.) The maximum altitude can be 2 found from Equation 2-11, h = ymax y0 = v0y /2g, since vy = 0 at this point in the trajectory. With the 2 value of v0 above, we nd h = v0 sin2 0 /2g = 2 2 (gR/ sin 20 ) sin 0 /2g = R sin 0 /2 sin 20 = 1 1 4 R tan 0 = 4 (1000 km) tan 15 = 67.0 km, in agreement with the text. (To answer just this question, one might have substituted numbers into the rst expression for h, i.e., (4.43 km/s)2 sin2 15 /2g, but working algebraically, the simpler expression from
2
Solution
The planes velocity is tangent to its circular path in the direction of motion and so is the tangential acceleration at . The radial (centripetal) acceleration ac is perpendicular to this, so the angle between the total acceleration and the velocity is = tan1 (ac /at ), inclined toward the center of the turn. For the linear motion over a circular distance of s = 3 (2r) = 4 1.5(7.1 km) = 33.5 km, Equation 2-11 can be used to 2 2 nd the constant at , at = (vf vi )/2s = [(740/3.6)2 2 2 2 2 (390/3.6) ](m /s )/2(33.5 km) = 0.456 m/s . We can 2 nd ac = v /r midway through the turn by using 2 Equation 2-11 again, since v 2 = vi + 2at (s/2) = 1 2 1 2 2 2 2 vi + 2 (vf vi ) = 2 (vi + vf ). Then
36
CHAPTER 4 Equation 2-11, evaluated at the highest point, 2 2 v0y = v0 sin2 0 = 2gymax. From Equation 4-10, 1 2 v0 sin 0 cos 0 = 2 gxmax . Dividing these, we nd tan 0 = 4ymax /xmax = 4(100 cm)/(80 cm) = 5, or 0 = tan1 5 = 78.7 . Substituting this angle into the rst equation (for example), we nd v0 = 2(9.8 m/s )(1 m)/ sin 78.7 = 4.51 m/s. (b) The time of ight for one of the balls can be calculated from Equation 4-8 (with y = y0 = 0) and the answer to part (a) above. Thus, t = 2v0y /g = 2 2gymax /g = 2 2ymax /g = 2 2(1 m)/(9.8 m/s2 ) = 0.904 s. Suppose the juggler throws a ball with the left hand while simultaneously catching another with the right. At this instant, the other two balls occupy positions in ight along the trajectory. During the time of ight of a given ball, three balls must be caught, namely, the given ball and the two already in the air. Thus, three catches must be made in 0.904 s, or one catch every 0.301 s. (This is also the time during which a ball must be transferred from the catching (right) hand to the throwing (left) hand. Thus, the total time for a given ball to circulate once around completely would be 0.904 s + 0.301 s 1.21 s.)
2
Problem 33 is obtained. Moreover, in this particular case, 2 sin 20 = 2 sin 2(15 ) = 1, so the numerical calculation is faster too.) The time of ight can be found from Equation 4-7, t = (xmax x0 )/v0x = R/v0 cos 0 = (1000 km)/(4.43 km/s) cos 15 = 234 s = 3.90 min, completing the verication of the text. (Again, alternate expressions could have been used, e.g., t = 2v0y /g, from Equation 4-6, or t = R/2g (cos 15 )1 .)
Problem
68. A jugglers hands are 80 cm apart, and the balls being juggled reach a maximum height of 100 cm above the jugglers hands. (a) At what velocity do the balls leave the jugglers hands? (b) If four balls are being juggled, how often must the juggler catch a ball?
Problem
69. A monkey is hanging from a branch a height h above the ground. A naturalist stands a horizontal distance d from a point directly below the monkey. The naturalist aims a tranquilizer dart directly at the monkey, but just as she res the monkey lets go. Show that the dart will nevertheless hit the monkey, provided its initial speed exceeds (d2 + h2 )g/2h.
Solution
Gravity accelerates the dart and the monkey equally, so both fall the same vertical distance from the point of aim (the monkeys original position) resulting in a hit, provided the initial speed of the dart is sucient to reach the monkey before the monkey reaches the ground. To prove this assertion, let the dart be red from ground level (y = 0) with speed v0 and direction 0 = tan1 (h/d) (line of sight from naturalist N to monkey M), while the monkey drops from height h at 1 t = 0. The vertical height of each is ymonkey = h 2 gt2 1 2 and ydart = v0y t 2 gt , where v0y = v0 sin 0 = 1 v0 h/ d2 + h2 . (The term 2 gt2 represents the eect of gravity, which appears the same way in both y-coordinate equations.) The dart strikes the monkey when ymonkey = ydart , which implies h = v0y t, or t = d2 + h2 /v0 . This must be less than the time required for the monkey to fall to the ground, which is 2h/g (from ymonkey = 0). Thus
Problem 68 Solution.
Solution
(a) Suppose the jugglers hands are located at the origin and 80 cm along the x-axis, as shown. From
CHAPTER 4 d2 + h2 /v0 < 2h/g or v0 > g(d2 + h2 )/2h. (This condition can also be understood from the horizontal range formula, Equation 4-10. The range of the dart has to be greater than the horizontal distance to the 2 2 monkey, d < v0 sin 20 /g = v0 2hd/(d2 + h2 )g.) 0 = tan1 a, and v0 = g(1 + a2 )/2b, so with the given numerical values, one nds 7.4a = and 7.4b = 9.5 2.1 2.1 9.5 3.2,
37
1 1 2.1 m 9.5 m
3.2,
or 0 = tan1 1.86 = 61.7 , and v0 = (9.8 m/s )(1 + 1.862)/2(0.160 m1 ) = 11.7 m/s.
2
Problem
71. A diver leaves a 3-m board on a trajectory that takes her 2.5 m above the board, and then into the water a horizontal distance of 2.8 m from the end of the board. At what speed and angle did she leave the board?
Problem 69 Solution.
Problem
70. A child tosses a ball over a at-roofed house 3.2 m high and 7.4 m wide, so it just clears the corners on both sides, as shown in Fig. 4-37. If the child stands 2.1 m from the wall, what are the balls initial speed and launch angle? Assume the ball is launched essentially from ground level.
y (x1, y1) (x2, y2)
Problem 71 Solution.
3.2 m x
(x0, y0) 2.1 m
7.4 m
(x3, y3)
Solution
Since we are given the maximum height (at which point vy = 0), Equation 2-11 can be used to nd the y component of the divers initial velocity, 2 0 = v0y 2g(ymax y0 ) or v0y = 2(9.8 m/s )(2.5 m) = 7.00 m/s. (We take the positive square root because the diver springs upward o the board.) The x-component of v0 can be found from Equation 4-7, once the time of ight is known. The latter is the positive root (the dive begins at t = 0) of the quadratic Equation 4-8, when y0 y = 3 m (a 3-m board is 3 m above the water level). Thus, t = [v0y +
2 v0y + 2g(y0 y)]/g = [7 m/s + 2
figure 4-37 Problem 70 Solution.
Solution
Choose the coordinate system shown in Fig. 4-37, such that the four given points on the trajectory are the origin (x0 , y0 ) = (0, 0), the rst corner (x1 , y1 ) = (2.1 m, 3.2 m), the second corner (x2 , y2 ) = (9.5 m, 3.2 m), and ground level at the right (x3 , y3 ) = (11.6 m, 0). Equation 4-9 for the trajectory, evaluated at any two points other than the origin, provides two equations which can be solved for v0 and 0 . Before substituting values, we may divide Equation 2 4-9 by x, and let a = tan 0 and b = g/2v0 cos2 0 . For example, selecting the two corner points, we obtain y1 /x1 = a x1 b, and y2 /x2 = a x2 b, with solutions (x2 x1 )a = (x2 y1 /x1 ) (x1 y2 /x2 ), and (x2 x1 )b = (y1 /x1 ) (y2 /x2 ). In terms of a and b,
49 m2 /s2 + 2(9.8 m/s2 )(3 m)]/(9.8 m/s2 ) = 1.77 s, and v0x = (x x0 )/t = 2.8 m/1.77 s = 1.58 m/s. From v0x and v0y we nd the magnitude v0 =
2 2 v0x + v0y = 7.18 m/s and direction
0 = tan1 (v0y /v0x ) = 77.3 .
38
CHAPTER 4 cos2 0 ] sin 0 + = 0,
Problem
72. In your calculus class, you may have learned that you can nd the maximum or minimum of a function by dierentiating and setting the result to zero. Do this for Equation 4-10, dierentiating with respect to 0 , and thus verify that the maximum range occurs for 0 = 45 .
= [sin 0 +
Solution
The derivative of Equation 4-10 with respect to 0 is 2 dx/d0 = 2(v0 /g) cos 20 . This is zero when 20 = 90 , or 0 = 45 as stated. (This is the only maximum, since launch angles are restricted to the range 2 0 < 0 < 90 , and d2 x/d0 < 0.)
Problem
73. A projectile is launched with speed v0 from the edge of a cli of height h; the ground below the cli is at. Using the technique of the preceding problem, show that the maximum range occurs when the launch angle is v0 . 0 = tan1 2 2gh + v0
2 where = sin2 0 + 2gh/v0 . The rst factor is never zero, so = cos2 0 / sin 0 = (1/ sin 0 ) sin 0 . Squaring and simplifying, we nd that 1 1 2gh (sin 0 ) ( . . .)2 = sin2 0 + 2 = 2 2 v0 sin 0 sin 0 2gh + sin2 0 , or 2 + 2 v0 1 1 =1+ , = tan2 0 sin2 0 v0 or tan 0 = , as stated. (With this value 2 v0 + 2gh for 0 , the maximum value of x turns out to be 2 2 (v0 /g) 1 + 2gh/v0 . For h = 0, these values of 0 and x reduce to the case discussed in the text.)
Problem
74. Two projectiles are launched with the same speed v0 , at angles 45 + and 45 . As Fig. 4-15 shows, they have the same horizontal range. Derive an expression for the dierence in their ight times.
Solution
The quadratic formula can be used to solve Equation 4-9 for the horizontal range of a projectile with positive v0x , whose trajectory begins at the origin and ends at the point (x, y) : x = 2 2 (v0 cos2 0 /g)[tan 0 tan2 0 2gy/v0 cos2 0 ]. When the origin is on a cli of height h above where the projectile lands, y = h, and only the solution with the positive sign before the square root corresponds to a positive range. (For y > 0, both solutions might be possible.) Thus, the horizontal range appropriate to the situation in this problem is
2 2 x = (v0 cos 0 /g)[sin 0 + sin2 0 + 2gh/v0 ]. We chose to multiply through one factor of cos 0 in order to simplify the 0 dependence of each term before dierentiating; other choices also work. Inspection of the trajectory shows that it is reasonable to expect one maximum value of x, with v0 constant, for 0 < 0 < 90 . It can be found by the method suggested, or by other methods, such as Lagrange multipliers. In taking the derivative, we use the Product rule, the Chain rule, and the derivatives of the sine, cosine, and square root ( z = z 1/2 ) given in Appendix A. Then,
Solution
The time of ight for a parabolic trajectory between points at the same altitude (y y0 = 0) is t = 2v0y /g (see Equation 4-6 with vy = v0y , or the solution to Problem 67). The dierence in ight-times for two trajectories with the same horizontal range and launch speeds, but dierent angles, is t = (2v0 /g)[sin(45 + ) sin(45 )]
1 = (2v0 /g)2 cos 2 (45 + + 45 ) 1 sin 2 (45 + 45 + ) = (4v0 /g) cos 45 sin = (2 2v0 /g) sin .
(We used an identity for the dierence of sines from Appendix A.)
Problem
75. A well-engineered ski jump is less dangerous than it looks because skiers hit the ground with very small velocity components perpendicular to the ground. Skiers leave the Olympic ski jump in Lake Placid, New York, at an angle of 9.5 below the horizontal. Their landing zone is a horizontal distance of 55 m from the end of the jump. The ground at the point is contoured so skiers trajectories make an angle of only 3.0 with the
g dx . . .]( sin 0 ) 2 d = [sin 0 + v0 0 + cos 0 cos 0 + sin 0 cos 0 ...
CHAPTER 4 ground on landing, as suggested in Fig. 4-38. What is the slope of the ground in the landing zone?
39
Solution
Suppose the particle passes the origin at time t = 0. 1 Then x = v0 t, y = 2 at2 , vx = v0 , and vy = at. The tangents of the angles made by the velocity and displacement vectors with the x-axis are vy /vx = at/v0 1 1 and y/x = 2 at2 /v0 t = 2 at/v0 = 1 vy /vx , respectively, 2 as asserted.
55 m 9.5
Problem
77. Derive a general expression for the ight time of a projectile launched on level ground with speed v0 and launch angle 0 .
3
Solution
Equation 4-8 gives the height of a projectile above its 1 launch site, y y0 = (v0y 2 gt)t. Here, we factored the time to show that the two solutions for zero height are the launch time t = 0, and the time of ight, given 1 by v0y 2 gt = 0, or t = 2v0y /g = 2v0 sin 0 /g.
figure 4-38 Problem 75.
Solution
The direction of the skiers velocity is = tan1 (vy /vx ), where angles are measured CCW from the x-axis, chosen horizontal to the right in Fig. 4-38 with the y-axis upward. In the landing zone, is in the fourth quadrant, which can be represented by a negative angle below the x-axis. The slope of the ground at this point can be represented by a similar angle g , and for the safety of ski jumpers, g = 3.0 . The slope vy /vx = (dy/dt)/(dx/dt) = dy/dx can be calculated by dierentiating the trajectory equation, but it is just as easy in this problem to use Equations 4-5, 4-6 and 4-7. Thus, vx = v0x = v0 cos 0 , and vy = v0y gt = v0 sin 0 gt. The time of ight can be eliminated, since x x0 = v0x t = 55 m is given, 2 so vy /vx = (v0y /v0x ) g(x x0 )/v0x = 2 2 tan 0 g(x x0 )/v0 cos 0 = tan(9.5 ) 2 (9.8 m/s )(55 m)/(28 cos(9.5 ) m/s)2 = 0.874. Finally, = tan1 (0.874) = 41.2 , and g = + 3.0 = 38.2.
Problem
78. An object moves at constant speed v in the x-y plane, describing a circle of radius r centered at the origin. It is on the positive x-axis at time t = 0. Show that the position of the object as a function of time can be written r = r[cos(vt/r) + sin(vt/r) where the argument of ], the sine and cosine is in radians. Dierentiate this expression once to obtain an expression for the velocity and again for the acceleration. Show that the acceleration has magnitude v 2 /r and is directed radially inward (that is, opposite to r).
Problem
76. A particle is moving along the x-axis with velocity v0 in the positive x direction. As it passes the origin, it begins to experience a constant acceleration a in the y direction. Show that, at any subsequent time, the tangent of the angle its velocity makes with the x-axis is twice the tangent of the angle its displacement vector makes with the x-axis.
Problem 78 Solution.
Solution
From the diagram, r = r(cos + sin In radians, ). = s/r, where s is the arc-length, and for constant speed, s = vt. Thus, r = r[cos(vt/r) + sin(vt/r) ].
40
CHAPTER 4 dened as R = [1 + (dy/dx)2 ]3/2 /(d2 y/dx2 ), where the positive (negative) sign is used if the arc length increases in the positive (negative) x direction. (See any comprehensive calculus text.) For the trajectory 2 of Equation 4-9, dy/dx = tan 0 x(g/v0 cos2 0 ), 2 d2 y/dx2 = g/v0 cos2 0 , and the positive sign in the expression for R is used when cos 0 0. Of course, dy/dx = vy /vx gives the direction of the instantaneous velocity, which is zero at the apex of the trajectory. 2 2 2 Then R = (d2 y/dx2 )1 = v0 cos2 0 /g = vx /g at the apex, as above. In terms of the horizontal range, the radius of curvature at an arbitrary point x (0 x xmax ) on the trajectory for which Equation 4-10 applies is R = [1 + (1 2x/xmax )2 tan2 0 ]3/2 (xmax /2 tan 0 ), since dy/dx = (1 2x/xmax ) tan 0 , and d2 y/dx2 = 2 tan 0 /xmax .
d The velocity v = dr/dt. Since dt (sin t) = cos t, d dt (cos t) = sin t, where (omega) = v/r is constant, we have v = v[ sin(vt/r) + cos(vt/r) ]. (Note that v r, i.e., v is tangent to the circle.) The acceleration a = dv/dt = (v 2 /r)[cos(vt/r) + sin(vt/r) has magnitude v 2 /r and direction opposite ] to r, i.e., toward the center of the circle. (This kind of acceleration is therefore called a centripetal, or center-seeking, acceleration.)
Problem
79. In the Olympic hammer throw, contestants whirl a 7.3-kg ball on the end of a 1.2-m-long steel wire before releasing it. In a particular throw, the hammer is released from a height of 1.3 m while moving in a direction 24 above the horizontal. If it travels 84 m horizontally, what is its radial acceleration just before release (see Fig. 4-39)?
Problem
81. Two golfers stand equal distances on opposite sides of a hole, as shown in Fig. 4-40. Golfer A hits his ball at a 50 angle to the horizontal. At the instant she hears As club hit the ball, Golfer B hits her ball at the same speed as A, but at a 40 angle. If the two balls reach the hole simultaneously, how far apart are the golfers? The speed of sound is 340 m/s.
Solution
Just before release, the hammer ball is traveling in a circle (approximately of radius r = 1.2 m), so its radial 2 (or centripetal) acceleration is ac = v0 /r, where v0 is the launch speed. We can determine v0 from Equation 4-9 and the data given for the throw, 0 = 24 , x = 84 m, y = 1.3 m, as in Example 4-5, thus: ac =
2 gx2 1 v0 2 = 892 m/s . = 2 (x tan y) r r 2 cos 0 0
900 figure 4.40 Problem 81 Solution
Problem
80. A projectile is launched at an angle 0 to the horizontal, with sucient speed to give it a horizontal range x. Show that the radius of curvature at the top of its trajectory is given by r = x/2 tan 0 .
Solution
Since both golfers and the hole are on level ground, we can use the result of Problem 77 to determine their separation, R. If golfer As ball is hit at t = 0, it will reach the hole at t = 2v0 sin 50 /g. Golfer Bs ball has a time of ight of 2v0 sin 40 /g, but it starts after a delay of R/(340 m/s), due to the sound travel-time. Since both balls arrive at the hole simultaneously (a double hole-in-one!), 2v0 sin 40 /g + R/(340 m/s) = 2v0 sin 50 /g, or R = (340 m/s)(2v0 /g) (sin 50 sin 40 ). We can eliminate v0 by using Equation 4-10, with range R/2, since the launch angles are complementary. Then, R/2 = 2 2 (v0 /g) sin 20 = (2v0 /g) sin 40 sin 50 , or sin 50 . Substituting 2v0 = gR/ sin 40 above, we nd R = (340 m/s) ( gR/ sin 40 sin 50 /g)(sin 50 sin 40 ), or (340 m/s)2 (sin 50 sin 40 )2 = 364 m. R= 2 (9.8 m/s ) sin 40 sin 50
Solution
At the apex of a parabolic trajectory, the acceleration is perpendicular to the velocity (v = vx , vy = 0) and is therefore entirely radial (a = ar , at = 0). In projectile motion, the acceleration has constant magnitude g, so at the apex, the radius of curvature has magnitude 2 2 R = v 2 /ar = vx /g = (v0 /g) cos2 0 . In terms of the horizontal range, Equation 4-10, this becomes R = (xmax / sin 20 ) cos2 0 = xmax /2 tan 0 . Actually, ar = g, and the radius of curvature is 2 negative, R = vx /g at the apex. It is only the magnitude of R that is discussed in the text and specied in this problem. A negative R means that the slope of the tangent decreases as the trajectory is traced in the direction of increasing arc length. For a plane curve y = f (x), the radius of curvature is
CHAPTER 4
41
Problem
82. A convertible is speeding down the highway at 130 km/h, when the driver spots a police airplane 600 m back at an altitude of 250 m. The driver decelerates at 2.0 km/h/s. (a) If the plane is ying horizontally at a steady 210 km/h, where should the plane be in relation to the car for the police ocer to drop a speeding ticket into the car? Assume the ticket is dropped with no initial vertical motion, and it is in a heavy capsule that experiences negligible air resistance. (b) How fast will the car be moving when the ticket reaches it?
ground, at ta is V = V0 + acar ta = 130 km/h + (2 km/h/s)(21.3 s) = 87.4 km/h = 54.3 mi/h, just under the urban highway speed limit.
Problem
83. A projectile is launched with initial speed v0 at an angle 0 to the horizontal. Find expressions for
Problem 82 Solution.
Solution
(a) Suppose that the road is horizontal and straight, and that the velocities of the car and plane are collinear. In the spirit of Section 3-5, let us work in a reference frame S attached to the car. At t = 0 (when the car spots the plane) S is 600 m from the origin of a frame S attached to the road. The initial horizontal velocity of the capsule relative to S is v0 = v0 V0 = 210 130 = 80 km/h. The horizontal acceleration of the capsule relative to S is a = ax acar = x 0 (2 km/h/s) = 2 km/h/s. (S is an accelerated frame, so we used the derivative of Equation 3-10 with dV /dt = acar .) The horizontal position of the capsule 1 relative to S is therefore x (t) = x + v0 t + 2 a t2 , x 0 where x0 = 600 m. The vertical motion of the capsule is the same in S and S , so if the capsule is dropped at time td , its height is y = 250 m for 0 t td , and 1 y = 250 m 2 g(t td )2 for td t. The capsule arrives at time ta , when y = 0, so ta td = 2y0 /g =
2(250 m)/(9.8 m/s2 ) = 7.14 s. When the capsule arrives, x (ta ) = 0 (its at the car), so 0 = x + v0 ta + 1 a t2 = (0.6 km) 0 2 x a (3600 s/h) + ta (80 km/h)+ 1 t2 (2 km/h/s), or 0 = 2 a 2160 s2 + ta 80 s + t2 . Taking the positive solution of a this quadratic, we nd ta = 402 + 2160 40 = 21.3 s. Therefore, td = 21.3 s 7.14 s = 14.2 s, and x (td ) = 600 m + (80 km/h)(14.2 s) + 1 2 2 (2 km/h/s)(14.2 s) = 229 m. (The plane should be 229 m behind the car when the capsule is dropped.) (b) Finally, the velocity of the car, relative to the
42
CHAPTER 4 the angle the trajectory makes with the horizontal (a) as a function of time and (b) as a function of position.
Solution
(a) The slope of the trajectory is vy /vx , and the angle it makes with the x-axis is = tan1 (vy /vx ). For projectile motion, vx = v0x = v0 cos 0 is a constant, and vy = v0y gt = v0 sin 0 gt; therefore, vy /vx = (v0y /v0x ) (gt/v0x ) = tan 0 gt/v0 cos 0 . (b) We can eliminate t from the expression for the slope by using x x0 = v0x t, or t = (x x0 )/v0x . Thus, 2 vy /vx = tan 0 g(x x0 )/v0 cos2 0 . (The expression for the angles is the inverse tangent of the slopes above.)

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Michigan State University - CEE - CE 221

CE 221Homework 4, due Wednesday September 29, 2010 Solve problems 2.83, 2.89, 2.90, 2.91, 2.96, 2.98, 2.101, 2.112, 2.114, and 2.1151

Michigan State University - CEE - CE 221

Michigan State University - CEE - CE 221

CE 221Homework 5, due Wednesday, October 6, 2010 Solve problems 3.4, 3.10, 3.15, 3.21, 3.37, 3.46, 3.55, 3.64, 3.86 and 3.88 in your book. The red and underlined problem numbers are for those problems to be solved in the recitations.1

Michigan State University - CEE - CE 221

CE 221 Homework 6 Due Wednesday October 13, 2010 Red colored problems to be solved in the recitation. Please add HW 6 problems to the list of problems for sample test 1. Problem 1 10 pointsForce F2 is located in the y-z plane at 30o above the y axis. The

Michigan State University - CEE - CE 221

CE 221Homework 7, due Wednesday October 20, 2010 Solve problems 4.20, 4.25, 4.34, 4.51, 4.59, 4.62 (b, c, d), 4.77, and 4.78 your book. The red and underlined problem numbers are for those problems to be solved in the recitations.1

Michigan State University - CEE - CE 221

CE 221Homework 8, due Wednesday October 27, 2010 Solve problems 4.63, 4.64, 4.74, 4.87, 5.5, 5.12, 5.13, 5.14, 5.27, and 5.40 in your book. The red and underlined problem numbers are for those problems to be solved in the recitations.1

Michigan State University - CEE - CE 221

Michigan State University - CEE - CE 221

MOMENT OF A FORCE MOMENTSCALAR AND VECTOR SCALAR FORMULATION FORMULATIONLEARNING OBJECTIVES LEARNING Be able to understand and define moment Be able to determine the moment of a force Bein 2-D and 3-D cases inPRE-REQUISITE KNOWLEDGE KNOWLEDGE Units

Michigan State University - CEE - CE 221

FORCE SYSTEM RESULTANTSMOMENT ABOUT AN AXIS MOMENTLEARNING OBJECTIVES LEARNING Be able to determine the moment of a Beforce about a specified axis using scalar and vector methods andPRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurements Trigono

Michigan State University - CEE - CE 221

FORCE SYSTEM RESULTANTSMOMENT OF A COUPLE MOMENTLEARNING OBJECTIVES LEARNING Be able to define a couple Be able to determine the moment of a Becouple couplePRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurements Trigonometry concepts Vector conc

Michigan State University - CEE - CE 221

FORCE SYSTEM RESULTANTS RESULTANTSEquivalent System EquivalentLEARNING OBJECTIVES LEARNING Be able to find an equivalent force-couple Besystem for a system of forces and couples systemPRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Trigon

Michigan State University - CEE - CE 221

FORCE SYSTEM RESULTANTS RESULTANTSReduction of a Simply Distributed Load ReductionLEARNING OBJECTIVES LEARNINGBe able to find an equivalent force for a Be simply distributed load simplyPRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Units

Michigan State University - CEE - CE 221

RIGID BODY EQUILIBRIUM EQUILIBRIUMRigid Body Equilibrium 2-D RigidLEARNING OBJECTIVES LEARNING Be able to recognize two-force Bemembers members Be able to apply equations of Be equilibrium to solve for unknowns equilibriumPRE-REQUISITE KNOWLEDGE PRE-

Michigan State University - CEE - CE 221

RIGID BODY EQUILIBRIUM 3-D EQUILIBRIUMEquilibrium of Rigid Body in 3-D EquilibriumLEARNING OBJECTIVES LEARNING Be able to identify support reactions in Be3-D and draw a free body diagram 3-D Be able to identify an indeterminate Be system system Be abl

Michigan State University - CEE - CE 221

STRUCTURAL ANALYSISMethod Of Joints MethodLEARNING OBJECTIVES LEARNING Be able to define a simple truss Be able to determine the forces in Bemembers of a simple truss using the method of joints the Be able to identify the zero-force Be members members

Michigan State University - CEE - CE 221

STRUCTURAL ANALYSISMethod of Sections MethodLEARNING OBJECTIVES LEARNING Be able to determine the forces in Bemembers of a simple truss using the method of sectionsPRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Trigonometry concepts Rect

Michigan State University - CEE - CE 221

STRUCTURAL ANALYSIS ANALYSISFrames and Machines FramesLEARNING OBJECTIVES LEARNING Be able to draw the free body Bediagram of a frame or machine and its members its Be able to determine the forces acting Be at the joints and supports of a frame or mac

Michigan State University - CEE - CE 221

INTERNAL FORCES INTERNALInternal Forces Developed in Structural Members MembersLEARNING OBJECTIVES LEARNINGBe able to use the method of sections for Be determining internal forces in 2-D load case determiningPRE-REQUISITE KNOWLEDGE PRE-REQUISITEUnits

Michigan State University - CEE - CE 221

Michigan State University Department of Civil and Environmental Engineering CE 221 Statics (3 credits) CESpring 2010 Maximum time allowed is 50 minutes Note: - This is a closed book and notes test. You are prohibited from using stored formulae and/or pro

Michigan State University - CEE - CE 221

INTERNAL FORCES FORCESShear Force and Bending Moment Equations and Diagrams EquationsLEARNING OBJECTIVES LEARNING Be able to draw shear force and Bebending moment diagrams for beams beamsPRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurement Tri

Michigan State University - CEE - CE 221