This preview shows page 1. Sign up to view the full content.
Unformatted text preview: CHAPTER 5 FORCE AND MOTION
ActivPhysics can help with these problems: All activities in Section 2, Forces and Motion Section 54: Newton's Second Law Problem
1. A subway train has a mass of 1.5x106 kg. What force is required to accelerate the train at 2.5 m/s2? Solution
Assume that the seat belt holds the passenger firmly to the seat, so that the passenger also stops in 0.14 s without incurring any secondary impact. Then the passenger's average acceleration is aav = (0  vo)/t, and the average net force on the passenger, while coming to rest, is Fav = maav = mvo/t = (60kg)(1l0/3.6)(m/s)/(0.14 s) = 13.1 kN, or about 1.5 tons. (Here, we used the onedimensional form of Newton's second law. The minus sign indicates that the force is opposite to the direction of the initial velocity. It is reasonable to assume that this is a component of the force exerted by the seatbelt.) Solution
F = ma = (1.5x106 kg)(2.5 m/s2) = 3.75 MN. (This is the magnitude of the net force acting; see Table 11 for SI prefixes.) Problem
2. A railroad locomotive with a mass of 6.1x 104 kg can exert a force of 1.2xl05 N. At what rate can it accelerate (a) by itself and (b) when pulling a L4x106kg train? Problem
5. In an xray tube, electrons are accelerated to speeds on the order of 108 mis, then slammed into a target where the); come to a stop in about 1O18S. Estimate the average stopping force on each electron. Solution
Ignoring the probable presence of other forces, we can apply Equation 53 to find (a) a = F/m = (1.2x105 N)/(6.1x104 kg) = 1.97 m/s2, and (b) a = (1.2x105N)/(1.46x106 kg) = 8.21 cm/s2• I Solution
The magnitude of the average force is P = mii = m l6.v/6.tl = (9.11 X 1031 kg)(108 m/s)/(1O18 s) ~ 9xlO5 N. Compared to the TV tube in Example 52, the electron in an xray tube experiences a force billions of times greater. It is a. result of the violence of this interaction that x rays, called bremsstrahlung, are emitted (see Problem 245). Problem
3. A small plane starts down the runway with acceleration 7.2 m/s2• If the force provided by its engine is 1.1x 104 N, what is the plane's mass? II
t Solution
If we assume that the runway is horizontal (so that the vertical force of gravity and the normal force of the surface cancel) and neglect aerodynamic forces (which are small just after the plane begins to move) then the net force equals the engine's thrust and is parallel to the acceleration. The horizontal component of Equation 53 gives the airplane's mass, m= F/a = (1.1xl04 N)/(7.2 m/s2) = 1.53x 103 kg. Problem
6. A 280kg crate is secured in a truck with ropes running horizontally forward and backward, as shown in Fig. 533. What is the maximum tension force these ropes must be able to withstand if the most rapid deceleration anticipated is 6.5 m/s2? Assume the tension is initially zero. I
I
f t ~ Problem
4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does a seatbelt exert on a 60kg passenger during this collision? FIGURE 5":33Problem 6. CHAPTER 5 71 Solution Suppose that the motion of the truck is also horizontal, positive to the right, in the direction shown in Fig. 533, and consider only the horizontal forces. The maximum anticipated net horizontal force on the crate is Fnet == mamax = (280 kg)(6.5 m/s2) = 1.82 kN, where the minus sign indicates a deceleration, or force to the left. The ropes can exert only tensile forces; therefore, the lefthand rope can only pull to the left (Ti < 0), and the righthand rope can only pull to the right (Tr > 0). If there is no friction between the crate and the truck bed, the ropes must supply all the horizontal force on the crate, so Fnet = Ti + Tr = Tr ITtI; that is, the maximum difference in tension between the ropes is 1.82 kN. (We wrote this equation in terms of the absolute values.because that is what is normally meant by the tension in a rope.) If the ropes are initially under zero tension, then all the tension when decelerating is in the lefthand rope, and tension in the righthand rope is zero. maav = m /lv/ /It = (0.170 kg)(50 m/s  0) + (2.5 ms)= 3.40 kN. Problem
9. By how much does the force required to stop a car increase if the initial speed is doubled and the stopping distance remains the same? Solution The average net force on a car of given mass is proportional to the average acceleration, Fav '" aav. To stop a car in a given distance, (x  xo), aav = (0  v5)/2(x  xo), so Fav '" v5. Doubling Vo quadruples the magnitude of Fav, a fact that is important to remember when .driving cat high speeds. Problem
10. A car moving at 70 km/h collides with a concrete bridge support. The bridge support is unaffected, but the front of the car is compressed by 0.94 m. What average force must a seat belt exert in order to restrain a 75kg passenger during this collision? Solution The average acceleration of the car, while stopping in a horizontal distance x  Xo, is aav = (0  vs) + 2(x  xo) = (70 m/3.6 s)2/2(0.94 m) = 201 m/s2, or about 21g. A properly functioning seat belt will constrain a passenger to the seat and prevent any secondary collision with the interior compartment of the automobile. Then the passenger's acceleration is also 201 m/s2, and the magnitude of the average horizontal net force on her/him is simply Fnet = maav = (75 kg)(201 m/s2) = 15.1 kN. This force is supplied mainly by the seat belt. Problem
7. Object A accelerates at 8.1 m/s2 when a 3.3N force is applied. Object B accelerates at 2.7 m/i when the same force is applied. (a) How do the masses of the two objects compare? (b) If A and B were stuck together and accelerated by the 3.3N force, what would be the acceleration of the composite object? Solution In this idealized onedimensional situation, the applied force of F = 3.3 N is the only force acting. (a) When applied to either object, Newton's secorid law gives F = mAaA and F = mBaB, so mB/mA = aA/aB = (8.1 m/s2)/(2.7 m/s2) = 3. (For constant net force, mass is inversely proportional to acceleration.) (b) When F is applied to the combined object, F = (mA + mB)a. Since F = mAaA, and mB = 3mA, one finds a = F/(mA + mB) = mAaA/4mA = ~(8.1 m/s2) = 2.03 m/s2. (Note: It was not necessary to calculate the masses, which are mA = (3.3 N) + (8.1 m/s2) = 0.407 kg, and mB = (3.3 N) + (2.7 m/s2) = 1.22 kg.) Problem
11. The maximum braking force of a 1400kg car is about 8.0 kN. Estimate the stopping distance when the car is traveling (a) 40 lan/hj (b) 60 km/hj (c) 80 km/hj (d) 55 mi/h. Solution The maximum braking acceleration is a = F/m = 8.0x103 N/1400 kg = 5.71 m/s2. (We expressed the braking force as negative because it is opposite to the direction of motion.) (a) On a straight horizontal road, acai traveling at a velocity of VOx = 40 km/h can stop in a minimum distance found from Equation 211 and the maximum deceleration just calculated: x  xo = v5x/2a = (40 m/3.6 s)2+ 2('5.71 m/i) = 10.8 m. For the other initial Problem
8. A hockey player strikes a 170g puck, accelerating it from rest to 50 m/s. If the hockey stick is in contact with the puck for 2.5 ms, what is the average force applied by the stick? Solution The stick is responsible for essentially all the average net force on the puck during impact, so Fav = 72 CHAPTER 5 velocities, the stopping distance is (b) 24.3 m, (c) 43.2 rri, and (d) 52.9 m. Problem
12. As a function of time, the velocity of an object of mass m is given by v = bt2i + (d + d)j, where b, c, and d are constants with appropriate units. What is the force acting on the object, as a function of time? average force of air on the parachute? Assume the parachute provides essentially all the stopping force. Solution
To stop a 380okg object traveling at 240 km/h in 170 m requires an average force of F = rna = ml v6l2(x  xo)/ = (3800 kg)(240 m/3.6 S)2..;2(170 m) = 49.7 kN. Solution
From the discussion leading to Equation 53, F net = mdv/dt = m(2bti+cj). Problem
15. A 1.25kg object is moving in the x direction at 17.4 m/s. 3.41 s later, it is moving at 26.8 m/s at 34.0° to the :vaxis. What are the magnitude and direction of the force applied during this time? Problem
13. A car moving at 50 km/hcollides with a truck, and the front of the car is crushed 1.1 m as it comes to a complete stop. The driver is wearing a seatbelt, but the passenger is not. The passenger, obeying Newton's first law, keeps moving and slams into the dashboard after the car has stopped. If the dashboard compressed 5.0 cm on impact, find and compare the forces exerted on the driver by the seat belt and on the passenger by the dashboard. Assume the two have the same 65kg mass. Solution
Newton's second law says that the average force acting is equal to the rate of change of momentum, F av m~v/~t (as explained following Equation 52). The initial velocity is 17.4i mis, and the final velocity is (26.8 m/s)(i cos 34° + j sin 34°) = (22.2i + 15.0j) mis, so Fav = (1.25 kg)[(22.2 17.4) i + 15.0j](m/s)/(3.41 s) = (1.77i+5.49j) N. This has magnitude 5.77 N and direction 72.2° CCW,to the xaxis. = ~. Solution
The seatbelt constrains the driver to have the same average acceleration as the car, while stopping. Assuming the passenger compartment stays intact and is stopped after moving a horizontal distance of ~Xd = 1.1 m, we can estimate the driver's average acceleration from Equation 211 as ad = v6l2~Xd. Then the magnitude of the average force stopping the driver is Fd = mladl =~(65 kg)(50 m/3.6 S)2..;(1.1 m) = 5.70 kN(over half a ton!), which is exerted mostly by the seat belt. A passenger striking a stationary dashboard with the same speed Vo, which yields a horizontal distance ~xp = 0.05 m before bringing her/him to a stop (probably permanently), experiences an average acceleration of ap = v5 ..;2~xp = (~Xd/~xp)ad = (Ll m/0.05 m)ad = 22.0ad. Therefore, the average force stopping the passenger is 22 times greater than that for the driver, or 22x5.70 kN = 125 kN. Buckleup for safetyit's a law of physics! Section 55: Mass and Weight: The Force of Gravity Problem
16. Show that the units of acceleration can be written N/kg. Why would it make sense to state 9 as 9.8 N/kg when talking about mass and weight? Solution
From Newton's second law, [acceleration] = [Forcel/[massj = N/kg ([... ] means "units of ... "). 9 = 9.8 N/kg is the conversion factor from mass to the everyday meaning of weight (Le., the force of a scale reading at reston the surface of the Earth). Problem
17. My spaceship crashes on one of the Sun's nine planets. Fortunately, the ship's scales are intact, and show that my weight is 532 N. If 1 know my mass to be 60 kg, where am I? Hint: Consult AppeIiaix E. Problem
14. A 3800kgjet touches down at 240 km/h on the deck of an aircraft carrier, and immediately deploys a parachute to slow itself down. If the plane comes to a stop in 170 m, what is the Solution
The surface gravity of the planet is 9 = W/m = 532 N/60 kg = 8.87 m/s2, precisely the value for Venus in Appendix E. CHAPTER 5 73 Problem
18. If I can barely lift a 35kg concrete block on Earth, how massive a block can I lift on the moon? value. What is the weight of a 68~kgastronaut in a shuttle at this altitude? Solution
The magnitude of the Earth's gravitational force on the astronaut at this altitude is Fgrav = mg(r) = (68 kg)(0.93x9.8 rn/s2) = 620N, where g(r) is the gravitational acceleration at the appropriate distance, r, from the center of the Earth. Of course, if the astronaut is orbiting with the shuttle, Le., in free fall around the Earth, his or her "weight" is zero. (An operational definition of ''weight'' is the force read on a scale at rest relative to the object being weighed.) Solution
To lift a 35kg block on Earth requires a muscular force of at least its weight, W = mg = (35 kg) x (9.8 m/s2) = 343 N. The same force on the moon could lift a mass m = W/gmoon = 343 N/(1.62 m/s2) = 212 kg. Problem
19. A cereal box says "net weight 340 grams." What is the actual weight (a) in SI units? (b) in ounces? Solution
(a) The actual weight (equal to the force of gravity at rest on the surface of the Earth) is mg = (0.340 kg)x 2 (9.81 m/s ) = 3.33 N. (b) With reference to Appendix C, (3.33 N)(0.2248x 16 oz/N) = 12.0 oz. (The word "net" in net weight means just the weight of the contents; gross weight includes the weight of the container, etc. This may be compared with the use of the word in net force, which means the sum of all the forces or the resultant force. A net weight, profit, or amount is the resultant after all corrections. have been taken into account.) Problem
23. A neutron star is a fantastically dense object with the mass of a star crushed into a region about 10 km in diameter. If my mass is 75 kg, and if I would weigh 5.8x1014 N on a certain neutron star, what is the acceleration of gravity 011 the neutron star? Solution
If we define weight on a neutron star analogously to its definition on Earth, the surface gravity of the neutron star is an enormous g = W/m = (5.8x 1014 N) + (75 kg) = 7.73x1012 m/s2, nearly 1012 times g. Problem
20. I weigh 735 N. What's my mass? Section 56: Problem Adding Forces Solution
m = Wig = 735 N/(9.81 m/s ) = 74.9 kg (for an object at rest on the surface of the Earth, where 2 9 = 9.81 m/s ). .
2 24. A 50kg parachute jumper descends at a steady 40 km/h. What is the force of air on the parachute? Solution
A steady descent means that the vertical acceleration is zero. From Newton's second law, the vertical net force is also zero. The force of gravity (equal to the weight of the parachutist) and the force of the air are the only vertical forces acting, so (with positive upward) Fair  W = 0, or Fair = (50 kg)(9.8 m/s2) = 490 N.
\ Problem
21. A bridge specifies a maximum load of 10 tons. What's the maXimum mass, in kilograms, that the bridge can carry? Solution
The conversion between mass and (ordinary) weight' is m = Wig. Because the English unit of mass (the slug) is rarely used, the direct equivalence between mass in SI units and weight (force) in English units is usually given, as in Appendix C. Thus 10 tons = 2x104 lb is equivalent to the weight of (2x104 Ib)(0.4536 kg/lb) = 9.07x103 kg. Problem
25. A rope can withstand a maximum tension force of 450 N before breaking. The rope is used to pull a 32kg bucket of water upward. What is the maximum upward acceleration if the rope is not to break? Problem
22. The gravitational acceleration at a typical space shuttle altitude is about 93 percent of its surface Solution
Assume that the only vertical forces acting on the bucket are the upward (positive) force exerted by the 74 CHAPTER 5 rope, .Prope :5 450 N, and the downward force of gravity, Fgrav = mg. (Air resistance is ignored.) The vertical component of Equation 53 applied to the bucket is Fnet = Frape + Fgrav = ma, so a = 2 (Frope  mg)/m :5 (450 N/32 kg)  9.8 m/s = 2 4.26 m/s , which is the maximum upward acceleration aA::hievable with this rope. (Note that if the tensile strength of the rope were less than the weight of the bucket, 314 N, the maximum acceleration would have been downward; see Problem 60.) or Flift = m(g + ay).(a) At constant altitude, ay = 0, and Flift = mg = (4.5 X 1Q5kg)(9.8 m/i) = 4.41 MN, the airplane's weight. (b) If ay = 1.1 m/i, then Flift = (4.41 MN)(l + 1.1/9.8) = 4.91 MN. Problem
29. An airplane encounters sudden turbulence, and you feel momentarily lighter. If your apparent weight seems to be about 70% of your normal weight, what are the magnitude and direction of the plane's acceleration? Problem
26. A 930kg motorboat aA::celerates away from a dock 2 at2.3 m/s • Its propeller provides a thrust force of 3.9 kN. What is the drag force exerted by the water on the boat? Solution
The vertical forces acting on you are gravity downward (mg) and the normal force of your seat (N 2: 0). The latter is what you experience as your apparent weight. During the turbulence, the vertical component of Newton's second law gives Fnet y = mg + N = mg + 70%mg = may; therefore ' 2 ay = 30%g = 2.94 m/s . If the horizontal components of the acceleration are zero, then the plane's aA::celeration 2.94 m/s2 downward. is Solution
The horizontal forces acting on the boat are the thrust and the oppositely directed drag force (take the positive :vaxis in the direction of the aA::celeration). Then the horiwntal component of Newton's second law gives Fnet,z = Fthrust,z + Fdrag,z = 3.9 kN + 2 Fdrag,z = maz = (930 kg)(2.3 m/s ), from which one finds Fdrag,z = 2.14 kN  3.9 kN = 1.76 kN. (As expected, the horizontal component of the drag force is negative, Le., opposite to the direction of the aA::celeration. ) Problem
30. A 74kg tree surgeon rides a "cherry picker" lift (Fig. 534) ,to reach the upper branches of a tree. What force does the bucket of the lift exert on the surgeon when the bucket is (a) at rest; (b) moving upward at a steady 2.4 m/s; (c) moving downward at a steady 2.4 m/s; (d) accelerating upward at 2 1.7 m/s ; (e) aA::celerating downward at 1.7 m/s2? Problem
27. An elevator aA::celerates ownward at 2.4 m/s2• d .What force doe; the floor of the elevator exert on a 52kg passenger? Solution
We assume that the only vertical forces acting on the tree surgeon are those given, namely, the force of gravity, mg acting downward, and the normal force of the bucket, N 2: 0 aA::ting upward. Then Fnet,y = mg + N = may, or N = m(g + ay). (a), (b) and (c) When the acceleration of the tree surgeon. is zero, the normal force is equal (in magnitude) to the weight, N = mg = (74 kg)(9.8 m/s2) = 725 N. (d) If ay = 2 +1.7 tn/s , N = (74 kg)(9.8 + 1.7) m/s2 = 851 N, while if (e) ay = 1.7 m/i,N = 599 N. Solution
The passenger also accelerates downward with 2 ay = 2.4 m/s (yaxis positive upward), so the vertical component of tJIe net force on the passenger is Fnet,y = may. The only vertical forces aA::ting n the o passenger are the force of gravity, Fg,y =  W = mg, and the normal force of the floor, Fnorm,y = N. Therefore, Fnet y = mg + N = may, or , 2 N = m(g + ay) = (52 kg)(9.8  2.4) m/s = 385 N. Problem
28. What is the vertical lifting force on a 747 jetliner when the plane is (a) flying at constant altitude and (b) aA::celerating pward at 1.1 m/s2? The u aircraft's mass is 4.5x105 kg. Problem
31. At liftoff, a space shuttle with 2.0x 106 kg total masS undergoes an upward aA::celeration f 0.60g. o (a) What is the total thrust force developed by its engines? (b) What force does the seat exert on a 60kg astronaut during liftoff? Solution
The vertical forces on the airplane are lift and gravity, so with the yaxis upward, Fnet,y = Flift mg= may, CHAPTER 5 75 Solution
(a) At liftoff, the only significant vertical forces on the space shuttle are gravity (Fg,y = mg downward) and the thrust (Fy > 0 upward). (Air resistance can be neglected because the initial velocity is zero.) Therefore, Fnet,y = Fy  mg = may, or Fy = 2 meg + ay) = mg(l + 0.6) = (2x106 kg)(9.8 m/s ) X (1.6) = 31.4 MN. (b) The vertical forces on an astronaut are her Ihis weight and the normal force of the seat back (N > 0 upward; the astronauts are seated facing upward at liftoff). The acceleration of the astronaut is the same as that of the shuttle, so with reasoning analogous to part (a) above, N = meg + ay) = (60 kg)(9.8 m/s2)(1.6) = 941 ~. Problem
34. A 2.50kg object is moving along the :z;..axis t a 1.60 m/s. As it passes the origin, two forces F1 and F2 are applied, both in the y direction (+ or ). Fl=15j N. The forces are applied for 3.00 s, after which the object is at the point x = 4.80 m, y = 10.8 m. Find F2. Solution
The motion is in the xy plane, and the object's acceleration a= (Fl +F2)/m has only a 'y component, ax = 0, ay = (Fly + F2y)/m. For constant acceleration, the given initial conditions (xo = Yo 0, Vox = 1.60 mis, vOy = 0) imply y(3 s) = 10.8 m = !ay(3 s)2 or ay = 2.4 m/s2. Then F2y , 2 may  Fly = (2.5 kg)(2.4 mls ) 15 N = 9 N, and F2= 9j N. = Problem
32. A ballet dancer (Fig. 535) executes a vertical jump during which the floor pushes up on his feet with a force 50% greater than his weight. What is his upward acceleration? = Section 57: Problem Newton's Third Law Solution
The vertical component (positive upward) of Newton's second law (with just the given forces acting) gives ay = Fnet,ylm = (1.50 mgmg)/m = 0.5g = 4.9 m/s2• 35. What upward force does a 5600kg elephant exert on Earth? Solution
The upward force that the elephant exerts on the Earth is equal in magnitude to the downward force of gravity that the Earth exerts on the elephant (Newton's third law). The latter is mg =(5600 kg) x 2 ' (9.8 mls ) = 54.9 kN. Problem
33. An elevator moves upward at 5.2 m/s. What is the minimum stopping time it can have if the passengers are to remain on the floor? Solution
I Problem There are two vertical forces on a passenger, gravity downward (equal to his or her weight) mg, and the upward normal force of the floor N ~ (we take the , yaxis positive upward). The latter is a contact force and always acts in a direction away from the surface of contact. (Otherwise" we would be dealing with an adhesive force.) The condition N = 0 is the limit that the surfaces remain in contact. Of course, as long as the passenger is in contact with the floor, his or her vertical acceleration is the same as the floor's and the elevator's. Using the y component of Newton's second law N  mg = may, we may express the condition for contact as N = m( ay + g) ~ 0, or since m is positive, ay ~ g. In words, the passenger remains in contact with the floor as long as the vertical acceleration of the elevator is either upward, or downward with magnitude less than g. To come to a stop (vy = 0) from an initial upward velocity (voy = 5.2 m/s) requires a time t = (0  voy)lay = voyl( ay), and therefore t ~ voylg = (5.2 m/s)/(9.8 m/s2) = 0.531 s is the condition for the passengers to stay on the floor. ° 36. Blocks of 1.0, 2.0, and 3.0 kg are lined up on a table, as shown in Fig. 536. A rightwardpointing 12N force is applied to the leftmost block. What force does the middle block exert on the rightmost one? FIGURE 536 Problem 36 Solution Solution
The analysis of the succeeding problem shows that the horizontal component of Newton's second law applied to each block separately gives Fapp + F2l = ml a, F12 + 76 CHAPTER 5 F32 = m2a, and F23 = m3a. (Here, F12 is the force block 1 exerts on block 2, etc., and all the blocks have the same acceleration a, positive to the right.) Adding all three equations and using Newton's third law (F12 + F21 = 0, etc.), one finds a = Fapp/(ml + m2 + m3)' Then F23 = m3Fapp/(ml + m2 + m3) = (3 kg) x (12 N)/(6 kg) = 6 N. Solution
A coordinate system orbiting with the astronaut and satellite can be treated like an inertial system if the force of gravity is ignored. The contact force which separates the objects acts with equal magnitude on each (Newton's third law) imparting accelerations of aA = F/mA and as = F/ms to the astronaut and satellite, respectively (Newton's second law). (We chose the positive direction parallel to the astronaut's motion.) (a) The accelerations act for a time t = 0.89 s and impart velocities VA= aAt = Ft/mA and Vs =: Ft/ms to each object. Using the data given, we find that VA = (120 N)(0.89 s)/(68 kg) = 1.57 mis, and Vs = 0.254 m/s. (b) After separating, the astronaut and satellite coast at constant velocity relative to their original point of contact in the' orbiting system. In one minute, they coast VAt = (1.57 m/s)(60 s) = 94.2 m and vst = (0.254 m/s) x (60 s) = 15.3 m, and so are 109 m apart. Problem
37. Repeat the preceding problem, now with the leftright order of the blocks reversed. (That is, find the force on the rightmost mass, now 1.0 kg.) Solution
Assume that the table surface is horizontal and frictionless so that the only horizontal forces are the . applied force and the contact forces between the blocks (positive to the right). The latter we denote by F12, etc., which means the force block 1 exerts on block 2, etc. Since the blocks are in contact, they all have the same acceleration a, to the right. Newton's second law applied to each block separately is Fapp + F23 = m3a, F32 + F12 = m2a, and F21 = ml a (where we just consider horizontal components). Similarly, Newton's third law applied to each pair of contact forces is F32 + F23 = 0, and F21 + F12 = O. Adding the secondlaw equations and using the thirdlaw equations, we find Fapp + (F23 + F32) + (F12 + F21) = Fapp = (ml + m2+ m3)a, or a = Fapp/(ml + m2. + m3)' Substituting this into the secondlaw equation for ml, we find F21 = mla = m1Fapp/(ml +m2+m3) = (1 kg)(12 N) + (1 + 2 + 3) kg = 2 N. This is the force that the middle block exerts on the block to its right; the other contact force, F23 = 6 N, can be found from a similar procedure. . Problem
39. I have a mass of 65 kg. If I jump off a 12ocm~high table, how far toward me does Earth move during the time I fall? Solution
As mentioned at the end of Section 57, during free fall, the accelerations of you and the Earth have magnitudes mg/m = 9 and mg/ME, respectively, in opposite directions. If you and the Earth both start from rest, the distance fallen by each is d = !gt2 and dE = !(mg/ME)t2 = (m/ME)d, where d+dE = 1.2 m. Solving for dE, we find dE = (m/ME)(1.2 mdE), or dE = (1.2 m)(m/ME)(1 + m/ME)l. Since m/ME = 65/5.97x1024 = 1.09x 1023 is so small, dE ~ (1.09xlO23)(1.2 m) ~ 1.3xlO23 m. (This is about 105 times smaller than the smallest physically meaningful distances studied to date.) Problem 37 Solution. Problem
38. A 68kg astronaut pushes off a 42okg satellite, exerting a force of 120 N for the 0.89 s they're in contact. (a) Find the speed of each after they've separated. (b) How far apart are they after 1.0 min? Problem 39 Solution. CHAPTER 5 77 Problem
40. A child pulls an llkg wagon with a horizontal handle whose mass is 1.8 kg, giving the wagon and handle an acceleration of 2.3 m/s2• (a) The tension at each end of the handle is different. Why? (b) Find the tension at each end of the handle. Solution
(a) The net horizontal force on the handle is just the differencein the (magnitude of the) tension at either end. (In the sketch, we show just the horizontal forces on the wagon and handle, with positive to the right.) Newton's second law equates this to the product of the handle's mass and its horizontal acceleration, T1  T2 = mHa. Since neither mH nor a is 0, one can conclude that T1 f:. ~2. (b) The net horizontal force on the wagon (by Newton's third law) is equal in magnitude to the tension in the handle at that end, and since the acceleration for both wagon and handle are the same, T2 = mwa. This and the equation in part (a) can be solved for the tensions, T2 = (11 kg)x (2.3 m/s2) = 25.3 N, and T1 = mHa + T2 = (mH + mw)a = (12.8 kg)(2.3 m/i) = 29.4 N.
action/reaction pair ~ (Note: the tension has the same magnitude at every point in a rope of negligible mass.) (a) Add all the equations of motion (the tensions cancel in pairs due to Newton's third law): Fth = (ml + m2 + ma)a = (2200 + 310 + 260) kg (1.9 m/s2) = 5.26 kN. (b) T1 = Fth  mla = (m2 + ma)a = (570 kg) x . (1.9m/s2) = 1.08 kN. (c) T2 = maa = (260 kg) x (1.9 m/s2) =494 N. (d) m2a = (310 kg)(1.9 m/s2) = 589 N. FIGURE 537 Problem 41 Solution. Problem
42. In a tractorpulling contes~, a 2300kg tractor pulls a 4900kg sledge with an acceleration of 0.61 m/s2• If the tractor exerts a horizontal force of 7700 N on the ground, determine the magnitudes of (a) the force of the tractor on the sledge, (b) the force of the sledge on the tractor, and (c) the frictional force exerted on the sledge by the ground. Solution
(b) The horizontal forces acting on the tractor and the sledge are shown in separate diagrams. (FST is the force of the sledge on the tractor, FCT is the force of the ground on the tractor, etc.) We assume that the tractor and sledge have the same acceleration, a = 0.61 m/s2, collinear to the forces. The horizontal component of Newton's second law (the equation of motion) for the tractor is FCT  FST = mTa (positive component in direction of motion). Thus, FST = 7700 N  (2300 kg)(0.61 m/s2) = 6.30 kN. The Problem 40 Solution. Problem
41. A 2200kg airplane is pulling two gliders, the first of mass 310 kg and the second of mass 260 kg, down the runway with an acceleration of 1.9 m/s2 (Fig. 537). Neglecting the mass of the two ropes and any frictional forces, determine (a) the horizontal thrust of the plane's propeller; (b) the tension force in the first rope; (c) the tension force in the second rope; and (d) the net force on the first glider. Solution
Assuming a level runway (as shown in Fig. 537), we may write the horizontal component (positive in direction of a) of the equations of motion (Newton's second law) for the three planes (all assumed to have the same a) as follows: Fth  T1 = mla(airplane), T1 T2 = m2a (first glider), T2 = m3a (second glider). Problem 42 Solution. 78 CHAPTER 5 direction of FST is opposite to a. (a) From Newton's third law, FTS == FST, so FTS is 6.30 kN in the direction of a. (c) The horizontal component of the equation of motion of the sledge is FTs  Fa~= msa, so that Fas = 6.30 kN  (4900 kg)(0.61 m/s ) = 3.31 kN. FGs, which is a frictional force, is directed opposite to a. Solution
At rest on the Earth's surface, the force of the spring is equal in magnitude to the weight of the fish, kx = mg. Therefore, x = mg/k = (6.7 kg)(9.8 m/s2);(340 N/m) = 19.3 em. Problem
47. A father pulls his 27kg daughter across frictionless ice, USulga horizontal spring with spring constant 160 N/m. If the spring is stretched 32 em from its equilibrium position, what is the child's acceleration? Section 58: Problem Measuring Force 43. What force is necessary to stretch a spring 48 em, if the spring constant is 270 N/m? Solution
If the spring obeys Hooke's law, the necessary applied force is kx = (270 N/m)(0.48 m) = 130 N. (We did not use the minus sign in Equation 59 because the force applied to the spring is the negative of the force exerted by the spring, by Newton's third law, i.e., Fapp = Fspr = kx.) Solution
The tension in the stretched spring is /kxl = (160 N/m)(0.32 in) = 51.2 N, and if we assume this is the only horizontal force acting on the child, it will impart an acceleration of a = F/m = 51.2 N/27 kg = 1.90 m/s2 to her, in the direction her father pulls. Problem
44. A 35N force is applied to a spring with spring constant k = 220N /m. How much does the spring stretch? Problem
48. You want to make a spring scale that reads directly in mass units. If a change of 1 em in the position of the scale's indicating needle is to correspond to a mass difference of 100 g, what should be the spring constant? Solution
From Hooke's law, a force of magnitude 35 N produces a stretch (or compression) of Ixl = IFI /k = (35 N);(220 N/m) = 15.9 em. Solution
The force exerted by a 100g mass when weighed is N. If this must produce a stretch of 1 em, a spring constant of k = IF/xl = 0.98 N/1 em = 98 N/m would be required.
my = 0.98 Problem
45. A spring sketches 22 em when a 40N force is applied. If a.6.1kg mass is suspended from the spring, how much will it stretch? Problem
49. A biologist is studying the growth of rats in an orbiting space ~tation. To determine a rat's mass, she puts it in a 320g cage, attaches a spring scale, and pulls so the scale reads 0.46 N. If the resulting acceleration of the rat and cage is 0.40 m/s2, what is the rat's mass? Solution
Hooke's law relates the magnitude of the applied force (the "reaction" to the spring force in Equation 59) to the magnitude of the stretch; hence k = 40 N ;'0.22 m = 182 N/m is the spring constant for this spring. If a 6.1 kg mass is suspended from the spring, it will exert a force equal to its weight, so the stretch produced in the spring will be x = (6.1 kg)x (9.8 m/s2)/(182 N/m) ~ 32.9 em. Solution
According to the scale, a force of 0.46 N applied to the cage and rat produces an acceleration of 0.40 m/s2, so their combined mass is F/a = (0.46 N)/(OAO m/s2) = 1.15 kg. ,:[,herat's mass is this minus the cage's, or 1150 320 = 830 g. Problem
46. A spring with spring constant k = 340N/m is used to weigh a 6.7kg fish. How far does the spring stretch? Problem
50. An elastic tow rope has a spring constant of 1300 N/m. It is connected between a truck and a 1900kg car. As the truck tows the car, the rope CHAPTER 5 stretches 55 cm. Starting from rest, how far do the truck and car move in 1 min? Assume the car experiences negligible friction. 79 Problem
52. An accelerometer consists of a spring of spring constant k = 1.25 N/m and unstretched length £0 = 10.0 cm fastened to a frictionless surface by a pivot that allows it to swivel in any direction in a horizontal plane. A 50.Ogmass is attached to the other end of the spring, as shown in Fig. 538. The whole system is mounted securely in an automobile. When the vehicle accelerates, the spring provides a force to keep the 50g mass accelerating with the vehicle; by measuring.the stretch of the spring, the acceleration can then be determined. To calibrate the accelerometer, circles marked with values of acceleration can be drawn on its frictionless surface. (a) How far apart should the circles be jf each represents an acceleration of0.250 m/52 larger than the next smaller circle? (b) What should be the radius of the circle marked 2.0 m/s2? (c) How do you read the direction of the car's acceleration from this device? Solution
Assume, also, that the car accelerates on level ground, and that the rope extends instantaneously to a fixed length. Then the tension in the rope, T = kx = (1300N/m)(0.55 m) = 715 N, is the only horizontal force on the car, if friction is negligible, sOa = T/m = 715 N/1900 kg = 0.376 m/s2. The distance traveled from rest in 1 minute is s = !at2 = 0.5(0.376 m/s2) x (60 s)2 = 677 m. Problem
 51. A 7.2kgmass is hanging from the ceiling of ali elevator by a spring of spring constant 150 N/m whose unstretched length is 80 cm. What is the overall length of the spring when the elevator (a) starts moving upward with acceleration 2 . 0.95 m/s ; (b) moves upward at a steady 14 m/s; (c) comes to a stop while moving upward at 14 mis, taking 9.0 s to do so? (d) If the elevator measures 3.2 m from floor to ceiling, what is the maximum acceleration it could undergo without the 7.2kg mass hitting the floor? Solution
There are two vertical forces acting on the mass, the spring force Fs (positive upward) and gravity mg. Newton's second law for the mass yieldsFs  mg = rna, or Fs = m(g + a). (a is the vertical acceleration of the elevator; we assume that the spring responds so quickly that the vertical acceleration of the mass is the same.) The stretch in the spring (assumed to obey Hooke's law) is y = Fs/k (y is negative measured downward from the lower end of the unstretched spring), so the total length of the spring is L = 80 em + Iyl = 80 em + m(g + a)/k. (a) If we substitute numerical values for m, k, and a, we find L = 0.80 m + (7.2 kg)(9.8 + 0.95)(m/s2)/(150 N/m) = 1.32 m. (b) When the velocity is constant, a = 0, and L = 1.27 m. (c) The acceleration (assumed constant) is a = D.v/D.t = (0 14 m/s)/9 s = 1.56 m/s2, which yields L = 1.20 m. (d) The total length of the spring has to be less than or equal to the height of the elevator compartment, or 3.2 m ~ L = 0.80 m + m(g + a)/k. This inequality can be solved for a: a ~ [(2.40 m) kim] 2  9 = [(2.40 m)(150 N/m)/7.2 kg!  9.8 m/s = 2. This is the ~aximum upward acceleration, 40.2 m/s provided the elastic behavior of the spring can be extrapolated to four times its unstretched length. FIGURE 538 Problem 52. Solution
(a) This accelerometer gives accurate readings on level ground. Then the only horizontal force on the mass is the spring force, with magnitude F = kx = ma, or x = (m/k)a. Thus & = (m/k)b.a, so for increments of D.a = 0.25 m/52, D.x = (0.05 kg/1.25 N/m) x (0.2~ m/s2) = 0.01 m = 1 em. (b) If a = 2 m/s2, x = (0.05 kg/1.25 N/m)(2 m/s2) = 8 em. Since the unstretched spring has length 10 em, the radius of this circle should be 10 em + x = 18 em. (c) If the mass is at rest relative to the car, a = acar is in the direction of thespfing force, tow~d the center of the circles~ (The case in which the spring is compressed, although possible, represents unstable equilibrium, as in Fig. 1418, which is improbable.) CHAPTER 5 Paired Problems Problem
53. Starting from rest, a 94Dkg racing car covers 400 m in 4.95 s. What is the average force acting on the car? Solution
The reasoning in the solution to the previous problem shows that the bumper deformation must be at least m /2Fmax = (1300 kg)(10 m/3.6s)2 /2(65 kN) = 7.72 em. jv81 Solution
The car's average acceleration (for straightline motion) is aay = 2(x  xo)/t2 (see Equation 210 with Vo = 0). Newton's second law gives the average net force on the car as Fay maav == 2(940 kg)(400 m) + (4.95 S)2 = 30.7 kN, in the direction of the motion. Problem
57. You step into an elevator, and it accelerates to a downward speed of 9.2 m/s in 2.1 s. How does your apparent weight during this acceleration time compare with your actual weight? = Solution Problem
54. Starting from rest, a 200g hockey puck moves 1.1 m during the 75 ms it's in contact with the stick. What average force does the stick exert on the puck? Your apparent weight is the force you exert on the floor of the elevator, equal and opposite to the upward force the floor exerts on you. The latter and gravity, which equals your actual weight, are the two forces which determine your vertical acceleration, ay. If we take the positive direction vertically upward, then Wapp  W = may = (W/g)ay, or Wapp/W = 1 + a'll/g. (We expressed your mass in terms of your actual weight in order to facilitate comparison with your apparent weight.) As expected, Wapp/W> 1 for upward acceleration ay > 0, and vice versa for downward acceleration as in this problem. If the elevator starts from rest, its acceleration is (vO)/t = (9.2 m/s)/(2.1 s) = 4.38 m/s2 and Wapp/W = 1  4.38/9.8 = 55.3%. Solution
As in the previous problem, Newton's second law and the kinematics of uniformly accelerated motion applied to the puck give Fay = maay = 2m(x  xo) 2 t = 2(0.2 kg)(l.1 m)/(75 ms) = 78.2 N. + Problem
55. In an eggdropping contest, a student 85g egg in a styrofoam block. If the egg is not to exceed 1.5 N, and if the the ground at 1.2 mis, by how much styrofoam crush? encases an force on the block hits must the Problem
58. A spacecraft blasts off from the moon's surface, achieving an upward speed of 2.5 kIll/s at the end of its rocket firing. During the firing, an astronaut on board feels she weighs exactly the same as she would on Earth. How long does the rocket firing last? Solution
In order that the average net force on the egg not exceed the stated limit, the magnitude of the deceleration should satisfy aav :5 Fmax/m = 1.5 N/0.085 kg = 17.6m/s2. From an initial speed of Vo = 1.2 mis, this implies a stopping distance x  xo satisfying aav = /0 /2(x  xo) :5 17.6 m/i, or x  xo 2: (1.2 m/s)2 /(~5.3 m/s2) = 4.08 em. (Note: in Solution
The vertical motion of the rocket is like the motion of the elevator in the previous problem, with ay = (2.5 km/s)/t > O. However, we must use the surface gravi~y of the moon in place of 9 and in interpreting apparent and actual weights. Thus, W= mgmoon and Wapp = mg (same as actual weight on Earth), so 9/ gmoon = 1 + ay / gmoon' Solving for ay and then t, after consulting Appendix E, we find t = (2.5 km/s) + (9.81 .1.62)(m/s2) = 305 s. vfil an actual case, the stopping force is the magnitude of the vector sum of the force of the styrofoam block and gravity acting on the egg.) Problem
56. In a frontend collision, a 1300kg car with shockabsorbing bumpers can withstand a maximum force of 65 kN before damage occurs. If the maximum speed for a nondamaging collision is 10 km/h, by how much must the bumper be able to move relative to the car? Problem
59. A 20kg fish at the end of a fishing line is being yanked vertically into a boat. When its acceleration reaches 2.2 m/s2, the line breaks, and CHAPTER 5 the fish goes free. What is the maximum tension the line can tolerate without breaking? Solution Consider the vertical motion of the fish, acted on by two forces,the tension in the line, T(positive upward), a.ndgravity, mg. Newton's second law gives T  mg == may, or T == m(g + ay). The maximum upward acceleration the line can tolerate when pulling on a 2Dkgfish is 2.2 m/s2; hence T < 20 kg (9.8 + 2.2) m/s2 == 240 N. Problem 60. Tarzan (mass 73 kg) slides down a vine that can withstand a maximum tension of 620 N before breaking. What minimum"acceleration must Tarzan have? Solution There are but two vertical forces acting on Tarzan (like the fish in the preceding problem). Thus, in order for the vine not to break, T == m(g + ay) ~ Tmax, which implies ay ~ (Tmax/m)  9 == (620 N/73 kg)9.8 m/s2 == 1.31 m/s2. That is, Tarzan must be accelerating downward with a magnitude at least 1.31 m/s2• (Tarzan's weight of 715 N is greater than the vine's tensile strength.) Problem 61. A 2.Dkgmass and a 3.0kg mass are on a horizontal frictionless surface, connected by a massless spring with spring constant k == 140 N/m. A 15N force is applied to the larger mass, as shown in Fig. 539. How much does the spring stretch from its equilibrium length? Solution The spring stretches until the acceleration of both masses is the same (positive in the direction of the applied force in Fig. 539). Since the spring is assumed massless, the tension in it is the same at both ends. (If this were not so, there would be a nonzero net horizontal force on the spring; hence its acceleration would be infinite, a == (F i= O)/(m == 0).) The magnitude of the spring tension is given by , 81 Hooke's law, Fa == k lxi, where Ixl is the stretch. The horizontal component of Newton's second law applied to each mass is Fapp  Fa == m3a and Fa == m2a, as indicated in the sketch. Eliminating the acceleration between. these equations gives us Fapp  Fs == m3(Fs/m2) or Fa == k Ixl == Fapp/(l + m3/m2), from which the stretch is easily found: Ixl == Fapp/k(l + m3/m2) == (15 N/140 N/m)/(l + 3/2) == 4.29 cm. Problem 62. Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k == 8.1 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 5.1 cm from its equilibrium length, what is the applied force? Solution The equations of motion (Newton's second law) for the two crates are the same as those of the masses in the previous problem, namely Fapp  Fs == M a (for the larger) and Fa == ma (for the smaller). The magnitude of the compression in the spring is still given by Hooke's law, Fs == klxl. Thus, Fapp == Fs(1 + M/m) == k Ixl(1 + M/m) == (8.1 kN/m) (0.051 m)(l +640/490) == 953 N. "t:, Supplementary Problems Problem 63. In throwing a 200g ball, your hand exerts a constant upward force of 9.4 N for 0.32 s. How high does the ball rise after leaving your hand? Solution The maximum height to which the ball rises after leaving your hand can be calculated from its initial velocity vOy (since vy == 0 when Y == Ymax in Equation 211, and we neglect air resistance), Ymax  Yo == v5y/29. To find vOy, the vertical forces acting on the ball dl1ring the 0.32 s it is thrown must be considered. These are the applied force of your hand, Fapp == 9.4 N (positive upward), and the weight of the ball, mg. They impart an upward acceleration of (Fapp mg)/m to the ball (Newton's second law), which r~ul~s in an initial upward speed for the throw of vOy == (Fapp  mg)t/m (definition of acceleration). Putting this together and substituting the numerical
:,'1 FIGURE 539 Problem 61 Solution. 82 CHAPTER 5 values given, we get
r;> .l'app Problem )2 t 2
2g Ymax  Yo = ( :;:;:; 9  =
Problem (9.4 N _ 9.8m)2 (0.32 s)2 0.2 kg s2 2(9.8 m/s2) = 7.23 m. 66. An elevator cable can withstand a maximum tension of 19,500 N before breaking. The elevator has a mass of 490 kg, and a maximum acceleration of 2.24 m/s2. Engineering safety standards require that the cable tension never exceed twothirds of the breaking tension. How many 65kg people can the elevator safely accommodate? 64. What downward force is exerted on the air by the blades of a 4300kg helicopter when it is (a) hovering at constant altitude; (b) dropping at 21 m/s with speed decreasing at 3.2 m/s2; (c) rising at 17 m/s with speed increasing at 3.2 m/s2; (d) rising at a steady 15 m/s; (e) 2rising . ... at 15 m/s with speed decreasing at 3.2 m/s ? Solution
The tension in the elevator cable is T = m(g + a) (positive up), where m is the total mass of the elevator and its passengers. For safety's sake, T ::;; ~x 19.5 kN = 13.0 kN, so m should be :::;13.0 kN :(9.8 + 2.24) m/,l = 1080 kg. Since m equals 490 kg + n(65 kg), where n is the number of passengers, n < (1080  490)/65 = 9.07, or n :::;9. Solution
From Newton's third law, the downward force exerted on the air by the helicopter is equal and opposite to the upward force on the helicopter (the engine's thrust). If we neglect air resistance, the thrust and gravity are the only vertical forces acting, so Newton's second law for the helicopter (positive component up) is Fth  mg = rna, or Fth = m(g + a). (a) Hovering means a = 0 (also v = 0, but v doesn't enter the equation of motion if air resistance is neglected), so 2 Fth = mg = (4300 kg)(9.8 m/s ) = 42.1 kN. (b) If v is decreasing downward, a 3.2 m/s2 upward, so Fth (4300 kg)(9.8 + 3.2) m/s2 = 55.9 kN. (c) a is the same as in part (b), and so is Fth. (d) If v = constant, a = 0, and Fth is the same as hovering. (e) If v is decreasing upward, a is downward, so Fth = (4300 kg)(9.8 3.2) m/s2 = 28.4 kN. Problem
67. You have a mass of 60 kg, and you jump from a 78cmhigh table onto a hard floor. (a) If you keep your legs rigid, you come to a stop in a distance of 2.9 cm, as your body tissues compress slightly. What force does the floor exert on you? (b) If you bend your knees when you land, the bulk of your body comes to a stop over a distance of 0.54 m. Now estimate the force exerted on you by the floor. Neglect the fact that your legs stop in a shorter distance than the rest of you. = = Solution
Suppose the motion is purely vertical. You would hit the floor with velocity v = ../2iTi (positive upward), where h = 78 cm. Your average acceleration while stopping is a = v2/2d, where d is the distance over which your body comes to rest. The net force exerted on you is ma = m(v2/2d) = m(2gh/2d) = mg(h/d), while the force exerted on you by the floor is F  mg = ma, or F = (1 + h/d)mg, expressed as a factor times your weight, mg = (60 kg)(9.8 m/s2) = 588 N. (a) With your legs kept rigid, F = (588 N)x (1 + 78/2.9) = 16.4 kN (within one order of magnitude of pro'ducing certain injury). (b) Under the conditions stated, a much safer value of F = (588 N)(l + 78/54) = 1.44 kN is sustained. Problem
65. What engine thrust (force) is needed to accelerate a rocket of mass m '(a) downward at 1.40g near Earth's surface; (b) upward at 1.40g near Earth's surface; (c) at 1.40g in interstellar space far from any star or planet? Solution
(a) Near the Earth's surface (constant g), but with the neglect of atmospheric friction, a vertical thrust produces an acceleration given by Newton's second law: Fth  mg = ma, or Fth = mg(l + a/g), (positive upward), where mg is the weight of the rocket. If a = 1.40 g, Fth = 0.40mg (the minus sign means Fth is downward). (b) If a = 1.40g, then Fth = 2.40mg.(c) Away from any significant gravitational attraction, Fth = ma = 1.40mg in the direction of a. Problem
68. An F14 jet fighter has a mass of 1.6x 104 kg and an engine thrust of 2.7x105 N. A 747 jumbo jet has a mass of 3.6x105 kg and a total engine thrust of 7.7x105 N. Is it possible for either plane to climb vertically, with no lift from its wings? If so, .what vertical acceleration could it achieve? CHAPTER 5 83 Solution
In a vertical climb, with no aerodynamic force acting (neither lift nor drag), the vertical thrust must exceed the aircraft's weight,Fth  mg > O.This is true for the F14, but not for the 747. The vertical acceleration for the F14 is ay = (Fth  mg)/m = (2.7x105 N + 2 . 2 1.6x104 kg)  9.8 m/s = 7.08 m/s . Problem
69. A spider of mass ms drapes a.silk thread of negligible mass over a stick with its far end a distance h off the ground, as shown in Fig. 540. The stick is lubricated by a drop of dew, so that there is essentially no friction between silk and stick. The spider waits on the ground until a fly of .mass m,(m, > ms) lands on the other end of the silk and sticks to it. The spider immediately begins to climb her end of the silk. (a) With what acceleration must she climb to keep the fly from falling? (b) If she climbs with acceleration as, at what height y above the ground will she encounter the fly? be different from the value found in part (a); here ::f. O. Starting at t = 0, the spider's height is Ys ~ast2, and the fly's height isy, = h+ !a,t2 = h + (atfas)Ys (we used Ys to eliminate t). The fly is encountered when y, = Ys = h + (atfas)Ys, or Ys = h/(l  a, las). (Note that if the fly is stationary, = 0 and Ys = h as expected.) We can find the ratio of the accelerations by eliminating T from the e<Iuationsof motion: (m,  ms)g = msas  m,a" or a, = a, Problem
70. As the Mars Pathfinder spacecraft fell toward its landing in 1997, a rocket fired for approximately 2.3 s to slow it from about 61 m/s to essentially zero speed, after which it dropped freely to the Martian surface. What force did the rocket engine exert on the 276kg spacecraft? (Don't forget gravity, which has essentially its Martian surface value.) Solution
Neglect any variation of mass due to the expenditure of fuel by the engine (see Equation 1010 with the addition of gravity) and ignore the resistance of the thin Martian atmosphere. The equation of motion, with just the vertical thrust Fth and gravity, is Fth mg = ma (positive up). Since 9 = 3.74 m/s2 on Mars, and if we assume a constant acceleration of a = [0  (61 m/s)IJ(2.3 s) :;= 26.5 m/s2,then the thrust was Fth = m(g + a) = (276 kg)(3.74 + 26.5) m/s2 = 8.35 kN. (In Pathfinder's complicated landing sequence, retrorocket ignition occurred at an altitude of 95 m, prior to the airbagcushioned impact d,escribed in Chapter 2, Problem 66.) FIGURE 540 Problem 69. Solution
Assuming vertical forces only, we can write the equations of motion of the spider and fly as T  msg = msas, and T  m,g = m,a" where Tis the tension in the thread, as and af. the accelerations ofthe spider and fly, and positive is up. (a) If "keep the fly from falling" means that the fly is stationary, then a, = O.Thus T = m,g, and as = (T  msg) + ms = g(m,  ms)/ms' (Note: If the spider starts on the floor, as > 0, which is possible only if > ms. If both were initially suspended, the spider could accelerate downwards and keep the fly stationary .even if mf < ms.) (b) It is important to realize that as can Problem
71.. Three identical massless springs of unstretched length .e and spring constant k are connected to thr~ equal masses m as shown in Fig. 541. A force is applied at the top of the upper spring to give the whole system the same acceleration a. Determine the length of each spring. m, 84 CHAPTER 5 (We did not round off because we need to subtract below.) (b) If just the barbell has a2 =f:. 0, then Fa  m2g = m2a2, and Fs  mIg  Fa =0 (we neglect the mass of any moving parts of the lifter's body). Thus a2 = (Fa  m2g)/m2 = [Fs  (ml + m2)g]/m2 = (2400 N  2254 N)/140 kg = 1.04 m/s2. (c) If the scale and the lifter are at rest relative to an orbiting spaceship, F gl = mlal, so Fs  Fa = 0 (we neglect any relative acceleration of the lifter). The barbell's equation of motion is Fa = m2a2  F g2 m2 X (a2  g) = m2a, where a is the barbell's acceleration relative to the spaceship. Then Fs = Fa = m2a = 2 (140 kg)(1.04 m/s ) = 146 N. Solution
Neglect the mass of the springs and let Tn, n = 1,2,3, be the tensions, starting from the bottom spring. With positive components upward, the equations of motion are T1  mg = ma, T2  T1  mg = ma, and Ts  T2  mg = ma. Therefore T1 = m(g + a),T2 = T1 + m(g + a) = 2m(g + a), and Ts = T2 + m(g + a) = 3m(g + a). The stretch in each spring is Xn Tn/k, and its length is e + Xn = e + nm(g + a)/k, where n = 1,2,3 is the spring's number. = = ••••
~ •• Ii  f: ••• ~1
~t
().. rrlf m2. Problem 72 Solution.
FIGURE 541 Problem 71 Solution. Problem Problem
72. A 90kg weightlifter is standing on a scale calibrated in newtons, holding a 140kg barbell. (a) What does the scale read? (b) The lifter then gives the barbell a constant upward acceleration. If the scale reads 2400 N, what is the acceleration of the barbell? (c) Suppose the lifter were in an orbiting spaceship with his feet contacting an identical scale. If he gave the barbell the same acceleration as in (b), what would the scale read? Assume in (b) and ,(c) that the lifter keeps his body rigid, bending only his arms, and neglect the mass of the arms and any downward acceleration of the lifter. 73. Two springs have the same unstretched length but different spring constants kl and k2• (a) If they are connected sidebyside and stretched a distance x, as shown in Fig. 542a, show that the force exerted by the combination is (k1 + k2)x. (b) If they are now connected endtoend and the combination is stretched a distance x (Fig. 542b), show that they exert a force klk2x/(kl + k2). (a) Solution
The forces acting on the lifter (ml) are the force of gravity, F gl, the force of the scale, F s, and the force of the barbell, Fa (this is the reaction to the force applied by the lifter to the barbell). The forces on the barbell (m2) are Fa and Fg2. Newton's second law gives Fs +Fg1  Fa =mIal, and Fa +Fg2 =m2a2. (a) If everything is at rest on the Earth's surface, al = a2 = 0, so, with positive components upward, Fs  mIg  Fa = 0, and Fa  m2g = O. Thus, Fs = (ml + m2)g = (90 + 140) kg(9.8 m/s2) = 2254 N. (b) FIGURE 542 Problem 73. Solution
(a) When connected in "parallel," F = FI + F2, and x = Xl = X2, where F and x are the (magnitude of the) CHAPTER 5 Te := mea, so that me == 726 N/(0.436 mil') := 1670 kg. The answers to (a) and (b) are different because part of the force exerted by the truck must overcome the inertia of the chain. 85 force and stretch of the combination, and subscripts 1 and 2 refer to the individual springs. From Hooke's law, H := klXl, and F2:= k2X2i therefore F := klXl + k2X2 ==(kl + k2)x. (If we write F := kllx, then kll := kl + k2.) (b) When connected in "series," the tension is the same in both springs, F := Fl := F2 (true for "massless" springs), .while the total stretch is the sum of the individual stretches, x := Xl + X2. Again using Hooke's Law, we find X ==Fd kl + F2/ k2 == F(k1l + k2"l), or F ==klk2x/(kl + k2). (If F:= ksx, then kl := kl s 1 4 iC m= 5"4. +
Q. k, , IT .. .4 ~ .436 rtf/s2 + kl) 2' Problem 74 Solution. Problem
74. A tow truck exerts a 75QN force on 54.4kg tow chain. The other end of the chain is attached to a car, and the whole system accelerates at 0.436 m/s2• Find (a) the force exerted on the truck by the chain, (b) the force exerted on the car by the chain, and (c) the mass of the car. Assume there are no horizontal forces acting on the car except for the chain. Why are your answers to (a) an~ (b) different? Problem
75. Although we usually write Newton's law for onedimensional motion in the form F == ma, the most basic version ofthe law reaas' F == d(mv)/dt. The simpler form holds only when the mass is constant. (a) Consider an object whose mass may be changing, and show that the rule for the derivative of a product (see Appendix A) can be used to write Newton's law in the form F := ma + v(dm/dt). (b) A railroad car is being pulled beneath a grain elevator that dumps in grain at the rate of 450 kg/s; What force must be applied to keep the car moving at a constant 2.0 m/s? Solution
(a) If the truck exerts a force of 750 N on the chain, then, from Newton's third law, the chain exerts an equal but opposite force on the truck. (b) If we assume that the chain and the acceleration are horizontal, and if TT := 750 N is the tension at the truck's end and Te that at the car's end, then (the horizontal component of) Newton's second law for the chain is TT  Te == rna, or Te == 750 N  (54.4 kg) x (0.436 m/s2) == 726 N. The direction of the force on the car is parallel to a. (Note: The weight of the chain is not negligible compared to the tension, so the assumption that it is horizontal is justified only for simplicity.) (c) The horizontal component of the equation of motion of the car is, by assumption, Solution
(a) Fnet == d(mv)/dt == m(dv/dt) + (dm/dt)v == ma + v(dm/dt). (b) We can apply Newton's second law in the above form to the horizontal motion of the rail car. At a steady speed, a == dv/dt == 0, so the net horizontal force that must act on the car, whose mass is increasing at a rate of dm/dt := 450 kg/s, is Fnet := v(dm/dt) == (2 m/s)(450 kg/s) := 900 N. If the grain does not exert any horizontal force on the car as it falls in, then all of the net force above must be applied to the car by a locomotive. ' ...
View
Full
Document
This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.
 Spring '08
 WORMER
 Physics, Force, Mass

Click to edit the document details