Tracey Pickering, Homework 26 Due: Apr 5 2006, noon Inst: Drummond This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9.8 m/s2 . A cylindrical stone column of diameter 2R = 1.26 m and height H = 3.34 m is transported in standing position by a dolly.
1
from the vertical. From the torque point of view, this apparent weight force applies at the center of mass of the column. The column is stable in the vertical position when the line of this force goes through the column's base
CM a Wapp
side view When the dolly accelerates or decelerates slowly enough, the column stands upright, but when the dolly's acceleration magnitude exceed a critical value ac , the column topples over. (For a > +ac the column topples backward; for a < -ac the column toppes forward.) Calculate the magnitude of the critical acceleration ac of the dolly. Correct answer: 3.69701 m/s2 . Explanation: In the non-inertial frame of the accelerating dolly, the column is subject to the horizontal inertial force Fin = -mg. Together, the gravity and the inertial force combine into the apparent weight force Wapp = m(g - a) in the direction = arctan a g
but when this line misses the base, the column topples over
CM Wapp
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond For the critical acceleration ac , the line goes through the edge of the base, hence the direction of the apparent weight force must deviate from the vertical by the angle c = arctan R hcm R = arctan H/2 2R = arctan H = 20.6688 . Explanation:
2
Let :
= 70 cm , m1 = 24 kg , and m2 = 18 kg .
For static equilibrium, net = 0. Denote x the distance from the left end point of the stick to the point where the cord is attached.
Consequently, the critical acceleration of the dolly is ac = g tan c 2R = g H = 3.69701 m/s2 . 002 (part 1 of 1) 10 points Two objects, of masses 24 kg and 18 kg, are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown. ABCDE F G
m1 g x - m2 g [ - x] = = 0 (m1 - m2 ) x = m2 m2 x= m1 - m 2 (18 kg) (70 cm) = 24 kg - 18 kg = 30 cm . Therefore the point should be point C . 003 (part 1 of 1) 10 points Two weights attached to a uniform beam of mass 23 kg are supported in a horizontal position by a pin and cable as shown in the figure. The acceleration of gravity is 9.8 m/s2 .
10 20 30 40 50 60 24 kg 18 kg
If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord? 1. A 2. E 3. D 4. F 5. C correct 6. G 7. B 2.6 m
23 kg
22
18 kg 4.3 m 24 kg What is the tension in the cable which supports the beam? Correct answer: 1.21343 kN. Explanation:
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond
3
7m
50 kg and 330 kg
64
Let :
m = 23 kg , M1 = 18 kg , M2 = 24 kg , L1 = 2.6 m , L2 = 4.3 m , = 22 .
Basic Concepts: Nr F =0 5m Nf Assume: The center-of-mass of the crane is half-way between the wheels (in the middle of the crane's body). Call the forces from the ground Nr at the rear wheels and Nf at the front wheels. What is the total normal force Ntotal from the ground? Correct answer: 3724 N. Explanation: Let : Solving for the tension T L2 + M1 g L1 + M2 g L2 2 T = L2 sin 4.3 m (23 kg)(9.8 m/s2 ) 2 = (4.3 m) sin (18 kg)(9.8 m/s2 )(2.6 m) + (4.3 m) sin (24 kg)(9.8 m/s2 )(4.3 m) + (4.3 m) sin = 1.21343 kN . mg M = 330 kg , m = 50 kg , = 7 m, d = 5 m , and = 64 .
= 0. Solution: The sum of the torques about the pivot is T L2 sin - mg L2 - M1 gL1 - M2 gL2 = 0 2
Newton's 2nd law for the crane with the crane's boom is Nr + N f - M g - m g = 0 , (1)
since we have static equilibrium. It can immediately be seen from this equation that the sum Ntotal = Nf + Nf = M +m g
004 (part 1 of 3) 10 points A crane of mass 330 kg supports a load of 50 kg. The crane's boom is 7 m long and the angle it makes with the horizontal is 64 . The distance between the front and rear wheels is 5 m. The acceleration of gravity is 9.8 m/s2 .
= (330 kg) + (50 kg) (9.8 m/s2 ) = 3724 N . 005 (part 2 of 3) 10 points Find the force Nr from the ground at the rear
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond wheels. Correct answer: 1316.28 N. Explanation: To find Nr , we need the torque equation for the crane with boom. If we choose the front wheel as our pivot point (i.e., our axis of rotation goes through the two front wheels), the torque of Nf is zero since it has no lever arm. The force Nr has a lever arm d with respect to the front wheels, and tries to rotate the crane clockwise around the pivot point. The line of application of the crane's weight M g goes through the middle of the crane, so the lever arm of the crane's weight is d/2, counterclockwise. The lever arm of the load m is found by drawing out the line of application, which is just the vertical down from the load. So the lever arm of the load m is cos , clockwise. With all this, the net torque is net = Nr d - M g d + m g cos 2
4
force Nr from the ground at the rear wheels change?
7m
50 kg 330 kg
64
Nr
5m
Nf
1. It vanishes: Nr = 0 2. It stays the same: Nr =Nr correct 3. It increases: Nr > Nr 4. It decreases: Nr < Nr 5. The crane tips over 6. Not enough information
if we choose clockwise to be positive. Since we have rotational equilibrium, net = 0 so Nr d - M g d + m g cos = 0 . 2 (2)
Note: Had we chosen clockwise to be negative, all terms would flip sign, but by multiplying both sides with (-1) we obtain the same equation). Therefore, the force Nr is Nr = M g d - m cos 2 d (5 m) = (330 kg) - (50 kg) (7 m) cos 64 2 (9.8 m/s2 ) (5 m) = 1316.28 N . 006 (part 3 of 3) 10 points Neglect the mass of the support wire. If the front axel acts as a hinge between the crane's body and its boom, and a support wire is added between the crane and the of top the crane's boom as in the figure, how does the
Explanation: All we need to do here is to realize that internal torques and internal forces cancel. That is, if we call the tension in the wire T , the force from the back of the crane is -T by the law of action and reaction. Therefore, we will add two forces to our old force Eq. 1, but they cancel since they are of the same magnitude but with different signs. We will add two terms to the torque Eq. 2, but they cancel by the same argument. So the net result is that the system is unchanged, and Nr = N r . 007 (part 1 of 3) 10 points A loaded trailer of weight W is pilled by a truck with force F as in the figure below.
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond
5
L d CM h n F
the center of mass must cancel out. cm = 0. In components, the force equation becomes Fx = ma = Fy + N - W = 0.(2) Wa , (1) g
w
The CM point on the figure indicates the location of the center of mass of the trailer and its load. Suppose the trailer accelerates forward (a > 0) or backwards (a < 0). What is the vertical component Fy of the force F on the trailer? 1. Fy = 2. Fy = 3. Fy = 4. Fy = 5. Fy = 6. Fy = 7. Fy = 8. Fy = 9. Fy = 10. Fy = W h W L 2W L W L W 2L W d W h W d 2W d W 2d aL g ah d- correct g ah d- g ad h- g ah d- g aL h- g ad L- g ah L- g aL h- g ah L- g d-
Also spelling out the torques around the center of mass, the torque equation becomes N d + Fx h - Fy (L - d) = 0. (3) Note that the trailer's weight does not contribute to this torque because it applies at the center of mass and hence has zero lever arm. Evaluating Fx according to eq. (1) and N according to eq. (2) and substituting them into eq. (3), we obtain (W - Fy ) d + Wa h - Fy (L - d) = 0. g
Simplifying the left hand side of this equation gives us W d + and therefore Fy = W L d- ah g . (4) Wa h - Fy L = 0, g
Alternatively, me may consider the trailer to be in full equilibrium in the accelerated frame, thus F+n+w+I = 0 (5)
Explanation: Note that the trailer accelerates but it does not tilt, so from the rotational point of view, it is in equilibrium. Therefore, F + w + n = ma = 0. but at the same time, the net torque around
where I = -ma is the inertial force. Also, we may use the zero-net-torque equilibrium condition O = 0 around any point O, provided we include the torque of the inertial force I applied at the center of mass. For simplicity, let point O be at the axle of the trailer's wheels; this gives zero lever arms
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond to the forces N and Fx and eliminates them from the torque equation. Consequently, O = W d - I h - F y L and therefore Fy = d h W - I. L L (6) L F The ladder is stable only when
6 1 . 2 tan
Finally, substituting I = ma = into eq. (6) we arrive at Fy = d ha W W- W = L Lg L d- ah g . (4) Wa g (7) N f = 0.4
h 45 mg b
Given = 0.4 , and = 45 . The ladder will be 1. unstable. correct 2. stable. 3. at the critical point.
008 (part 2 of 3) 10 points Now let L = 6.04 m, h = 2.36 m, W = 23000 N and g = 9.8 m/s2 . At what value of d would the vertical force Fy on the trailer vanish for the forward acceleration a1 = 2.2 m/s2 ? Correct answer: 0.529796 m. Explanation: In light of eq. (4), zero vertical force Fy requires d = ah = 0.529796 m. g
Explanation: Basic Concepts Evaluate the right-handside of the inequality given
009 (part 3 of 3) 10 points For the same value of d and other parameters of the trailer, what would be the vertial force Fy when the trailer decelerates at a2 = -2.2 m/s2 ? Correct answer: 4034.87 N. Explanation: According to eq. (4) Fy = W L d- ah g = 4034.87 N.
1 = 0.5 . 2 tan 45
Since = 0.04 < 0.5 , the inequality is violated. Thus, the ladder is unstable . 011 (part 1 of 1) 10 points A light string has its ends tied to two walls separated by a distance equal to four-fifth the length L of the string as shown in the figure. A 74 kg mass is suspended from the center of the string, applying a tension in the string. The acceleration of gravity is 9.8 m/s2 .
010 (part 1 of 1) 10 points A ladder is leaning against a smooth wall. There is friction between the ladder and the floor, which may hold the ladder in place.
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond 4L 5 y y= 1 [x - (3.2 m)]2 (10.24 m)
7
L 2
2
1m
74 kg What is the tension in the two strings of L length tied to the wall? 2 Correct answer: 604.333 N. Explanation: The angle can be found from 4L 4 cos = 10 = L 5 2 = 0.8 , so = arccos(0.8) = 36.8699 . 3.2 m x
From the free body analysis of the vertical forces, we have m g = 2 T sin , so mg T = 2 sin (74 kg) (9.8 m/s2 ) = 2 sin(36.8699 ) = 604.333 N .
1m
012 (part 1 of 1) 10 points Pat builds a track for his model car out of wood. The track is 5 cm wide (along the z coordinate), 1 m high (along the y coordinate) and 3.2 m long (along the x coordinate starting from x = 0). The runway is cut such that it forms part of the left-hand side of a parabola, see figure below.
L
5 cm
Locate the horizontal position of the center of gravity of this track. Correct answer: 0.8 m. Explanation: Let : a = 3.2 m , b = a2 = 10.24 m , z = 5 cm , and = surface density . Basic Concepts: rCM = y y= 1 [x - (3.2 m)]2 (10.24 m) r dV = dV r dV M
(1)
(2)
5 cm dx x 3.2 m x
Solution: Let represent the mass-perface area. A vertical strip at position x, with (x - a)2 , has mass width dx and height b (x - a)2 dm = dx . b
Pickering, Tracey Homework 26 Due: Apr 5 2006, noon Inst: Drummond The total mass is
3.2 m
8
M= = = b b
dm =
x=0 3.2 m
(x - a)2 dx b
(x2 - 2 a x + b) dx
0m x3 3.2 m
3
- a x2 + b x
0m
.
The x coordinate of the center of gravity is xcg = 1 x dm M 3.2 m 1 x (x - a)2 dx = M x=0 b 3.2 m x (x - a)2 dx = b M x=0 3.2 m = (x3 - 2 a x2 + b x) dx bM 0 m x4 2 a x 3 b x2 - + bM 4 3 2
3.2 m b x2 3.2 m 0m
=
x4 2 a x 3 - + 3 2 = 4 3 x - a x2 + b x 3 8.73813 m2 = 10.9227 m = 0.8 m .
0m