(tp8327) pampalone HW#6 swaminathan (10500) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 2 s. A passenger in the elevator is holding a 5.5 kg bundle at the end of a vertical cord. The acceleration of gravity is 9.8 m/s2 . What is the tension in the cord as the elevator accelerates? Correct answer: 56.65 N. Explanation: T g mg aelevator Correct answer: 43.4721 . Explanation: Observe the free-body diagram below.
2
1
Consider the 672 N weight held by two cables shown below. The left-hand cable had tension 490 N and makes an angle of 2 with the ceiling. The right-hand cable had tension 510 N and makes an angle of 1 with the ceiling.
49
2
0 N
1
0 51
N
672 N a) What is the angle 1 which the righthand cable makes with respect to the ceiling?
Let h be the distance traveled and a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression and solve for acceleration of the elevator: 1 1 h = v0 t + a t2 = a t2 2 2 2h a= 2 . t
F
2
F1
1
Wg
=
The equation describing the forces acting on the bundle is Fnet = m a = T - m g T = m (g + a) 2h =m g+ 2 t = (5.5 kg) 9.8 m/s2 + = 56.65 N . 2 (1 m) (2 s)2
Note: The sum of the x- and y-components of F1 , F2 , and Wg are equal to zero.
Given :
Wg = 672 N , F1 = 510 N , F2 = 490 N .
and
Basic Concepts: Fx = 0
002 (part 1 of 2) 10.0 points Hint: sin2 + cos2 = 1 .
x x F1 = F2 F1 cos 1 = F2 cos 2
(1)
pampalone (tp8327) HW#6 swaminathan (10500)
2 2 F1 cos2 1 = F2 cos2 2
2
(2)
and
Fy = 0
y y y F1 + F2 + F3 = 0 F1 sin 1 + F2 sin 2 - F3 = 0 F1 sin 1 = -F2 sin 2 + F3 2 2 F1 sin2 1 = F2 sin2 2 2 -2 F2 F3 sin 2 + F3 , since (3) F3 sin 3 = F3 sin 270 = -F3 , and F3 cos 3 = F3 cos 270 = 0 .
it. The F force acts from above on the left at an angle of with the horizontal. The 6.7 N force acts from above on the right at an angle of 42 with the horizontal. The 6.3 N force acts from below on the right at an angle of 44 with the horizontal. The 5.7 N force acts from below on the left at an angle of 45 with the horizontal.
42
6.
Solution: Since sin2 + cos2 = 1 and adding Eqs. 2 and 3, we have
2 2 2 F2 = F1 - 2 F1 F3 sin 1 + F3 2 2 F 2 + F3 - F2 sin 1 = 1 2 F1 F3 2 2 2 F3 + F1 - F2 1 = arcsin 2 F1 F3 (672 N)2 + (510 N)2 = arcsin 2 (510 N) (672 N) (490 N)2 - 2 (510 N) (672 N) = 43.4721 .
knot
44
6.
3
N
F
7
N
Note: F, are not to scale. What is the magnitude of the force F ? Correct answer: 6.74018 N. Explanation: Note: Standard angular measurements are from the positive x-axis in a counter-clockwise direction. Given : F1 1 F2 2 F3 3 F4 4 = 6.7 N , = 180 - 42 = 138 , = 6.3 N , = 180 + 44 = 224 , = 5.7 N , = 360 - 45 = 315 , = F , and = .
5. 7 N
45
003 (part 2 of 2) 10.0 points b) What is the angle 2 which the left-hand cable makes with respect to the ceiling? Correct answer: 40.9952. Explanation: Using Eq. 1, we have cos 2 = F1 cos 1 F2 F1 cos 1 2 = arccos F2 510 N = arccos cos 43.4721 490 N = 40.9952 .
Observe the free-body diagram below where the vectors are decomposed into components along the x- and y-axes.
42
6.
7
N
4 6. 7
N
35. 6
004 (part 1 of 2) 10.0 points The knot at the junction is in equilibrium under the influence of four forces acting on
44
3 6.
N
5. 7 N
45
pampalone (tp8327) HW#6 swaminathan (10500) Note: The vectors along the xand y-coordinates add to zero. Basic Concepts: Fx = 0
x x x x F1 + F2 + F3 + F4 = 0 F1 cos 1 + F2 cos 2 x +F3 cos 3 + F4 = 0
3
006 10.0 points A 4.1 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 32 with the vertical The acceleration of gravity is 9.8 m/s2 . (1)
F =0
y y y y F1 + F2 + F3 + F4 = 0 F1 sin 1 + F2 sin 2 y +F3 sin 3 + F4 = 0
y
32
a
4.1 kg
(2)
Solution: Using Eqs. 1 and 2, we have
x F4 = -F1 cos 1 - F2 cos 2 - F3 cos 3 (1) = -(6.7 N) cos 138 - (6.3 N) cos 224 - (5.7 N) cos 315 = 5.4804 N , and y F4 = -F1 sin 1 - F2 sin 2 - F3 sin 3 (2) = -(6.7 N) sin 138 - (6.3 N) sin 224 - (5.7 N) sin 315 = 3.92368 N , so
Find the acceleration of the car. (Hint: aobject = acar ) Correct answer: 6.12372 m/s2 . Explanation: Given : m = 4.1 kg , = 32 , and g = 9.8 m/s2 . T cos T
F4 = =
y x (F4 )2 + (F4 )2
T sin mg Vertically Fy = T cos - m g = 0 T cos = m g . Horizontally, Fx = T sin = m a . Dividing Eqs 1 and 2, we have a T sin = T cos g a tan = g a = g tan = 9.8 m/s2 tan 32 = 6.12372 m/s2 . (2) (1)
(5.4804 N)2 + (3.92368 N)2
= 6.74018 N . 005 (part 2 of 2) 10.0 points What is the angle of the force F as shown in the figure above? Correct answer: 35.6007. Explanation: 4 = arctan
y F4 x F4 3.92368 N = arctan 5.4804 N = 138 , from the positive x-axis = 35.6007 .
pampalone (tp8327) HW#6 swaminathan (10500) 007 (part 1 of 3) 10.0 points The suspended 2.7 kg mass on the right is moving up, the 1.4 kg mass slides down the ramp, and the suspended 7.4 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0.17 . The acceleration of gravity is 9.8 m/s2 . The pulleys are massless and frictionless. a T3 T1
4 a
a
T1 N
T3
m3 g m2 g
N
m1 g
For the mass m1 , T1 acts up and the weight m1 g acts down, with the acceleration a directed upward Fnet1 = m1 a = T1 - m1 g (1)
1. 4
kg 0. 1 7
33
=
7.4 kg
2.7 kg
For the mass on the table, the parallel component of its weight is m g sin and the perpendicular component of its weight is m g cos . (N = m g cos from equilibrium). The acceleration a is directed down the table, T3 and the parallel weight component m2 g sin act down the table, and T1 and the frictional force N = m2 g cos act up the table Fnet2 = m2 a (2) = T3 + m2 g sin - T1 - m2 g cos .
What is the acceleration of the three block system? answer: Correct 4.4849 m/s2 . Explanation:
For the mass m3 , T3 acts up and the weight m3 g acts down, with the acceleration a directed downward Fnet3 = m3 a = m3 g - T3 . Adding Eqs. (1), (2), & (3) yields (3)
Let : m1 m2 m3 Basic Concept:
= 2.7 kg , = 1.4 kg , = 7.4 kg , = 33 .
(m1 + m2 + m3 ) a = m3 g + m2 g sin - m2 g cos - m1 g . and Solving for a, we have a= [m2 sin - m2 cos + (m3 - m1 )] g m1 + m2 + m3 (1.4 kg) (9.8 m/s2 ) sin 33 = 2.7 kg + 1.4 kg + 7.4 kg (0.17) (1.4 kg) (9.8 m/s2 ) cos 33 - 2.7 kg + 1.4 kg + 7.4 kg (7.4 kg - 2.7 kg) (9.8 m/s2 ) + 2.7 kg + 1.4 kg + 7.4 kg = 4.4849 m/s2 . 008 (part 2 of 3) 10.0 points
Fnet = m a = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T1 be the tension in the right string and T3 the tension in the left string. Consider the free body diagrams for each mass
pampalone (tp8327) HW#6 swaminathan (10500) What is the tension in the cord connected to the 2.7 kg block? Correct answer: 38.5692 N. Explanation: Using Eq. 1, we have T1 = m1 g + m1 a = (2.7 kg) (9.8 m/s2 + 4.4849 m/s2 ) = 38.5692 N . 009 (part 3 of 3) 10.0 points What is the tension in the cord connected to the 7.4 kg block? Correct answer: 39.3318 N. Explanation: Using Eq. 3, we have T3 = m3 g - m3 a = (7.4 kg) (9.8 m/s2 - 4.4849 m/s2 ) = 39.3318 N . 010 (part 1 of 2) 10.0 points Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton's Laws, and (3) solve the equations of motion. A block starts from rest at a height of 5.3 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s2 .
d
5
m = 6.2 kg , = 0.23 , = 30 , and vf = final speed .
N Ff
30
mg
h
30
The normal force to the inclined plane is N = m g cos . The sum of the forces parallel to the inclined plane is Fnet = m a = m g sin - m g cos a = g sin - g cos Since
2 2 vf = v0 + 2 a x = 2 a d
(1)
6. 2 =
kg 0. 2 3
30
along the plane and the distance moved along the plane is h (2) d= sin therefore vf = = = 2ah 2 g h (sin - cos ) = sin sin 2 g h (1 - cot )
What is the speed of the block at the bottom of the ramp? Correct answer: 7.90551 m/s. Explanation: Let : h = 5.3 m ,
(3)
2 (9.8 m/s2 ) (5.3 m) [1 - (0.23) cot 30 ]
= 7.90551 m/s . 011 (part 2 of 2) 10.0 points
pampalone (tp8327) HW#6 swaminathan (10500) If the block continues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground until coming to rest? Correct answer: 13.8636 m. Explanation: N 270 N Wstaging Explanation: 380 N 140 N
6
Ff
From the equilibrium rule, Fi = F + Fr - Wstaging - Wman
mg Here the normal force to the x axis is N = m g. The sum of the forces parallel to the x axis is Fnet = m a = m g Since
2 2 vf = v0 + 2 a x 2 v0 = -2 a x = -2 g x
Wstaging
= 0 , so = +F + Fr - Wman = (380 N) + (140 N) - (270 N) = 250 N .
then
2 -v0 2g (7.90551 m/s)2 |x| = 2 (9.8 m/s2 ) (0.23) = 13.8636 m .
013 10.0 points A 37.4 kg lamp is hanging from wires as shown in figure below. The acceleration of gravity is 9.81 m/s2 . d
x=
T1 2 T3
T2 2 slip-ring
012 10.0 points A staging that weighs Wstaging supports a painter weighing 270 N. The reading on the left scale is 380 N and the reading on the right scale is 140 N.
37.4 kg 380 N 140 N Find the tension T3 in the vertical wire. Correct answer: 366.894 N. Explanation: What is the weight of the staging? Correct answer: 250 N. m = 37.4 kg and
2
g = 9.81 m/s .
pampalone (tp8327) HW#6 swaminathan (10500) Consider the forces acting on the lamp: y T3 lamp x Fnet = m2 g = (5.66101 kg) (9.8 m/s2 ) = 55.4779 N . From Newton's second law, Fnet = m2 g = (m1 + m2 ) a . Solving for a, a= m2 g m1 + m2 5.66101 kg (9.8 m/s2 ) = 2.50393 kg + 5.66101 kg = 6.79465 m/s2 .
7
The net force on the system is simply the weight of m2 .
W The vertical string supports the full weight W, so T3 - W = 0 T3 = W = mg = (37.4 kg) (9.81 m/s2 ) = 366.894 N . 014 10.0 points A block of mass 2.50393 kg lies on a frictionless table, pulled by another mass 5.66101 kg under the influence of Earth's gravity. The acceleration of gravity is 9.8 m/s2 . 2.50393 kg
Analyzing the horizontal forces on block m1 , we have Fx : T = m1 a = (2.50393 kg) (6.79465 m/s2 ) = 17.0133 N .
5.66101 kg What is the magnitude of the tension T of the rope between the two masses? Correct answer: 17.0133 N. Explanation: Given : m1 = 2.50393 kg and m2 = 5.66101 kg . T T m1 g m2 m2 g a
015 (part 1 of 3) 10.0 points A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are s = 0.6 and k = 0.51, respectively. The acceleration of gravity is 9.8 m/s2 .
13
kg
27
a m1 N
What is the frictional force acting on the 13 kg mass? Correct answer: 57.8384 N. Explanation:
pampalone (tp8327) HW#6 swaminathan (10500)
8
N Ff
is the kinetic friction Fk = k N = k M g cos . Newton's equation for the block then becomes 27 M a = M g sin - Ff = M g sin - k M g cos and a = g [sin - k cos ] = (9.8 m/s2 ) [sin 35 - (0.51) cos 35 ] = 1.52693 m/s2 .
mg
The forces acting on the block are shown in the figure. Since the block is at rest, the magnitude of the friction force should be equal to the component of the weight on the plane of the incline Ff = M g sin = (13 kg) (9.8 m/s2 ) sin 27 = 57.8384 N . 016 (part 2 of 3) 10.0 points What is the largest angle which the incline can have so that the mass does not slide down the incline? Correct answer: 30.9637 . Explanation: The largest possible value the static friction force can have is Ff,max = s N , where the normal force is N = M g cos . Thus, since Ff = M g sin , M g sin m = s M g cos m tan m = s m = tan-1 (s ) = tan-1 (0.6) = 30.9637 . 017 (part 3 of 3) 10.0 points What is the acceleration of the block down the incline if the angle of the incline is 35 ? Correct answer: 1.52693 m/s2 . Explanation: When exceeds the value found in part 2, the block starts moving and the friction force