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Practice 2.2A
Push/Pull Problems
Goals




Draw free body diagrams correctly for push/pull problems
Know the difference between kinematic and static friction (and how to use them)
Be able to solve push/pull problems using only variables
How pushing/pulling alter the normal force and the friction acting on the object
1.
A student pulls a wooden box along a rough horizontal floor at constant speed by means of a
force P as shown to the right. Which of the following must be true?
(A) P > f and N < W.
(B) P > f and N = W.
(C) P = f and N > W.
(D) P = f and N = W.
(E) P < f and N = W.
Since P is at an upward angle, the normal force is decreased as P supports some of the weight.
Since a component of P balances the frictional force, P itself must be larger than f.
2.
A 20.0 kg box remains at rest on a horizontal surface while a person pushes directly to the right on the box
with a force of 60 N. The coefficient of kinetic friction between the box and the surface is μk = 0.20. The
coefficient of static friction between the box and the surface is μs = 0.60. What is the magnitude of the force
of friction acting on the box during the push?
(A) 200 N (B) 120 N (C) 60 N (D) 40 N (E) 0 N
The maximum value of static friction in this case is μsFN = 120 N. Since the person is pushing with only 60
N of force, the box remains at rest.
3. A block with initial velocity 4.0 m/s slides 8.0 m across a rough horizontal floor before coming to
rest. The coefficient of friction is:
(A) 0.80 (B) 0.40 (C) 0.20 (D) 0.10 (E) 0.05
4.
A push broom of mass m is pushed across a rough horizontal floor by a force of magnitude T directed at
angle θ as shown above. The coefficient of friction between the broom and the floor is μ. The frictional
force on the broom has magnitude
(A) μ(mg + Tsinθ) (B) μ(mg – Tsinθ) (C) μ(mg + Tcosθ) (D) μ(mg – Tcosθ) (E) μmg
Ff = μFN where FN is found from ΣFy = 0 = (FN – mg – Tsinθ)
Free Response
Problem 1
1. A disturbing weight hangs suspended as shown in the drawing. Find the tensions in the two strings.
F
 0  T1 sin   T2 sin   w  0
F
 0  T1 cos   T2 cos 
y
x
T1  T2
cos 790
cos 20.30
T1  T2
cos 
cos 
119 kg
 0.2034 T2
T1
T1 sin   T2 sin   mg
T2 
79.0
20.3
mg
0.2034sin   sin 
 0.2034 T2  sin   T2 sin   mg
m

119 kg  9.8 2 
s 


0
0.2034sin 20.3  sin 790
20.3


T2

1110 N
mg
T1  0.2034 T2
 0.2034 1110 N  
226 N
79.0
Solution