Problem 6.5 A large box of mass M is pulled across a horizontal, frictionless su
ID: 1331967 • Letter: P
Question
Problem 6.5
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T . A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are s and k , respectively.
Part A
Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.
Express your answer in terms of the variables M , m , s , and appropriate constants.
Part B
A horizontal rope pulls a 12 kg wood sled across frictionless snow. A 4.4 kg wood box rides on the sled. What is the largest tension force for which the box doesn't slip? Assume that s =0.50 .
Express your answer to two significant figures and include the appropriate units.
Problem 7.34
The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F causes both blocks to cross a distance of 5.0 m , starting from rest.
(Figure 1) (Note: top block weight = 4.0kg; bottom block = 3.0kg. Moving to right)
Part A
What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
A)
Tmax = maximum static frictional force
Tmax = fs = us*m*g
B)
normal force N = (12+4.4)10 = 164
tension force = us*N = 0.5*164 = 82 N
C)
F = m*a
for top m1 = 4 kg
(0.2*(4+3)*9.8) + (0.6*4*9.8) = 3*au1*m1*g = m1*a
a = u1*g = 0.6*9.8 = 5.88 m/s2
for lower block
f2 = u2*(m1+m2)*g
F - (f2 + f1 ) = m2*a
F - u2*(m1+m2)*g - u1*m1*g = m2*a
F - (0.2*(4+3)*9.8) - (0.6*4*9.8) = 3*5.88
F = 54.88 N
x = v*t + 0.5*a*t^2
5 = 0 + 0.5*5.88*t^2
t = 1.3 s
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