when the elevator accelerates downward at a rate of 3.2 m/s? ..P14 A 10 kg block
ID: 1780991 • Letter: W
Question
when the elevator accelerates downward at a rate of 3.2 m/s? ..P14 A 10 kg block is placed on top of a 35 kg block (Figure 5.71). A force of 350 N is applied to the right on the lower block, and the upper block slips on the lower block (accelerating less than the lower block). The coefficient of kinetic friction between the upper block and the lower block is 0.2, and the coefficient of kinetic friction between the lower block and the floor is 0.7. (a) What is the acceleration of the upper block? (b) What is the acceleration of the lower block? (c) How big would the coefficient of static friction between the upper and lower block have to be so that the upper block would not slip on the lower block? m2 m1 Figure 5.71Explanation / Answer
(a)
for the upper block
force acting F2 = uk2*m1*g = m2*a2
0.2*10*9.8 = 10*a2
acceleration a2 = 1.96 m/s^2
==================
(b)
for the lower block
F - fk2 - fk1 = m1*a
350 - (0.2*10*9.8) - (0.7*35*9.8) = 35*a1
aceeleration a1 = 2.58 m/s^2
==================
(c)
for the upper not slip
fs1 = us*m2*g = m2*a .........(1)
for the lower block
F - uk1*m1*g = m1*a.........(2)
from 1 & 2
us*m2*g/(F - uk1*m1*g) = m2/m1
us*10*9.8/(350 - (0.7*35*9.8)) = 10/35
coefficient of static friction us = 0.32 <<<<<-------------ANSWER
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