Question

A block of mass
M1 = 2.7 kg
rests on top of a second block of mass
M2 = 4.7 kg,
and the second block sits on a surface that is so slippery that the friction can be assumed to be zero (see the figure below).

(a) If the coefficient of static friction between the blocks is
¦ÌS = 0.21,
how much force can be applied to the top block without the blocks slipping apart?

Incorrect: Your answer is incorrect.
N

(b) How much force can be applied to the bottom block for the same result?

Answer 1

(a)
Top block:
F-friction=M1a
F=2.7a+(0.21*2.7*9.8)

From bottom block:
friction=M2a
a=(0.21*2.7*9.8)/4.7

F=(0.21*2.7*9.8*2.7)/4.7+(0.21*2.7*9.8)=8.749N


(b) Bottom block:
F-friction=M2a
F=4.7a+(0.21*2.7*9.8)

Top block: 0.21*2.7*9.8=2.7a
a=0.21*9.8

F=(0.21*4.7*9.8)+(0.21*2.7*9.8)=15.23N