Posted by **April** on Friday, March 2, 2012 at 11:35pm.

A block of mass M1=2.9 kg rests on top of a second block of mass2=5.5 kg, and the second block sits on a surface that is so slippery that the friction can be assumed to be zero.

Question:

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

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

- Physics -
**drwls**, Saturday, March 3, 2012 at 8:21am
(a) 0.21*M1*g = 5.968 N

(b) The maximum friction force that M2 can apply to M1 without M1 slipping is also 5.968 N. That will cause M1 AND M2 to accelerate at rate

a = 5.968 N/M1 = 2.058 m/s^2

The force that would have to be applied to M2 to make this happen is

(M1 + M2)*a = 8.4 * 2.058 = 17.29 N

- Physics -
**Hayden**, Tuesday, January 28, 2014 at 8:23pm
The answer to part a above is incorrect. You must remember that the force of friction resisting the applied force on M1 also applies an equal and opposite force on M2. The applied force must overcome /both/ of these.

The sum of forces on the X axis for M2 is just this equal and opposite force (Force of friction). Setting this to ma will allow you to solve for an acceleration value.

You can use this acceleration value in solving for the applied force on M1. (T - Ffriction = ma).

Hope this helps.

- Physics -
**Kaity**, Friday, September 12, 2014 at 3:08am
(a) Top block:

F-friction=M1a

F=2.9a+(0.21*2.9*9.8)

Bottom block:

friction=M2a

a=(0.21*2.9*9.8)/5.5

F=(0.21*2.9*9.8*2.9)/5.5+(0.21*2.9*9.8)=9.12N

(b) Bottom block:

F-friction=M2a

F=5.5a+(0.21*2.9*9.8)

Top block: 0.21*2.9*9.8=2.9a

a=0.21*9.8

F=(0.21*5.5*9.8)+(0.21*2.9*9.8)=17.29N

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