# Minimum Force such that box m does not slide down

1. Feb 23, 2015

### p671

1. The problem statement, all variables and given/known data

2. Relevant equations

F = ma; f = µN
3. The attempt at a solution
Let F' be the action-reaction force between box m and box M
f be the friction on small box m
For small box m:
F' = ma
f - mg = 0
µF' - mg = 0 since f = µF'
µma - mg = 0 since F' = ma
a = g / µ = 9.81 m/s2 / 0.2 = 49.05 m/s2
which is not equal to the given acceleration of 2.0 m/s2

2. Feb 23, 2015

### BvU

Hello p, welcome to PF

This is a most peculiar exercise. What you calculate is correct: an acceleration of no less than 5g is required to make m stick to M with so much friction that it doesn't slide down. I find the exercise wording/picture combination very unsatisfactory: It isn't clear at all where the given 2 m/s2 comes from, and -- as you so justly put it -- 2 m/s2 is not 5g at all.

If an answer is absolutely required to get to the next level or something like that, then perhaps it's allowed to make some assumptions: for example that the 2 m/s2 is already there (some jet engine inside M that we can't see) and that the horizontal component of F is supposed to provide the remaining a' = 47.05 m/s2 to the ensemble of (m + M) . You apply Fx = (m+M) a' and some trigonometry to go from Fx to |F|.

Good luck, and maybe you can let us know what came out ?

3. Feb 23, 2015

### p671

Alright. Thanks! :)