# Two blocks and friction

1. Jan 19, 2004

### HW_unlimited

A block M1 of mass 16.5 kg sits on top of a larger block M2 of mass 26.5 kg which sits on a flat surface. The kinetic friction coefficient between the upper and lower block is 0.405. The kinetic
friction coefficient between the lower block and the flat surface is 0.105. A horizontal force F = 94 N pushes against the upper block, causing it to slide. The friction force between the blocks then causes the lower block to slide also. Find the magnitude of the acceleration of the upper block.

94N -----> [ M1 ]
[ M2 ]
-----------------------------

i found the net force acting on m1 but i dont know how to go from there ...im just stuck on this one

2. Jan 19, 2004

### ShawnD

It sounds a lot more complicated than it really is.

First draw a FBD of M1. Aside from gravity and its reaction, there are really only 2 forces: the applied force and the friction between M1 and M2. Find the friction force

$$f = \mu F_N$$

$$f = (0.405)(16.5kg)(9.81\frac{m}{s^2})$$

$$f = 65.56N$$

So here's what the FBD would look like:

94N----->[M1]<--------65.56N

The resultant force would be 28.44N which makes the acceleration 1.724 m/s^2

You were probably trying to factor in the mass of M2 as well. Since the question only asks for M1, you only need to look at M1.