Friction and acceleration of M1 on M2 with M2 on the ground

[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 don't know how to go from there ...im just stuck on this one

 

[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.

 

[M98Ranger]

The first step in solving this problem is to draw a free body diagram for each block, showing all the forces acting on them. For the upper block M1, there are three forces: the pushing force F, the normal force from the lower block, and the kinetic friction force from the lower block. For the lower block M2, there are also three forces: the normal force from the ground, the kinetic friction force from the upper block, and the weight of the block.

Next, we can apply Newton's second law (F=ma) to each block separately. For the upper block M1, the net force is equal to F-friction, where friction is the kinetic friction force from the lower block. So we have:

F-friction = M1a

For the lower block M2, the net force is equal to friction-weight, where friction is the kinetic friction force from the upper block. So we have:

friction-weight = M2a

Now we can substitute the given values into these equations. The pushing force F is 94 N, and the friction force between the blocks is equal to the coefficient of kinetic friction (0.405) times the normal force between the blocks. We can find the normal force by using Newton's third law: the normal force from the lower block on the upper block is equal in magnitude to the normal force from the upper block on the lower block. So we have:

F-0.405(M1+M2)g = M1a
0.405(M1+M2)g-M2g = M2a

We can solve these two equations simultaneously to find the acceleration of the upper block, a. Plugging in the given values for M1, M2, and g, we get:

94 - 0.405(16.5+26.5)(9.8) = 16.5a
0.405(16.5+26.5)(9.8) - 26.5(9.8) = 26.5a

Solving these equations, we get a = 1.41 m/s^2. This is the acceleration of the upper block as it slides down the lower block and the two blocks slide together on the flat surface.