Accelerating stacked blocks minimum coefficient of static friction to prevent slipping.

In this video, we get to use static and kinetic friction in the same problem. Given a horizontal force on two accelerating stacked blocks on a level surface, we find the minimum coefficient of static friction to prevent slipping (if the acceleration is too fast, the top block slips off the back of the bottom block).

We start by collecting the two masses into a single mass in order to compute the acceleration of the combined mass due to the external force. We include a kinetic friction force at the base of the bottom block. Completing the force diagram, we include the weight of the stacked blocks and the normal force on the bottom block. The normal force is mg because the normal force does whatever is has to do in order to guarantee a zero acceleration perpendicular to the surface. Applying Newton's second law and expressing the kinetic friction as mu_k*n, we solve for the acceleration of the blocks.

Next, we analyze the top block. Assuming the block doesn't slip, it should be accelerating at the value we computed in the first part of the problem. This accelation must be due to the friction force speeding up the block: the friction force points in the direction of acceleration. Next, we apply Newton's second law to the top block, and we sub in the expression for maximum static friction force. We are assuming the static friction is at its maximum value, so this is a coefficient of static friction barely big enough to hold the block in place. We notice that mass cancels out of the equation, and we solve symbolically for the minimum coefficient of static friction mu=a/g. Plugging in our acceleration, we find the minimum static friction coefficient to keep the block locked in place, and we've got the stacked blocks minimum coefficient of static friction.

[fourfour]

If we have a situation where both blocks have same acceleration, then the static friction force between the two will be 0?

Like if we assume a force of 15N is applied on Block 1, and has a mass 5kg, and a block placed over it with a mass 3kg and force of 9 N is applied on, so in this case both will have a = 3m/s^2, so what will happen here?
If:
1. The table is smooth.
2. The table is not smooth.

It would be really helpful if you can make a video on this.

[alibadreldeen]

Hi there! Why are the two masses added when calculating Fs between the blocks? I.e. Fs = μs mg => μs x (1.5kg+2.5kg) x g - thank you in advance :)