Veirian

Figure 3 depicts a mass (A) of 20 kg resting on another mass (B) of 100kg. Mass B rolls
without friction over the ground, but the interface between the masses is rough, μ=0.5.
Determine the accelerations of the masses if the force P in the cable is: (i) 60 N, (ii) 40 N

https://imgur.com/a/GmZXZ

I drew the FBD for mass A and got this:
y: -W + N = 0 => N=W=196.2

x: 2P - Fri = ma = 20a
And from both equations I got a=1.096 m/s^2 , which is correct

But I can't seem to find the acceleration for the second mass.
Would the force acting on mass B be the same as the friction force on A (but in opposite direction), since that is the force that will be moving the object? For the FBD, the weight of A would be acting downwards, as well as the weight of B. What are the reaction forces gonna be? I am really confused on how the two objects will interact since I've never done similar dynamics problems before.
Thanks for anyone who helps

benit13

The acceleration of B occurs due to friction between A and B. Therefore, you need to calculate the frictional force, which is:

$\displaystyle F_r \le \mu R$

where R is the normal force at the interface. R can be found because you know the mass of A and R must be equal to the weight of A (from Newton's third law).

Note that this is an inequality, so if you find that $\displaystyle \mu R$ is greater than the pulling force, you should reduce it to be the same as the pulling force.

Since there's no other horizontal forces acting on the object, that should be all you need for Newton's second law to get the acceleration.

Veirian

Thanks, it does make sense.
However for the second part of the question where the pulling force is 2x40 = 80 and it's less than the friction force. That means that the two objects won't move relative to each other (as if they are stuck together) but still move relative to the ground, right? From the answers they both have the same acceleration so that must be true?