HELP A dynamic question with friction

[Kudo Shinichi]

HELP!A dynamic question with friction

Homework Statement

Two blocks lie on a table as shown and are connected by an inextensible cord around a frictionless, massless pulley. A force F=50N is applied to the pulley. The coefficients of friction between the blocks and the table are mu(s)=0.20, and mu(k)=0.10, and the masses are m1=10kg,m2=20kg.
a)determine the accelerations of the blocks and the pulley
b)the applied force is instantaneously increased to F=100N. What are the accelerations now?
c)What are the accelerations if the force drops instantaneously from F=100N to 50N

Homework Equations

F(friction)=mu(k) times F(normal)
Acceleration= net horizontal force/total mass
F(normal)=total mass times gravity
net horizontal force= force applied - frictional force

The Attempt at a Solution

a)F(normal)=(10+20)(9.8)=294N
net horizontal force=50N-29.4N=20.6N
Acceleration=20.6/30=0.69m/s^2
b)net horizontal force=100N-29.4N=70.6N
Acceleration=70.6/30=2.35m/s^2
c)I am not really sure how to do this because if the force drops to 50N then isn't it same as the question a). If it is same as the question a) then the answer will be 0.69m/s^2.

Thank you for helping me.

 

[PhanthomJay]

In part a), you have assumed by using mu(k) that the blocks are moving under the 50N applied force. Are they??

Part b)looks OK.

In part c), the blocks are still in motion.

 

[Kudo Shinichi]

In part a), you have assumed by using mu(k) that the blocks are moving under the 50N applied force. Are they??

Part b)looks OK.

In part c), the blocks are still in motion.

So for part a I will use mu(s) and find the acceleration. The answer for part a should be the answer for part c is it what you are saying?

a) mu(s)Fn=(0.2)(294)=58.8N
ax=58.8N/30=1.96m/s^2
c)ax=20.6/30=0.69m/s^2

 

[PhanthomJay]

So for part a I will use mu(s) and find the acceleration

but if it is not moving, there will not be any acceleration.

The answer for part a should be the answer for part c is it what you are saying?

No.

a) mu(s)Fn=(0.2)(294)=58.8N

this is the max available static friction force

ax=58.8N/30=1.96m/s^2

what happened to the 50N force? That's part of the net force. The static friction force can be less than 58.8N, as required for equilibrium.

c)ax=20.6/30=0.69m/s^2

This part is correct. You just need to relook at part(a).