Force with friction, pulleys and cords

[saructk]

Homework Statement

Homework Equations

F=ma
f_k=μR

The Attempt at a Solution

What's wrong in my work? Is a(B) not equal to a(A)? Why?

 

[swsw]

There is no guarantee to my answer but here goes:
Friction of ground on B is μ_k*N (where N is normal force of ground on block B which is equals to weight of B, 40Newtons)
This gives friction as 8N.
The force on A would be 100N (weight of A)

This is an atwood machine problem. First think of block A and B as on the same plane, A on left and B on right. Force on A to the right (weight) is 100N and friction on B (to the left) is 8N, giving a net force of 92N to right.

Using F=ma, find a (acceleration of the two blocks, it has to be the same for each block) [This is taking m as total mass of blocks A and B which is 140N/9.81 (taking g as 9.81)].
My acceleration is about 0.64466ms^-2.
Using V=U+at ( V is final speed, U is initial speed, a is acceleration and t is time)
speed after 0.5s is 2.32233ms^-1.

Hope this helps. I cannot guarantee that my solution is correct, but in my opinion it should be.

 

[saructk]

There is no guarantee to my answer but here goes:
Friction of ground on B is μ_k*N (where N is normal force of ground on block B which is equals to weight of B, 40Newtons)
This gives friction as 8N.
The force on A would be 100N (weight of A)

This is an atwood machine problem. First think of block A and B as on the same plane, A on left and B on right. Force on A to the right (weight) is 100N and friction on B (to the left) is 8N, giving a net force of 92N to right.

Using F=ma, find a (acceleration of the two blocks, it has to be the same for each block) [This is taking m as total mass of blocks A and B which is 140N/9.81 (taking g as 9.81)].
My acceleration is about 0.64466ms^-2.
Using V=U+at ( V is final speed, U is initial speed, a is acceleration and t is time)
speed after 0.5s is 2.32233ms^-1.

Hope this helps. I cannot guarantee that my solution is correct, but in my opinion it should be.

The ans is 3.585 ms^-1


Does the cord in another side(not B side) affect the net force?

btw thanks swsw

 

[swsw]

The reason that I felt the cord on the other side will not affect the net force is because as block B slides, the cord will move towards the other side. 2 m per sec is rather fast and I thought that the movable pulley will slide along the cord and hence the cord attached to wall might not affect the net force, but turns out I may be wrong.

I also made a mistake in my previous working, its not 0.64466, it is 6.4466.

 

[swsw]

Perhaps the reason why a(B) is not equals to a(A) in your original working is because there is a movable pulley. Force would be halved but distance doubled to do the same amount of work, so A would move through 0.5X the distance of B?

 

[swsw]

From my friend:

The acceleration of both blocks cannot be the same. Because one of the features of movable pulley is that you must apply force over 2d distance to move the load by d distance. This is due to the fact the force is distributed with the wall, so you will need to move over twice the distance to complete the same amount of work.

Conversely, if block A were to travel by a distance d, block B must travel over distance 2d in the same time interval. B needs to have 2X the speed of A at all times, hence 2X the acceleration of A at all times.

The equations would be
T-friction = (40/9.81)*2a
and
(100/9.81)*a=100-2T
Solve simultaneously and you should get the answer.