1. Smooth Pulleys

1..A particle of mass 10kg is at rest on a plane inclined at an angle (arc tan 3/4)Tan-1 3/4 to the horizontal.Contact is rough where µ(Mu)is equal to .25. It is connected by a string passing over a smooth,fixed pulley to a second particle of mass 20kg hanging freely.the system is released from rest.
Find (i)Acceleration
(ii)Tension in the string
(iii)Force on the pulley

2. Originally Posted by Darragh92
1..A particle of mass 10kg is at rest on a plane inclined at an angle (arc tan 3/4)Tan-1 3/4 to the horizontal.Contact is rough where µ(Mu)is equal to .25. It is connected by a string passing over a smooth,fixed pulley to a second particle of mass 20kg hanging freely.the system is released from rest.
Find (i)Acceleration
(ii)Tension in the string
(iii)Force on the pulley
Have you drawn a large, clearly labelled and multi-coloured diagram that shows all the forces acting on the two objects (including the direction of those forces)?

Do you realise that the 10 kg object will move up the slope and the 20 kg object will move downwards?

Have you resolved (in directions parallel and perpendicular to the slope) the forces acting on the 10 kg mass?

Do you realise that the magnitude of the acceleration for each object is the same? The 20 kg object moves downwards with an acceleration of a m/s^2. The 10 kg object has an acceleration of a m/s^2 up the slope and an acceleration of 0 m/s^2 perpendicular to the slope.

Have you applied Newtons Laws of Motion to each object (for the 10 kg object you apply the Laws in the two directions)?

3. Originally Posted by Darragh92
1..A particle of mass 10kg is at rest on a plane inclined at an angle (arc tan 3/4)Tan-1 3/4 to the horizontal.Contact is rough where µ(Mu)is equal to .25. It is connected by a string passing over a smooth,fixed pulley to a second particle of mass 20kg hanging freely.the system is released from rest.
Find (i)Acceleration
(ii)Tension in the string
(iii)Force on the pulley
See the diagram below.

We have to break this into two parts: A Free Body Diagram for the 10 kg object and a FBD for the 20 kg object.

10 kg FBD:
We have a +x direction up the slope and a +y direction out of the plane. The object will be moving up the slope. There is a weight w1 acting straight down, a normal force N acting in the +y direction, a tension T acting in the +x direction, and a friction force f acting in the -x direction. Newton's 2nd in both coordinate directions give
$\sum F_x = T - f - w_1~sin(\theta) = m_1a$

$\sum F_y = N - w_1~cos(\theta) = 0$

and we have that $f = \mu N$.

We can find N from the y equation and insert it into the x equation, but that leaves two unknowns a and T. We need another equation.

So we turn to the second FBD:
20 kg FBD:
We have a +y direction downward. (I always try to choose a positive direction either the direction of motion or direction of the acceleration.) There is a tension T acting in the -y direction and a weight w2 acting in the +y direction. The tension T and acceleration a of this mass must be the same as in the 10 kg FBD, so
$\sum F_y = -T + w_2 = m_2a$

This gives a second equation to solve simultaneously with the first.

Give this a try. If you need more details, just let me know.

-Dan