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# Mechanics 1 (Dynamics + Tension) watch

1. Two particles A and B of mass 5kg and 3kg respectively are connected by a light inextensible string. Particle A lies on a rough horizontal table and the string passes over a smooth pulley which is fixed at the edge of the table. Particle B hangs freely. The coefficient of friction between A and the table is 0.5. The system is released from rest.

a) Acceleration, figured out to be 0.613 ms^-2 (3 s.f).

b) Tension in the string, figured out to be 27.6N.

c) The magnitude of the force exerted on the pulley by the string - can't work out.

I've drawn a diagram which must be right considering part a) and b) are correct but I don't know how to work out part c).
2. (Original post by kai4321)
I've drawn a diagram which must be right considering part a) and b) are correct but I don't know how to work out part c).
The Force on the pulley will be the vector sum of the two tensions, one acting horizontally, and one acting vertically.
3. (Original post by ghostwalker)
The Force on the pulley will be the vector sum of the two tensions, one acting horizontally, and one acting vertically.
Could you explain that a bit more please?
4. Do you know how to find the sum of two forces acting in different directions?
5. (Original post by ghostwalker)
Do you know how to find the sum of two forces acting in different directions?
No.
6. (Original post by kai4321)
No.
Have you covered velocities, or displacements, as vectors?
7. You realise it's R right?
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8. (Original post by ghostwalker)
Have you covered velocities, or displacements, as vectors?
I've done velocity and displacement if you mean doing that in Kinematics of a particle.

(Original post by sheligalo)
You realise it's R right?
Never knew but how do I work out R then?
9. (Original post by kai4321)
Never knew but how do I work out R then?
How would you normally work out the resultant? If the method you use is by drawing a right angled triangle then im pretty sure that technique can be used here.. i.e. R = tan(downwards tension)/(horizontal tension)
10. (Original post by sheligalo)
How would you normally work out the resultant? If the method you use is by drawing a right angled triangle then im pretty sure that technique can be used here.. i.e. R = tan(downwards tension)/(horizontal tension)
No, sorry - I don't get it. I mean I get what you're asking but I need to get 39.0 and what I'm working out gives me TINY answers.

Someone I know did it by doing root 2 x 27.6^2 and you get 39.0 from that but I don't get why he did that. Btw, 27.6 is the tension.
11. (Original post by kai4321)
No, sorry - I don't get it. I mean I get what you're asking but I need to get 39.0 and what I'm working out gives me TINY answers.

Someone I know did it by doing root 2 x 27.6^2 and you get 39.0 from that but I don't get why he did that. Btw, 27.6 is the tension.
Shoot, i told you the above point wrong..

i.e. pythagoras

Assuming these 2 tensions are the same this simplifies to your friends answer. Sorry for the mistake!

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