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    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).
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    (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.
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    (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?
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    Do you know how to find the sum of two forces acting in different directions?
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    (Original post by ghostwalker)
    Do you know how to find the sum of two forces acting in different directions?
    No.
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    (Original post by kai4321)
    No.
    Have you covered velocities, or displacements, as vectors?
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    You realise it's R right?
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    (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?
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    (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)
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    (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.
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    (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..

    R=\sqrt{(horizontal tension)^2+(vertical tension)^2} i.e. pythagoras

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