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Thread: Pulley and Tension Madness! (Load weight)

  1. #1
    Jan 2010

    Pulley and Tension Madness! (Load weight)

    1. The problem statement, all variables and given/known data
    See figure.

    2. Relevant equations
    Sum of Fy = 0
    Sum of Fx = 0
    Tension is constant across the cord, regardless of the number of pulleys.

    3. The attempt at a solution
    My first thought after looking at the B A E connection to load P was that the hypotenuse of the 60 degree triangle must be 850 newtons since tension is constant across the rope. Is there any flaws in my reasoning? (I can't seem to get the correct answer so there must be )

    Also doesn't the tension in the cord ABCD depend on load P as well? The heaver P is, the greather the tension in the cord will have to be to keep the cord-pulley system from moving. I'm not sure what forces I'm suppose to use to obtain a tension across ABCD.

    Any help is greatly appreciated.
    Attached Thumbnails Attached Thumbnails Pulley and Tension Madness! (Load weight)-pulleyq2.jpg  
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    The tension in the rope EAB is 850 N. Let T (newtons) denote the tension in the rope ABCD. Then the forces acting on the pulley at B are (i) 850 + T in the direction of A, (ii) T in the direction of C, and (iii) P vertically downwards. If the system is in equilibrium then these forces must balance each other out. So, resolve the forces horizontally to see that $\displaystyle (850+T)\cos60^\circ = T\cos30^\circ$. Then resolve them vertically to get $\displaystyle (850+T)\cos30^\circ + T\cos60^\circ = P$. Solve the first equation to find T, then solve the second equation to find P.
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