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I'm trying to do this problem using components but I'm having trouble. I see no reason why I can't use this to solve. Maybe I got one of my variable wrong? Can some one help me figure out what I'm doing wrong?

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3 Answers 3

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Let $\sum M_A = 0$, $T = 500*3.464/6 = 288.7$ lbs

$\sum F_x = 0$, $R_{Ax} = -T = -288.7$ lbs

$\sum F_y = 0$, $R_{Ay} = -500$ lbs

Note, when rolling up, the contact with point "B" is lost, thus $R_B$ is zero.

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  • $\begingroup$ isn't the reaction at A variable depending where you are in the lift/pull? If this is the case, wouldn't the maximum reaction be sqrt(2*500^2)? $\endgroup$
    – Forward Ed
    Commented Feb 13, 2022 at 3:22
  • $\begingroup$ @ForwardEd Yes, the moment arm of rope changes, with the shortest at the beginning (max effort), then on final, the arm equals the diameter that requires the least effort. $\endgroup$
    – r13
    Commented Feb 13, 2022 at 4:43
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Whatever you have done is correct. The only thing extra you have to do is set reaction @B to '0' because when the ball lifts up from the ground, there will not be any contact with point B (or any point on the ground below for that matter). So, the expression for tension T becomes, $$T = \frac{Mg.x}{6}$$ x is equal to 3.464 ft in your case.

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What you need to is set the reaction at B as zero. This reduces the number of unknowns to 3 and you will be able to drop one of the 4 equations constraints (I would drop the equilibrium of moments around B).

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