20. [Ipt A uniform horizontal bar of length L 3.00 m and weight 207 N is pinned
ID: 1788279 • Letter: 2
Question
20. [Ipt A uniform horizontal bar of length L 3.00 m and weight 207 N is pinned to a vertical wall and supported by a thin wire that makes an angle of e 25 with the horizontal. A mass M, with a weight of 643 N, can be moved anywhere along the bar. The wire can withstand a maximum tension of 754 N. What is the maximum possible distance from the wall at which mass M can be placed before the wire breaks? Submit All Answers Answer: 21. [lpt] With s placed at this maximum distance, what is the horizontal component of the force exerted on the bar by the pin at 4? Answer: Submit All AnswersExplanation / Answer
20) Resolve the torques about point A. We have the vertical sin() component of the tension rotating counter-clockwise, the weight of the bar rotating clockwise, and the weight of the box rotating clockwise. Recall that gravity acts at the center of mass, which is located at L / 2. Thus, balancing these gives:
T (tension) = T (bar) + T (box)
L T sin() = (L / 2) Wb + x Wm
x = [L T sin() - (L / 2) Wb] / [Wm]
Where Wb is the weight of the bar, Wm is the weight of the mass, and T is the maximum value of the tension possible. Subbing in the numbers:
x_{max} = [(3 m)(754 N)(sin(25)) - (3 / 2)(207 N)] / [643 N]
x_{max} 1 m
21) Balance the forces in the horizontal direction:
F (horizontal bar) = F (horizontal tension)
F (horizontal bar) = T cos()
F (horizontal bar) = (754 N) cos(25)
F (horizontal bar) 683.36 N
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