Finding maximum distance until wire will break...
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Finding maximum distance until wire will break...

[From: ] [author: ] [Date: 11-06-21] [Hit: ]
but it may be cos...A uniform horizontal bar of length L=4m and weight 245 N is pinned to a vertical wall and supported by a thin wire that makes an angles of theta=28 degrees with the horizontal A mass M=325N, can be moved anywhere along the bar.The wire can withstand a maximum tension of 565N.......
Ok here is the question, i'm confused as to which equation to use, and i'm almost sure you will use sin, but it may be cos...

A uniform horizontal bar of length L=4m and weight 245 N is pinned to a vertical wall and supported by a thin wire that makes an angles of theta=28 degrees with the horizontal A mass M=325N, can be moved anywhere along the bar. The wire can withstand a maximum tension of 565N. What is the maximum possible distance from the wall at which mass M can be placed before the wire breaks?

So this is force question...

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Summing torques = 0 about the pivot point:

L*T*sin(theta) = (L/2)*Wb + x*Wm
where Wb = weight of board = 245N, Wm = weight of M = 325N, T = max. permitted wire tension, N, and x = distance of M from pivot.

Solve for x = [L*T*sin(theta) - L*Wb/2]/Wm = 1.76m from wall.
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