In the figure, suppose the length L of the uniform bar is 2.6 m and its weight i

Question

In the figure, suppose the length L of the uniform bar is 2.6 m and its weight is 230 N. Also, let the block's weight W = 280 N and the angle ? = 37?. The wire can withstand a maximum tension of 400 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical
components of the force on the bar from the hinge at A?

In the figure, suppose the length L of the uniform bar is 2.6 m and its weight is 230 N. Also, let the block's weight W = 280 N and the angle ? = 37?. The wire can withstand a maximum tension of 400 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?

Explanation / Answer

Summing torques = 0 about the pivot point:

L*T*sin(theta) = (L/2)*Wb + x*Wm
where Wb = weight of board = 230, Wm = weight of M = 280N, T = max. permitted wire tension, N, and x = distance of M from pivot.

Solve for x = [L*T*sin(theta) - L*Wb/2]/Wm =1.167 m from wall.

b) Balance the forces in the horizontal direction:

F (horizontal bar) = F (horizontal tension)
F (horizontal bar) = T cos(theta)
F (horizontal bar) = (400 cos(37)
F (horizontal bar) = 319.45 N

c) Balance the forces in the vertical direction:

F (vertical bar) + F (vertical tension) + F (bar weight) + F (mass weight) = 0
F (vertical bar) + T sin(theta) - Wb - Wm = 0
F(vertical bar) + (400 N) sin(37) - 280 N - 230 N = 0
F (vertical bar) = 269.3 N

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