In the figure, suppose the length L of the uniform bar is 3.5 m and its weight i

Question

In the figure, suppose the length L of the uniform bar is 3.5 m and its weight is 170 N. Also, let the block's weight W = 250 N and the angle theta = 25 degree. The wire can withstand a maximum tension of.430 N. Free Body Sum Forces If you have bar, it will apply on center., What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontaLand (c) vertical components of the force on the bar from the hinge at A? tau = FR sin theta sigma tau = - W_Block X_max - W_BAR (L/2) + T_max L sin theta = 0 F_h: sigma F_x = F_h - T_max cos theta = 0 F_v = sigma F_g = F_v - W_Block + T_max sin theta = 0

Explanation / Answer

a) considering tension in wire is 430 N.

Balancing moment (r x F) about point A,

( x * 250 ) + (L/2 * 170 ) - ( L * Tsin@) = 0

250x + (3.5/2 * 170) - (3.5 * 430 sin25) = 0

x = 1.35 m

b) balancing forces in horizontal direction,

Fx - Tcos25 = 0

Fx = 430 cos25 = 389.7 N

c) balancing forces in vertical direction,

Fy + 430sin25 - 170 - 250 = 0

Fy = 238.27 N

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