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 v = 25 degree. The wire can withstand a maximum tension of 430 N. If you have bar, it will apple on center. What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the horizontal and vertical components of the force on the bar from the hinge at A? tau = FR si theta siga tau = - W_Block.X_max - W_BAR(L/2) + T_max L sin theta = 0 F_n = sigma F_x = F_h - T_ax cos theta = 0 F_r = sigma F_y = F_v- w_Block + T_max sin theta = 0

Explanation / Answer

W1 = 170 N

w2 = 250 N

Tension in the wire T = 430 N

l1 = L/2

l2 = x

in equilibrium net torque about the point A = 0

T*l*sintheta = w1*l/2 + w2*x

430*sin25 = (170*0.5)+(250*x)

x = 0.39 m

++++++++++++++++++++

along horizantal Fnet = 0

Fh - T*cos25 = 0

Fh = 430*cos25 = 389.7 N

__________________

along vertical

Fnet = 0

Fv + T*sin25 = w1 + w2

Fv+(430*sin25) = 170 + 250

Fv = 238.27 N

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