Set the moment about the hinge equal to zero, with a cable tension equal to the maximum value, 520 N. The torque due to the weight will be 468x Newton-meters.
Solve for x. Once you know x, apply a force balance on the entire beam to get the vertical and horizontal forces at the hinge
In Fig. 12-46, suppose the length L of the uniform bar is 2.95 m and its weight is 164 N. Also, let the block's weight W = 468 N and the angle θ = 33.8°. The wire can withstand a maximum tension of 520 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?its like a right angle triangle, with beam at the base, wall is along y axis ,block is between wall and centre of mass of beam and cable is the hypotenuse
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