As seen in the figure, a thin horizontal bar AB of negligible weight and length

Question

As seen in the figure, a thin horizontal bar AB of negligible weight and length L = 14.2 m is pinned to a vertical wall at A and supported at B by a thin wire BC that makes an angle ? = 37° with the horizontal. A weight W =500 N can be moved anywhere along the bar; its position is defined by the distance x from the wall to its center of mass. As a function of x, find the tension in the wire, and the horizontal and vertical components of the force exerted on the bar by the pin at A.
What are the values for x = 10.65 m? Tension?

Horizontal component by pin?

Explanation / Answer

the system is in equlibrium so net torque will be 0

torque about pivot point = 500 * 10.65 - T * sin(37) * 14.2

500 * 10.65 - T * sin(37) * 14.2 = 0

tension in the cable T = 623.115 N

sum of horizontal force = 0

T * cos(37) = horizontal component by pin

horizontal component by pin = 623.115 * cos(37)

horizontal component by pin = 497.6417 N

vertical component by pin = 500 - T * sin(37)

vertical component by pin = 500 - 623.115 * sin(37)

vertical component by pin = 125 N

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