A uniform 600 kg beam, 6 m long, is freely pivoted at P. The beam is supported i

Question

A uniform 600 kg beam, 6 m long, is freely pivoted at P. The beam is supported in a horizontal position by a light strut, 5 m long, which is freely pivoted at Q and is loosely pinned to the beam at R. A load of mass is suspended from the end of the beam at S. A maximum compression of 18,000 N in the strut is permitted, due to safety. In Figure 10.11, under maximum load, the x-component of the force exerted on the beam by the pivot at P is closest to: A uniform 600 kg beam, 6 m long, is freely pivoted at P. The beam is supported in a horizontal position by a light strut, 5 m long, which is freely pivoted at Q and is loosely pinned to the beam at R. A load of mass is suspended from the end of the beam at S. A maximum compression of 25,000 N in the strut is permitted, due to safety. In Figure 10.11, under maximum load the y-component of the force exerted on the beam by the pivot at P is

Explanation / Answer

Number 33)

The angle in the strut is found from the tangent function

tan(angle) = 3/4

Angle = 36.9o

The x component will be 18000(sin 36.9)

Fx = 10800 N

Number 34)

By the sum of torques...

M(9.8)(6) + (600)(9.8)(3) = 25000(3)(cos 36.9)

M = 720 kg

The forces in the y direction will be Fy + (720)(9.8) + (600)(9.8) = 25000(cos 36.9)

Fy = 7060 N

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