A 100 kg nonuniform boom, 6.0 m long, is loosely pinned at the pivot at P. A 700

Question

A 100 kg nonuniform boom, 6.0 m long, is loosely pinned at the pivot at P. A 700 kg block is suspended from the end of the boom at A. The boom forms a 30 degree angle with the horizontal, and is supported by a cable, 4.0 m long, between points D and B. Point B is 4.0 m from P, and point D is 4.0 m above P. The center of mass of the boom is at point C, which is 2.0 m from P. In Figure 10.14, the y-component of the pivot force on the boom at P is A 40 kg uniform ladder, 5.0 m long, is placed against a smooth wall at a height of h = 4.0 m. The base of the ladder rests on a rough horizontal surface whose coefficient of static friction is 0.40. An 80 kg block is suspended from the top rung of the ladder, just at the wall. In Figure 10.12, the force exerted on the wall by the ladder is closest to:

Explanation / Answer

Number 41)

By the sum of torques...

The 700 kg mass torque = (700)(9.8)(cos 30)(6) - Clockwise

The 100 kg mass of the beam torque = (100)(9.8)(cos 30)(2) - Clockwise

The torque from the cable = Tcos30(4) - Counter Clock Wise

Set the clockwise equal to the counter clockwise

Tcos30(4) = (100)(9.8)(cos 30)(2) + (700)(9.8)(cos 30)(6)

T = 10780

Then by the sum of forces in the y direction...

Tsin(30) +Fy = 100(9.8) + (700)(9.8)

10780(sin 30) + Fy = (800)(9.8)

Fy = 2450 N

That rounds to 2500 N using two significant figures

Number 36)

Solve by the sum of torques with pivot point at the ground

The Force from the Wall is 4 m from the ground, and the weight of the ladder is 1.5 m horizontally from the base point and the 80 kg mass is 3 m horizontally from the base.

The Torque from the Wall force = F(4) Counterclockwise

The torque from the Weight of the ladder is (40)(9.8)(1.5) - Clockwise

The torque from the mass is (80)(9.8)(3) - Clockwise

Thus F(4) = (40)(9.8)(1.5) + (80)(9.8)(3)

F = 735 N

Round to 740 for two significant figures

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