A person with mass 55.0 kg stands d = 2.50 m away from the wall on a x = 5.80 m

Question

A person with mass 55.0 kg stands d = 2.50 m away from the wall on a x = 5.80 m beam, as shown in the figure below. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in the wire. (Indicate the direction with the sign of your answer.)

T = N Rx = N Ry = N A person with mass 55.0 kg stands d = 2.50 m away from the wall on a x = 5.80 m beam, as shown in the figure below. The mass of the beam is 40.0 kg. Find the hinge force components and the tension in the wire. (Indicate the direction with the sign of your answer.) T = N Rx = N Ry = N

Explanation / Answer

The components of the tension are Tcos30, to the right, and Tsin30, upward.

Ry is upward. Rx is to the left.

Sum horizontal forces.

- Rx + Tcos30 = 0

Sum vertical forces.

Ry - mg - Mg + Tsin30 = 0, where m is the mass of the man and M is the mass of the beam.

Sum moments about the right end of the beam.

Clockwise moments are positive, counter-clockwise moments are negative.

- Mg(2.90) - mg(3.30) + Ry(5.80) = 0

Solve for Ry

- (40.0)(9.8)(2.90) - (55.0)(9.8)(3.30) + Ry(5.80) = 0

- 1136.8 - 1778.7 + Ry(5.80) = 0

2915.5 = Ry(5.80)

Ry = 502.7 N (upward)

Go back and solve for T.

Ry - mg - Mg + Tsin30 = 0

502.7 - (55.0)(9.8) - (40.0)(9.8) + Tsin30 = 0

502.7 - 539 - 392 + Tsin30 = 0

- 728.3 + Tsin30 = 0

T = 728.3/sin30 = 1456.6 N (up to the right)

Go back and solve for Rx.

- Rx + Tcos30 = 0

- Rx + 1456.6cos30 = 0

Rx = 1261.4 N (as drawn to the left)

Input negative 1261.4 N for Rx

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