Question

Goal Solve an equilibrium problem with nonperpendicular torques. Problem A uniform horizontal beam 5.00 m long and weighting 3.00 × 10² N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Fig. 8.12a). If a person weighing 6.00 × 10² N stands 1.50 m from the wall, find the magnitude of the tension T? in the cable and the force R? exerted by the wall on the beam. Figure 8.12 (a) A uniform beam attached to a wall and supported by a cable. (b) A free-body diagram for the beam. (c) The component of the free-body diagram. Strategy The second condition of equilibrium, ??? = 0, with torques computed around the pin, can be solved for the tension T in the cable. The first condition of equilibrium, ?F?? = 0, gives two equations and two unknowns for the two components of the force exerted by the wall, R? and R?. Solution From Figure 8.12, the forces causing torques are the wall force R?, the gravity forces on the beam and the man, w? and w?, and the tension force T?. Apply the condition of rotational equilibrium. ??? = ?? + ?? + ?? + ?? = 0 Compute torques around the pin at O, so ?? = 0 ??? = 0 - w?(L/2) - w?(1.50 m) + TL sin(53.0°) = 0

          Goal Solve an equilibrium problem with nonperpendicular torques.

Problem A uniform horizontal beam 5.00 m long and weighting 3.00 × 10² N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Fig. 8.12a). If a person weighing 6.00 × 10² N stands 1.50 m from the wall, find the magnitude of the tension T? in the cable and the force R? exerted by the wall on the beam.

Figure 8.12 (a) A uniform beam attached to a wall and supported by a cable. (b) A free-body diagram for the beam. (c) The component of the free-body diagram.

Strategy The second condition of equilibrium, ??? = 0, with torques computed around the pin, can be solved for the tension T in the cable. The first condition of equilibrium, ?F?? = 0, gives two equations and two unknowns for the two components of the force exerted by the wall, R? and R?.

Solution

From Figure 8.12, the forces causing torques are the wall force R?, the gravity forces on the beam and the man, w? and w?, and the tension force T?. Apply the condition of rotational equilibrium.

??? = ?? + ?? + ?? + ?? = 0

Compute torques around the pin at O, so ?? = 0

??? = 0 - w?(L/2) - w?(1.50 m) + TL sin(53.0°) = 0

Goal Solve an equilibrium problem with nonperpendicular torques.

Problem A uniform horizontal beam 5.00 m long and weighting 3.00 × 10² N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal (Fig. 8.12a). If a person weighing 6.00 × 10² N stands 1.50 m from the wall, find the magnitude of the tension T? in the cable and the force R? exerted by the wall on the beam.

Figure 8.12 (a) A uniform beam attached to a wall and supported by a cable. (b) A free-body diagram for the beam. (c) The component of the free-body diagram.

Strategy The second condition of equilibrium, ??? = 0, with torques computed around the pin, can be solved for the tension T in the cable. The first condition of equilibrium, ?F?? = 0, gives two equations and two unknowns for the two components of the force exerted by the wall, R? and R?.

Solution

From Figure 8.12, the forces causing torques are the wall force R?, the gravity forces on the beam and the man, w? and w?, and the tension force T?. Apply the condition of rotational equilibrium.

??? = ?? + ?? + ?? + ?? = 0

Compute torques around the pin at O, so ?? = 0

??? = 0 - w?(L/2) - w?(1.50 m) + TL sin(53.0°) = 0

Added by Catherine G.

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