SolutionET; = TR + TB + TM + Tr = 0From Figure 8.12, the forces causing torques are thewall force R, the gravity forces on the beam and theman, WB and wM, and the tension force T. Apply thecondition of rotational equilibrium.Compute torques around the pin at 0, so TR = 0Et; = 0 - WB(L/2) - Wm(1.50 m) + TL sin(53.0°) = 0%3D%3D(zero moment arm). The torque due to the beam'sweight acts at the beam's center of gravity.Substitute L = 5.00 m and the weights, solving for T.-(3.00 x 102 N)(2.5 m) - (6.00 x 102 N)(1.50 m) +(Tsin53.0°)(5.00 m)T == 0NNow apply the first condition of equilibrium to thebeam.EF = Rx + Tcos53.0° = 0 (2)%3Dх2Fy = Ry - WB - WM + Tsin53.0° = 0 (2)Substituting the value of T found in the previous stepand the weights, obtain the components of R.Rx=RyNRemarks Even if we selected some other axis for the torque equation, the solution would be the same. Forexample, if the axis were to pass through the center of gravity of the beam, the torque equation would involveboth T and Ry. Together with equations (1) and (2), however, the unknowns could still be found - a goodexercise. Example 8.7 Walking a Horizontal BeamGoal Solve an equilibrium problem with nonperpendicular torques.Problem A uniform horizontal beam 5.00 m long and weighting 3.00 x 102 N is attached to a wall by a pinconnection 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 themagnitude of the tension T in the cable and the force R exerted by the wall on the beam.R53.0(300 N600 N(b)R,T sin 53.0°RxT cos 53.0°53.0%4.50 m→(300 N5.00 m-2.50 m-600 N(a)(c)Figure 8.12 (a) A uniform beam attached to a wall and supported by a cable. (b) A free-body diagram for thebeam. (c) The component of the free-body diagram.Strategy The second condition of equilibrium, Et; = 0, with torques computed around the pin, can be solvedfor the tension T in the cable. The first condition of equilibrium, EF, = 0, gives two equations and twounknowns for the two components of the force exerted by the wall, Rx and Ry.

Answered: Image /qna-images/answer/1530af70-f11e-406a-b23e-cae973905cbc.jpg

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. R 53.0 300 N 600 N (b) R, T sin 53.0° R T cos 53.0° 53.0% 4.50 m→ 300 N 2.50 m- 600 N 5.00 m (a) (c)

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. R 53.0 300 N 600 N (b) R, T sin 53.0° R T cos 53.0° 53.0% 4.50 m→ 300 N 2.50 m- 600 N 5.00 m (a) (c)

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter8: Rotational Equilibrium And Rotational Dynamics

Section: Chapter Questions

Problem 82AP

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