EXAMPLE 8.8 Walking a Horizontal Beam GOAL Apply the two conditions of equilibrium. (a) A uniform beam attached to a wall and supported by a cable. (b) A force diagram for the beam. (c) The component form of the force diagram. 51 i" 53.o>\\ l 300 N 53.07, 4 5.00 m-{ 500 N [1 [3 R1 Tsin 530 RX Tcos 530 1.50 m * 300 N 600 N 2.50 m -{ E (D PROBLEM A uniform horizontal beam 5.00 m long and weighing 3.00 x 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 (Figure (.30). If a person weighing 6.00 x 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. STRATEGY See Figures (a)-(c) (Steps 1 and 2). The second condition of equilibrium, Zri = 0, with torques computed around the pin, can be solved for the tension T in the cable. The first condition of v equilibrium, 2F, = 0, gives two equations and two unknowns for the two components of the force exerted by the wall, R and R . x y SOLUTION From the figure, the forces causing _, torques are the wall force R, the gravity forces on the beam and the man, \"'3 and wM, and the tension force > T. Apply the condition of rotational equilibrium (Step 3). Compute torques around the pin at 0, so TR = 0 (zero moment arm). The torque clue to the beam's weight acts at the beam's center of gravity. Substitute L = 5.00 m and the weights, solving for T. Now apply the first condition of equilibrium to the beam (Step 4). Substituting the value of T found in the previous step and the weights, obtain _, the components of R (Step 5). LEARN MORE M (a! II TR+TB+TM+TT=0 ET 0 wB(L/2) wM(1.so m) + TL sin (53) = 0 (3.00 x 102 N)(2.50 m) (6.00 x 102 N)(1.50 m) + (Tsin 53.0)(5.00 m) = o T=413N (1) EFX = RX Tcos 53.0 = 0 (2) EFy = Ry WB WM + Tsin 53.0 = R = 5.7Ox102N RX: 249N y REMARKS Even if we selected some other axis for the torque equation, the solution would be the same. For example, if the axis were to pass through the center of gravity of the beam, the torque equation would involve both T and Ry. Together with Equations (1) and (2), however, the unknowns could still be founda good exercise. In this example, notice the steps of the Problem-Solving Strategy could be carried out in the explicit recommended order. QUESTION What happens to the tension in the cable if the man in Figure (a) moves farther away from the wall? 0 The tension would increase. 0 The tension would decrease. PRACTICE IT Use the worked example above to help you solve this problem. A uniform horizontal beam 5.00 m long and weighting 2.94 x 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 530" with the horizontal (Figure (a)). If a person weighing 5.86 x 102 N stands 1.40 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. EXERCISE HINTS: GETTINGSTARTED | I'M STUCK! A person with mass 55.0 kg stands at = 2.30 m away from the wall on a x = 5.75 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.) __L3'3 _______ I d x_'{ Q)