EXAMPLE 9–5 | Forces on a beam and supports. A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support column (Fig. 9–8). Calculate the force on each of the vertical support columns. APPROACH We analyze the forces on the beam (the force the beam exerts on each column is equal and opposite to the force exerted by the column on the beam). We label these forces F, and Fg in Fig. 9–8. The weight of the beam itself acts at its center of gravity, 10.0 m from either end. We choose a con- venient axis for writing the torque equation: the point of application of FA (labeled P), so F, will not enter the equation (its lever arm will be zero) and we will have an equation in only one unknown, FR. CG (1500 kg)ỹ 10.0 m -5.0 m-5.0 m- (15,000 kg)ỹ SOLUTION The torque equation, E7 = 0, with the counterclockwise direction as positive, gives FIGURE 9-8 A 1500-kg beam supports a 15,000-kg machine. Example 9–5. Er = -(10.0 m)(1500 kg)g (15.0 m)(15,000 kg)g + (20.0 m)F3 = 0. Solving for FB, we find Fg = (12,000 kg)g = 118,000 N. To find FA, we use EF, = 0, with +y upward: = FA - (1500 kg)g – (15,000 kg)g + Fg EFy = 0. Putting in Fg = (12,000 kg)g, we find that FA (4500 kg)g 44,100 N.

EXAMPLE 9–5 | Forces on a beam and supports. A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support column (Fig. 9–8). Calculate the force on each of the vertical support columns. APPROACH We analyze the forces on the beam (the force the beam exerts on each column is equal and opposite to the force exerted by the column on the beam). We label these forces F, and Fg in Fig. 9–8. The weight of the beam itself acts at its center of gravity, 10.0 m from either end. We choose a con- venient axis for writing the torque equation: the point of application of FA (labeled P), so F, will not enter the equation (its lever arm will be zero) and we will have an equation in only one unknown, FR. CG (1500 kg)ỹ 10.0 m -5.0 m-5.0 m- (15,000 kg)ỹ SOLUTION The torque equation, E7 = 0, with the counterclockwise direction as positive, gives FIGURE 9-8 A 1500-kg beam supports a 15,000-kg machine. Example 9–5. Er = -(10.0 m)(1500 kg)g (15.0 m)(15,000 kg)g + (20.0 m)F3 = 0. Solving for FB, we find Fg = (12,000 kg)g = 118,000 N. To find FA, we use EF, = 0, with +y upward: = FA - (1500 kg)g – (15,000 kg)g + Fg EFy = 0. Putting in Fg = (12,000 kg)g, we find that FA (4500 kg)g 44,100 N.

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter12: Static Equilibrium And Elasticity

Section: Chapter Questions

Problem 34P: In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the...

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