Explanation Given data: Refer to the Figure P8.43 in the textbook for the standing of a student on the beam. Mass of the beam ( m B ) is 100 kg. Mass of the student ( m S ) is 80 kg. From the Figure, the distance between the two supports is 3.0 m and the distance between the student and support 1 is 2.0 m. Therefore, the distance of the student from the support 1 ( d 1 ) is 2.0 m. The distance between the two supports ( d ) is 3.0 m. Formula used: Write the expression for equilibrium condition of the torque as follows. ∑ τ = 0 (1) Here, ∑ τ is the sum of the torques acting on any object. Write the expression for weight of the object as follows. w = m g (2) Here, m is the mass of the object, and g is the acceleration due to gravity, which is 9.8 m s 2 . Write the expression for torque in terms of force and distance as follows. τ = F d (3) Here, F is the force, and d is the distance. Write the expression for equilibrium condition of the forces as follows. ∑ F = 0 (4) Here, ∑ F is the sum of the forces acting on any object. From the given data, the following forces acts on the beam. Weight of the beam acts at the center of the gravity (center of the beam, 1.5 m) towards in the downward direction. Weight of the student acts at the standing position of the student (2.0 m from the support 1) towards in the downward direction. The normal forces exerted by each beam towards in the upward direction. From the analysis, modify the expression in equation (4) for the sum of the forces in the vertical component ( y -component) as follows. − w B + ( − w S ) + F 1 + F 2 = 0 (5) Here, w B is the weight of the beam in the downward direction, w S is the weight of the student in the downward direction, F 1 is the normal exerted by the support 1 in the upward direction, and F 2 is the normal exerted by the support 2 in the upward direction. Calculation of weight of the student: Substitute 80 kg for m , and 9.8 m s 2 for g in equation (2), w S = ( 80 kg ) ( 9.8 m s 2 ) = 784 kg ⋅ m s 2 = 784 N { ∵ 1 kg ⋅ m s 2 = 1 N } Calculation of weight of the beam: Substitute 100 kg for m , and 9.8 m s 2 for g in equation (2), w B = ( 100 kg ) ( 9.8 m s 2 ) = 980 kg ⋅ m s 2 = 980 N Rewrite the expression in equation (5) as follows

College Physics: A Strategic Approach (3rd Edition)

3rd Edition

ISBN: 9780321879721

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 8, Problem 43GP

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To find: The upward force exerted by each support on the beam.

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Chapter 8, Problem 42P

Chapter 8, Problem 44GP

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A 3.0-m-long rigid beam with a mass of 100 kg is supported at each end, as shown. An 80 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?

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Chapter 8 Solutions

College Physics: A Strategic Approach (3rd Edition)

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The upward force exerted by each support on the beam.

Solution Summary: The author explains the upward force exerted by the support 1 and support 2 on the beam.

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To find: The upward force exerted by each support on the beam.

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Chapter 8 Solutions