Explanation Given data: Refer to the Figure P8.55 in the textbook for the walking position of the boy on the beam. Mass of the beam ( m b ) is 40 kg. Mass of the boy ( m B ) is 20 kg. Length of the beam ( L b ) is 5.0 m. From the figure, the distance between two posts ( 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 torque in terms of force and distance as follows. τ = F d (2) Here, F is the force, and d is the distance. Write the expression for equilibrium condition of the forces as follows. ∑ F = 0 (3) Here, ∑ F is the sum of the forces acting on any object. Write the expression for weight of the object as follows. w = m g (4) Here, m is the mass of the object, and g is the acceleration due to gravity, which is 9.8 m s 2 . Explanation: Consider the beam as rigid body. For the beam not to fall over, the beam must be in translational ( ∑ F = 0 ) and rotational equilibrium ( ∑ τ = 0 ) condition. Consider the boy walks along the beam a distance of x meters from the left end of the beam. The following are the forces act on the given system. The force due to weight of the beam, which acts at the center of the gravity (center of the beam, 2.5 m from any end of the beam) towards in the downward direction. The force due to weight of the boy, which acts at the standing position of the boy ( x m from the left end of the beam) towards in the downward direction. The normal forces exerted by each support (post), which are acted in the upward direction. From the analysis, draw the given system as shown in Figure 1. In the Figure 1, F → 1 is the normal force exerted by the left post (support), F → 2 is the normal force exerted by the right post, w → b is the force due to weight of the beam, w → B is the force due to weight of the boy, and x is the distance walked by the boy from the left end of the beam. From the analysis, modify the expression in equation (3) for the sum of the forces in the vertical component ( y -component) as follows. − w B + ( − w b ) + F 1 + F 2 = 0 (5) Calculation of force due to weight of the boy: Modify the expression in equation (4) to find the force due to weight of the boy as follows. w B = m B g Substitute 20 kg for m B , and 9.8 m s 2 , w B = ( 20 kg ) ( 9.8 m s 2 ) = 196 kg ⋅ m s 2 = 196 N { ∵ 1 kg ⋅ m s 2 = 1 N } Calculation of weight of the beam: Modify the expression in equation (4) to find the force due to weight of the beam as follows. w b = m b g Substitute 40 kg for m b , and 9.8 m s 2 , w b = ( 40 kg ) ( 9

MODIFIED MASTERING COLLEGE PHYSICS 18WK.

4th Edition

ISBN: 9780136782216

Author: Knight

Publisher: PEARSON

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Chapter 8, Problem 55GP

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To find: The distance that the boy can get to the right end of the beam without it tipping.

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Chapter 8, Problem 53GP

Chapter 8, Problem 56GP

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MODIFIED MASTERING COLLEGE PHYSICS 18WK.

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