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

College Physics: A Strategic Approach Plus Mastering Physics with Pearson eText -- Access Card Package (4th Edition) (What's New in Astronomy & Physics)

4th Edition

ISBN: 9780134641492

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

Publisher: PEARSON

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Question

Chapter 8, Problem 55GP

To determine

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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College Physics: A Strategic Approach Plus Mastering Physics with Pearson eText -- Access Card Package (4th Edition) (What's New in Astronomy & Physics)

Ch. 8 - Prob. 1CQ Ch. 8 - Could a ladder on a level floor lean against a...Ch. 8 - Prob. 3CQ Ch. 8 - Prob. 4CQ Ch. 8 - Prob. 5CQ Ch. 8 - Prob. 6CQ Ch. 8 - Prob. 7CQ Ch. 8 - A spring exerts a 10 N force after being stretched...Ch. 8 - Prob. 9CQ Ch. 8 - A typical mattress has a network of springs that...

Ch. 8 - Take a spring and cut it in half to make two...Ch. 8 - A wire is stretched right to its breaking point by...Ch. 8 - Prob. 13CQ Ch. 8 - Prob. 14CQ Ch. 8 - Steel nails are rigid and unbending. Steel wool is...Ch. 8 - Two children hold opposite ends of a lightweight,...Ch. 8 - Prob. 19MCQ Ch. 8 - Prob. 20MCQ Ch. 8 - Prob. 21MCQ Ch. 8 - Prob. 22MCQ Ch. 8 - Prob. 23MCQ Ch. 8 - Prob. 24MCQ Ch. 8 - Prob. 25MCQ Ch. 8 - Prob. 26MCQ Ch. 8 - You have a heavy piece of equipment hanging from a...Ch. 8 - Prob. 1P Ch. 8 - Prob. 2P Ch. 8 - Prob. 3P Ch. 8 - Prob. 4P Ch. 8 - Youre carrying a 3.6-m-long, 25 kg pole to a...Ch. 8 - Prob. 6P Ch. 8 - Prob. 7P Ch. 8 - Prob. 9P Ch. 8 - Prob. 11P Ch. 8 - Prob. 13P Ch. 8 - Prob. 14P Ch. 8 - Prob. 15P Ch. 8 - The stability of a vehicle is often rated by the...Ch. 8 - Prob. 18P Ch. 8 - A car manufacturer claims that you can drive its...Ch. 8 - Prob. 20P Ch. 8 - An orthodontic spring, connected between the upper...Ch. 8 - Prob. 22P Ch. 8 - Experiments using optical tweezers measure the...Ch. 8 - Prob. 24P Ch. 8 - One end of a 10-cm-long spring is attached to the...Ch. 8 - Prob. 26P Ch. 8 - A spring has an unstretched length of 10 cm. It...Ch. 8 - A spring stretches 5.0 cm when a 0.20 kg block is...Ch. 8 - You need to make a spring scale to measure the...Ch. 8 - Prob. 30P Ch. 8 - A force stretches a wire by 1.0 mm. a. A second...Ch. 8 - Prob. 32P Ch. 8 - What hanging mass will stretch a 2.0-m-long,...Ch. 8 - An 80-cm-long, 1.0-mm-diameter steel guitar string...Ch. 8 - A mineshaft has an ore elevator hung from a single...Ch. 8 - The normal force of the ground on the foot can...Ch. 8 - A three-legged wooden bar stool made out of solid...Ch. 8 - Prob. 38P Ch. 8 - Prob. 39P Ch. 8 - Prob. 40P Ch. 8 - A glass optical fiber in a communications system...Ch. 8 - The Achilles tendon connects the muscles in your...Ch. 8 - Prob. 44P Ch. 8 - Prob. 47P Ch. 8 - Prob. 48P Ch. 8 - Prob. 49P Ch. 8 - Prob. 51GP Ch. 8 - Prob. 52GP Ch. 8 - Prob. 53GP Ch. 8 - Prob. 55GP Ch. 8 - Two identical, side-by-side springs with spring...Ch. 8 - Prob. 57GP Ch. 8 - Prob. 58GP Ch. 8 - Prob. 59GP Ch. 8 - A 25 kg child bounces on a pogo stick. The pogo...Ch. 8 - Prob. 61GP Ch. 8 - In the hammer throw, an athlete spins a heavy mass...Ch. 8 - There is a disk of cartilage between each pair of...Ch. 8 - Orb spiders make silk with a typical diameter of...Ch. 8 - Larger animals have sturdier bones than smaller...Ch. 8 - Prob. 67GP Ch. 8 - Prob. 68GP Ch. 8 - Prob. 69MSPP Ch. 8 - Prob. 70MSPP Ch. 8 - Prob. 71MSPP Ch. 8 - Prob. 72MSPP

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