Explanation Given data: Refer to the Figure P8.1 in the textbook for the standing of a student on a very light, rigid board. Mass of the student ( m ) is 64 kg. From the Figure, the distance between the two scales is 2.0 m and the distance between the student and left scale is 1.5 m. Therefore, the distance of the student from the left scale ( d left ) is 1.5 m. The distance between the two scales ( d ) is 2.0 m. Formula used: Write the expression for static 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. Explanation: As the board is very light and rigid, consider the board as massless. Therefore, it does not contribute to the scale reading. Calculation of weight of the student: Substitute 64 kg for m , and 9.8 m s 2 for g in equation (2), w = ( 64 kg ) ( 9.8 m s 2 ) = 627.2 kg ⋅ m s 2 ≃ 627 N { ∵ 1 kg ⋅ m s 2 = 1 N } From the given data, it is clear that, the sum of the readings is equal to the weight of the person. Therefore, write the expression for readings of the scale as follows. n left + n right = w (4) Here, n left is the reading on the left scale, n right is the reading on the right scale, and w is the weight of the student. Substitute 627 N for w in equation (4), n left + n right = 627 N (5) Consider the board rests on the left scale and compute the torques around point on the left scale. From the Figure, the weight of the student creates a clockwise (negative) torque and the normal force of the right scale creates a counter clockwise (positive) torque. Therefore, modify the expression in equation (1) as follows. τ right + ( − τ w ) = 0 (6) Here, τ right is the torque created by the normal force of the right scale, and τ w is the torque created by the weight of the student

4th Edition

ISBN: 9780137574728

Author: Knight

Publisher: PEARSON

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Center of Gravity of a system

The centre of gravity is a property that is based on the geometry of a rigid body or any particle. It is described as an equilibrium point or a hypothetical location where the total weight of the body is concentrated.

Question

Chapter 8, Problem 1P

To determine

To find: The reading on each of the scales.

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Chapter 8, Problem 27MCQ

Chapter 8, Problem 2P

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