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Ch 11: Equilibrium & Elasticity
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 11: Equilibrium & ElasticityProblem 7a
Chapter 11, Problem 7a

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 N. What is the weight of the motor, and where along the board is its center of gravity located?

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1
Understand that the board is in static equilibrium, meaning the sum of forces and the sum of torques (moments) around any point must be zero.
Calculate the total weight of the motor by summing the forces exerted by the two people: 400 N + 600 N. This is because the upward forces must balance the downward force of the motor's weight.
Choose a pivot point to calculate the torques. A convenient choice is one end of the board, say where the 400 N force is applied, to simplify calculations.
Set up the torque equation around the chosen pivot point. The torque due to the 600 N force is 600 N multiplied by the full length of the board (2.00 m). The torque due to the motor's weight is its weight (W) multiplied by the distance (d) from the pivot to the center of gravity.
Solve the torque equation for the distance (d) to find the center of gravity: 600 N * 2.00 m = W * d. Use the previously calculated weight of the motor to find d.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium of Forces

In physics, equilibrium refers to a state where all forces acting on an object are balanced, resulting in no net force and no acceleration. For the motor to be in equilibrium on the board, the sum of the upward forces (400 N and 600 N) must equal the weight of the motor. This ensures that the motor remains stationary and does not move vertically.
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Center of Gravity

The center of gravity of an object is the point where its entire weight is considered to be concentrated. It is crucial for determining how the weight is distributed along the board. In this problem, the center of gravity can be found by setting the torques about any point to zero, ensuring rotational equilibrium, and solving for the position where the torques due to the lifting forces balance.
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Torque and Rotational Equilibrium

Torque is the rotational equivalent of force, calculated as the product of force and the perpendicular distance from the pivot point. For rotational equilibrium, the sum of the torques around any point must be zero. In this scenario, calculating torques about one end of the board helps determine the position of the motor's center of gravity by ensuring the clockwise and counterclockwise torques balance.
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Related Practice
Textbook Question

A diving board 3.00 m long is supported at a point 1.00 m from the end, and a diver weighing 500 N stands at the free end (Fig. E11.11). The diving board is of uniform cross section and weighs 280 N. Find the force at the support point.


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Textbook Question

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weighs 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing 740 N climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. What is the maximum friction force that the ground can exert on the ladder at its lower end?

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Textbook Question

A uniform 300-N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges if the upward force is applied at the center.

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Textbook Question

Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

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Textbook Question

A 350-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to 400.0 N. You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

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Textbook Question

A uniform 300-N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges if the upward force is applied at the center of the edge opposite the hinges.

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