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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
Chapter 11, Problem 11a

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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1
Identify the forces acting on the diving board: the weight of the diver (500 N) acting downward at the free end, the weight of the board (280 N) acting at its center of gravity (1.5 m from the pivot), and the reaction force at the support point.
Set up the torque equilibrium equation about the support point. The sum of the torques around the support point must be zero for the board to be in equilibrium.
Calculate the torque due to the diver's weight: Torque_diver = 500 N * 2.00 m.
Calculate the torque due to the board's weight: Torque_board = 280 N * 0.50 m (since the center of gravity is 0.50 m from the support point).
Set the sum of the torques equal to zero and solve for the reaction force at the support point: Reaction_force * 0 m = Torque_diver + Torque_board. Rearrange to find the reaction force.

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Key Concepts

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

Torque

Torque is the rotational equivalent of force, calculated as the product of force and the distance from the pivot point. In this problem, the diver and the board create torques around the support point, which must be balanced to find the force at the support. Understanding torque is crucial for analyzing rotational equilibrium in systems like the diving board.
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Equilibrium

Equilibrium in physics refers to a state where all forces and torques are balanced, resulting in no net motion. For the diving board, both translational and rotational equilibrium must be considered. The sum of forces and the sum of torques around the support point must be zero to solve for the force exerted at the support.
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Center of Mass

The center of mass is the point where the mass of an object is concentrated and acts as if all the mass were located there for the purpose of calculating torque. For the uniform diving board, its center of mass is at its midpoint. This concept helps in determining the torque due to the board's weight, which is essential for solving the problem.
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Related Practice
Textbook Question

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find the tension in the cable.


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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

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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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

Find the tension T in each cable and the magnitude and direction of the force exerted on the strut by the pivot in each of the arrangements in Fig. E11.13. In each case let w be the weight of the suspended crate full of priceless art objects. The strut is uniform and also has weight w. Start each case with a free-body diagram of the strut.

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Textbook Question

Find the tension T in each cable and the magnitude and direction of the force exerted on the strut by the pivot in each of the arrangements in Fig. E11.13. In each case let w be the weight of the suspended crate full of priceless art objects. The strut is uniform and also has weight w. Start each case with a free-body diagram of the strut.

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