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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 4b

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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Identify the forces acting on the trapdoor: the weight of the trapdoor (300 N) acts downward at its center of gravity, and the upward force is applied either at the center or at the edge opposite the hinges.
For part (a), when the upward force is applied at the center, calculate the torque about the hinge. The torque due to the weight is \( \tau = \text{weight} \times \text{distance from hinge} \). Since the weight acts at the center, the distance is half the width of the door.
Set the net torque to zero to find the upward force needed to begin opening the door. The equation is \( F_{up} \times \frac{w}{2} = 300 \times \frac{w}{2} \), where \( F_{up} \) is the upward force and \( w \) is the width of the door.
For part (b), when the upward force is applied at the edge opposite the hinges, calculate the torque about the hinge again. The torque due to the weight remains the same, but the distance for the upward force is now the full width of the door.
Set the net torque to zero for this scenario: \( F_{up} \times w = 300 \times \frac{w}{2} \). Solve for \( F_{up} \) to find the force needed. Additionally, consider the forces at the hinge, which include both vertical and horizontal components due to the applied force and the weight of the door.

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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, defined as the product of force and the distance from the pivot point (lever arm). It determines how effectively a force can cause an object to rotate around an axis. In this problem, calculating torque helps determine the force needed to open the trapdoor by considering the distance from the hinge where the force is applied.
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Center of Mass

The center of mass is the point in an object where its mass is evenly distributed and about which it balances in all directions. For a uniform object like the trapdoor, the center of mass is at its geometric center. Understanding this concept is crucial for calculating the torque when the force is applied at different points on the trapdoor.
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Equilibrium of Forces

Equilibrium occurs when all forces and torques acting on an object are balanced, resulting in no net motion. To find the net upward force needed to open the trapdoor, one must consider the equilibrium of forces and torques, ensuring that the applied force counteracts the gravitational force and any additional forces exerted by the hinges.
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Related Practice
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 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

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Textbook Question

A uniform rod is 2.00 m long and has mass 1.80 kg. A 2.40 kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 m from the left-hand end of the rod?

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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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