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Ch 11: Equilibrium & Elasticity
Chapter 11, Problem 11

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 (a) if the upward force is applied at the center and (b) 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. These include the weight of the trapdoor acting downward through its center of mass and the upward force applied either at the center or at the center of the edge opposite the hinges.
Calculate the torque due to the weight of the trapdoor about the hinge. Torque (\(\tau\)) is given by the product of the force and the perpendicular distance from the point of rotation (hinge) to the line of action of the force. The weight acts at the center, so the distance is half the length of the trapdoor.
Calculate the torque due to the applied upward force. If the force is applied at the center, the distance from the hinge is again half the length of the trapdoor. If the force is applied at the center of the edge opposite the hinges, the distance is the full length of the trapdoor.
Set the sum of the torques equal to zero for equilibrium to find the magnitude of the upward force needed to just start opening the trapdoor. This involves solving the equation \(\tau_{weight} + \tau_{applied} = 0\).
Determine the total force exerted on the door by the hinges. This force must balance both the vertical forces (the weight and the upward force) and provide a counter-torque to keep the door from rotating.

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Key Concepts

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

Torque

Torque is a measure of the rotational force applied to an object around a pivot point. It is calculated as the product of the force applied and the distance from the pivot point to the line of action of the force. In this scenario, understanding torque is essential to determine how the applied upward force affects the trapdoor's ability to rotate about its hinges.
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Equilibrium

Equilibrium refers to a state where the sum of forces and the sum of torques acting on an object are both zero, resulting in no net movement. For the trapdoor, analyzing the forces acting on it, including the weight of the door and the applied upward force, is crucial to determine the conditions under which it will begin to open.
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Force Distribution

Force distribution involves understanding how forces are transmitted through an object and how they affect different points, such as hinges. In this problem, the location of the applied force (center vs. edge) will influence the distribution of forces at the hinges, affecting the total force exerted on the door and its ability to rotate.
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Related Practice
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 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. (a) 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 (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.
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Textbook Question
Suppose that you can lift no more than 650 N (around 150 lb) unaided.

(a) How much can you lift using a 1.4-m-long wheelbarrow that weighs 80.0 N and whose center of gravity is 0.50 m from the center of the wheel (Fig. E11.16)? The cen-ter of gravity of the load car-ried in the wheelbarrow is also 0.50 m from the center of the wheel. (b) Where does the force come from to enable you to lift more than 650 N using the wheelbarrow?

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Textbook Question
A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it (Fig. E11.20). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

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Textbook Question
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center.

Find (a) the tension in the cable.
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