Ch 11: Equilibrium & Elasticity

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 11: Equilibrium & Elasticity

Problem 14b

Chapter 11, Problem 14b

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find the horizontal and vertical components of the force exerted on the beam at the wall.

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Identify the forces acting on the beam: the weight of the beam (190 N) acting at its center, the tension in the cable, and the reaction forces at the wall (horizontal and vertical components).

Set up the equilibrium conditions for the beam. Since the beam is in static equilibrium, the sum of forces in both the horizontal and vertical directions must be zero, and the sum of torques about any point must also be zero.

Choose the point where the beam is attached to the wall as the pivot point to calculate torques. This will eliminate the reaction forces at the wall from the torque equation, simplifying the calculation.

Write the torque equation about the pivot point: \( \tau = 0 = (300 \text{ N} \times 4.00 \text{ m}) + (190 \text{ N} \times 2.00 \text{ m}) - T \times 3.00 \text{ m} \), where T is the tension in the cable.

Solve the torque equation for the tension T in the cable. Then, use the equilibrium conditions for forces to find the horizontal and vertical components of the reaction force at the wall. The horizontal component is equal to the horizontal component of the tension, and the vertical component is the sum of the vertical component of the tension and the weight of the beam.

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Key Concepts

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

Rotational Equilibrium

Rotational equilibrium occurs when the sum of all torques acting on a system is zero, ensuring the system is not rotating. For a beam hinged to a wall, this means the torques due to the beam's weight and any other forces must balance. Calculating these torques involves considering the force magnitudes, their directions, and the distances from the pivot point.

Components of Force

Forces can be broken down into horizontal and vertical components, which are essential for analyzing equilibrium in two dimensions. The horizontal component acts parallel to the ground, while the vertical component acts perpendicular. In this problem, the force exerted by the wall on the beam can be resolved into these components to ensure both translational and rotational equilibrium.

Center of Gravity

The center of gravity of an object is the point where its weight is evenly distributed in all directions. For a uniform beam, this point is at its geometric center. Knowing the center of gravity is crucial for calculating the torque due to the beam's weight, as it determines the lever arm distance from the pivot point, affecting the rotational equilibrium.

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Related Practice

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

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 9.00-m-long uniform beam is hinged to a vertical wall and held horizontally by a 5.00-m-long cable attached to the wall 4.00 m above the hinge (Fig. E11.17). The metal of this cable has a test strength of 1.00 kN, which means that it will break if the tension in it exceeds that amount. What is the heaviest beam that the cable can support in this configuration?

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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.

Suppose that you can lift no more than 650 N (around 150 lb) unaided.

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 center of gravity of the load carried in the wheelbarrow is also 0.50 m from the center of the wheel.

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