Ch 11: Equilibrium & Elasticity

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 11: Equilibrium & Elasticity

Problem 22

Chapter 11, Problem 22

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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Begin by sketching a free-body diagram of the beam. Identify all forces acting on the beam: the weight of the beam (1.40 kN) acting downward at its center of gravity, the tension T in the cable acting perpendicular to the beam, the force exerted by the pivot with horizontal (Fh) and vertical (Fv) components, and the 5.00 kN force exerted by the lighting equipment at the lower left end of the beam.

Apply the conditions for equilibrium. Since the beam is in static equilibrium, the sum of all forces and the sum of all torques (moments) about any point must be zero. Choose the pivot point to sum torques, as this will eliminate the unknown forces at the pivot from the torque equation.

Write the torque equation about the pivot point. The torque due to the beam's weight is (1.40 kN) * (2.00 m) * cos(25°), the torque due to the lighting equipment is (5.00 kN) * (4.50 m) * cos(25°), and the torque due to the tension in the cable is T * (3.00 m). Set the sum of these torques to zero to solve for T.

Write the force equilibrium equations. For the vertical forces, sum the vertical components: Fv - 1.40 kN - 5.00 kN + T * sin(25°) = 0. For the horizontal forces, sum the horizontal components: Fh - T * cos(25°) = 0.

Solve the system of equations. Use the torque equation to find the tension T in the cable. Substitute T into the force equilibrium equations to find the horizontal (Fh) and vertical (Fv) components of the force exerted by the pivot.

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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, resulting in no angular acceleration. In this problem, the beam is in rotational equilibrium, meaning the torques due to the beam's weight, the tension in the cable, and the force from the lighting equipment must balance. Calculating these torques involves considering the distances from the pivot and the angles at which forces are applied.

Torque

Torque is a measure of the rotational force applied to an object, calculated as the product of the force and the perpendicular distance from the pivot point. In this scenario, the torque due to the beam's weight, the tension in the cable, and the lighting equipment's force must be considered. The direction of each torque (clockwise or counterclockwise) is crucial for determining the net torque on the beam.

Free-Body Diagram

A free-body diagram is a graphical representation used to visualize the forces acting on an object. For the beam, this includes the gravitational force at its center of gravity, the tension in the cable, and the force exerted by the pivot. Drawing a free-body diagram helps in identifying all forces and their points of application, which is essential for setting up the equations of equilibrium.

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Textbook Question

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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

A 15,000-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25° angle with the crane (Fig. E11.18). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane. When the crane is raised to 55° above the horizontal holding an 11,000-N pallet of bricks by a 2.2-m, very light cord, find the tension in the cable. Start with a free-body diagram of the crane.

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Textbook Question

A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 700 N is applied to each end of the wire. What minimum diameter is required for the wire?

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. Find the horizontal and vertical components of the force the hinge exerts on the beam. Is the vertical component upward or downward?

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Textbook Question

A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0 kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for nylon?

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