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 17b

Chapter 11, Problem 17b

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?

Verified step by step guidance

First, identify the forces acting on the beam. The beam is in static equilibrium, so the sum of forces and the sum of torques must be zero. The forces include the weight of the beam acting at its center of gravity, the tension in the cable, and the reaction forces at the hinge.

Calculate the angle θ between the cable and the horizontal using trigonometry. The cable forms a right triangle with the wall and the beam. Use the tangent function: tan(θ) = opposite/adjacent = 4.0 m / 5.0 m.

Determine the torque due to the weight of the beam. The weight acts at the center of the beam, which is 4.5 m from the hinge. The torque τ due to the weight is τ = (weight of the beam) * (distance from hinge) * cos(θ).

Calculate the maximum tension the cable can withstand, which is given as 1.00 kN. The torque due to the tension in the cable is τ = (tension) * (distance from hinge to cable attachment) * sin(θ).

Set the sum of torques around the hinge to zero to solve for the weight of the beam. The equation is: (weight of the beam) * 4.5 m * cos(θ) = (1.00 kN) * 9.0 m * sin(θ). Solve this equation for 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.

Torque and Rotational Equilibrium

Torque is the rotational equivalent of force, calculated as the product of force and the perpendicular distance from the pivot point. For an object in rotational equilibrium, the sum of the torques around any pivot point must be zero. In this problem, the beam is in equilibrium, so the torques due to the beam's weight and the tension in the cable must balance each other.

Tension in the Cable

Tension is the force exerted by a rope or cable when it is used to transmit force. The cable in this scenario has a maximum tension capacity of 1.00 kN. To prevent the cable from breaking, the tension caused by the weight of the beam must not exceed this limit. Calculating the tension involves considering the angle of the cable and the forces acting 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, the diagram would include the gravitational force acting at its center of mass, the tension in the cable, and the reaction forces at the hinge. Analyzing these forces helps in setting up the equations needed to solve for the maximum weight the beam can have without exceeding the cable's tension limit.

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Free-Body Diagrams

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

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.

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 the horizontal and vertical components of the force exerted on the beam at the wall.

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

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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