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Ch 11: Equilibrium & Elasticity
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 11: Equilibrium & ElasticityProblem 20a
Chapter 11, Problem 20a

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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Begin by drawing a free-body diagram of the crane. Identify all forces acting on the crane: the weight of the crane (15,000 N) acting downward at its center of gravity, the tension in the cable, and the weight of the pallet of bricks (11,000 N) acting downward at the end of the cord.
Calculate the torque due to the weight of the crane. Torque is given by \( \tau = r \cdot F \cdot \sin(\theta) \), where \( r \) is the distance from the pivot, \( F \) is the force, and \( \theta \) is the angle between the force and the lever arm. For the crane's weight, \( r = 7.0 \) m and \( \theta = 55° \).
Calculate the torque due to the weight of the pallet of bricks. The distance from the pivot to the point where the bricks are hanging is \( 16 \) m (length of the crane) plus \( 2.2 \) m (length of the cord), and the angle is also \( 55° \).
Calculate the torque due to the tension in the cable. The cable is attached \( 3.0 \) m from the upper end of the crane, so the distance from the pivot is \( 16 \) m - \( 3.0 \) m = \( 13 \) m. The angle between the cable and the crane is \( 25° \), but the angle for torque calculation is \( 90° - 25° = 65° \).
Set up the equation for rotational equilibrium: the sum of the torques around the pivot must be zero. Solve for the tension in the cable using the equation \( \tau_{cable} = \tau_{crane} + \tau_{bricks} \).

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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 the context of the crane, this means that the torques due to the crane's weight, the weight of the pallet, and the tension in the supporting cable must balance each other. Calculating these torques involves considering the forces' magnitudes, directions, and distances from the pivot point.
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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 to the line of action of the force. In this problem, the torque due to the crane's weight, the pallet's weight, and the tension in the cable must be considered to determine the system's equilibrium. The direction of each torque (clockwise or counterclockwise) is crucial for setting up the equilibrium equation.
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Free-Body Diagram

A free-body diagram is a graphical representation used to visualize the forces acting on an object. For the crane, this involves depicting the gravitational forces on the crane and the pallet, the tension in the cable, and the reaction at the pivot. This diagram helps in setting up the equations for translational and rotational equilibrium, allowing for the calculation of unknown forces such as the tension in the cable.
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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

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

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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 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

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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