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Chapter 11: Problem 20

A \(15,000 \mathrm{~N}\) crane pivots around a friction-free axle at its base and is supported by a cable making a \(25^{\circ}\) angle with the crane (Fig. E11.20). The crane is \(16 \mathrm{~m}\) long and is not uniform, its center of gravity being \(7.0 \mathrm{~m}\) from the axle as measured along the crane. The cable is attached \(3.0 \mathrm{~m}\) from the upper end of the crane. When the crane is raised to \(55^{\circ}\) above the horizontal holding an \(11,000 \mathrm{~N}\) pallet of bricks by a \(2.2 \mathrm{~m},\) very light cord, find (a) the tension in the cable and (b) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

Short Answer

Expert verified
The tension in the cable is approximately \(11894 \mathrm{~N}\) while the horizontal and vertical components of the force that the axle exerts on the crane are approximately \(6784 \mathrm{~N}\) and \(20922 \mathrm{~N}\) respectively.

Step by step solution

01

Identify Forces

Identify all the forces acting on the crane. There are four major forces acting on the crane at pivot - the weight of the crane \(W_1 = 15,000 \mathrm{~N}\), the weight of the pallet \(W_2 = 11,000 \mathrm{~N}\), the tension in the cable \(T\), and the reaction at the pivot, \(R\) which can be resolved into horizontal and vertical components.
02

Finding Tension in the Cable

Use the principle of moments to calculate the tension in the cable. The clockwise moments about the pivot must balance with the anti-clockwise moments for the crane to stay in equilibrium. Consider \(W_2\) to provide clockwise moment and \(T\) and \(W_1\) to give anti-clockwise moments. Since the net torque is zero, the equation becomes \( W_2 * 16 \sin(55) = W_1 * 7\sin(55) + T * 13\sin(25) \). Rearrange this equation to find \(T = \frac{W_2*16-W_1*7}{13 * \cos(25)}\), and proceed to insert the given values.
03

Resolve forces at the pivot

To find the forces at the pivot, use the equations of equilibrium that state the sum of forces in any direction should be zero. Resolve \(T\) into horizontal and vertical components: \(T_h = T * \cos(25)\) and \(T_v = T * \sin(25)\). Then, by summing forces in the horizontal and vertical directions, obtain the reaction forces \(R_v\) and \(R_h\) at the pivot. Write out the equations to find the reaction pair: \(R_h = T_h\) and \(R_v = T_v + W_1 + W_2\). Insert the calculated and given values to find the horizontal and vertical reactions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque
Torque is a crucial concept in understanding static equilibrium, especially when dealing with objects like cranes. It refers to the rotational effect of a force applied at a distance from a pivot point. The mathematical formula for torque (\( \tau \)) is given by:
  • \( \tau = r \times F \times \sin(\theta) \)
where \(r\) is the distance from the pivot point, \(F\) is the force, and \(\theta\) is the angle between the force and the lever arm.
In the exercise, the crane's equilibrium involves ensuring that the sum of torques around the pivot is zero. This condition means that the clockwise torques balance the counterclockwise torques, preventing it from rotating.
To find the tension in the cable, we calculate the torques produced by the weight of the crane and the pallet that wants to rotate the crane clockwise. At the same time, the tension in the cable acts to rotate it counterclockwise, stabilizing the structure.
Free-Body Diagram
A Free-Body Diagram (FBD) is a simplified representation of an object that details all the forces acting on it. For the crane problem, drawing an FBD helps visualize the forces and their directions.
In this scenario, you start by identifying key forces:
  • The weight of the crane, \(W_1 = 15,000 \text{ N}\), acting downward through the center of gravity.
  • The weight of the pallet, \(W_2 = 11,000 \text{ N}\), hanging at the end.
  • The tension in the cable, \(T\), angled at \(25^{\circ}\).
  • The reactions at the pivot, which include both vertical and horizontal components.
The FBD allows us to apply equilibrium equations efficiently, systematically breaking down the problem to find unknown forces like the tension and reaction forces at the pivot.
Force Components
When dealing with forces at an angle, it's essential to break them into horizontal and vertical components for easy analysis. This process is crucial in problems involving static equilibrium.
In our exercise, the tension in the cable, \(T\), has components:
  • Horizontal: \( T_h = T \times \cos(25^{\circ}) \)
  • Vertical: \( T_v = T \times \sin(25^{\circ}) \)
These components help in applying the equilibrium condition that the sum of forces in both horizontal and vertical directions must be zero.
Understanding force components allows us to solve for the reaction at the pivot. By resolving forces, we can determine the horizontal reaction force \(R_h = T_h\) and the vertical reaction force \(R_v = T_v + W_1 + W_2\). This breakdown makes complex problems more manageable.

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Most popular questions from this chapter

You open a restaurant and hope to entice customers by hanging out a sign (Fig. \(\mathbf{P} 1 \mathbf{1 . 5 3}\) ). The uniform horizontal beam supporting the sign is \(1.50 \mathrm{~m}\) long, has a mass of \(16.0 \mathrm{~kg},\) and is hinged to the wall. The sign itself is uniform with a mass of \(28.0 \mathrm{~kg}\) and overall length of \(1.20 \mathrm{~m}\). The two wires supporting the sign are each \(32.0 \mathrm{~cm}\) long, are \(90.0 \mathrm{~cm}\) apart, and are equally spaced from the middle of the sign. The cable supporting the beam is \(2.00 \mathrm{~m}\) long. (a) What minimum tension must your cable be able to support without having your sign come crashing down? (b) What minimum vertical force must the hinge be able to support without pulling out of the wall?

A ladder carried by a fire truck is \(20.0 \mathrm{~m}\) long. The ladder weighs \(3400 \mathrm{~N}\) and its center of gravity is at its center. The ladder is pivoted at one end (A) about a pin (Fig. E11.7); ignore the friction torque at the pin. The ladder is raised into position by a force applied by a hydraulic piston at \(C\). Point \(C\) is \(8.0 \mathrm{~m}\) from \(A,\) and the force \(\overrightarrow{\boldsymbol{F}}\) exerted by the piston makes an angle of \(40^{\circ}\) with the ladder. What magnitude must \(\boldsymbol{F}\) have to just lift the ladder off the support bracket at \(B ?\) Start with a free-body diagram of the ladder.

You are a construction engineer working on the interior design of a retail store in a mall. A 2.00 -m-long uniform bar of mass \(8.50 \mathrm{~kg}\) is to be attached at one end to a wall, by means of a hinge that allows the bar to rotate freely with very little friction. The bar will be held in a horizontal position by a light cable from a point on the bar (a distance \(x\) from the hinge) to a point on the wall above the hinge. The cable makes an angle \(\theta\) with the bar. The architect has proposed four possible ways to connect the cable and asked you to assess them: $$ \begin{array}{lllll} \text { Alternative } & \text { A } & \text { B } & \text { C } & \text { D } \\\ \hline x(\mathrm{~m}) & 2.00 & 1.50 & 0.75 & 0.50 \\ \theta(\text { degrees }) & 30 & 60 & 37 & 75 \end{array} $$ (a) There is concern about the strength of the cable that will be required. Which set of \(x\) and \(\theta\) values in the table produces the smallest tension in the cable? The greatest? (b) There is concern about the breaking strength of the sheetrock wall where the hinge will be attached. Which set of \(x\) and \(\theta\) values produces the smallest horizontal component of the force the bar exerts on the hinge? The largest? (c) There is also concern about the required strength of the hinge and the strength of its attachment to the wall. Which set of \(x\) and \(\theta\) values produces the smallest magnitude of the vertical component of the force the bar exerts on the hinge? The largest? (Hint: Does the direction of the vertical component of the force the hinge exerts on the bar depend on where along the bar the cable is attached?) (d) Is one of the alternatives given in the table preferable? Should any of the alternatives be avoided? Discuss.

A lead sphere has volume \(6.0 \mathrm{~cm}^{3}\) when it is resting on a lab table, where the pressure applied to the sphere is atmospheric pressure. The sphere is then placed in the fluid of a hydraulic press. What increase in the pressure above atmospheric pressure produces a \(0.50 \%\) decrease in the volume of the sphere?

A uniform \(300 \mathrm{~N}\) 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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