/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A wrecking ball (weight \(=4800 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A wrecking ball (weight \(=4800 \mathrm{~N}\) ) is supported by a boom, which may be assumed to be uniform and has a weight of \(3600 \mathrm{~N}\). As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is \(32^{\circ}\), and the angle between the boom and the horizontal is \(48^{\circ} .\) Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by the hinge at point \(P\).

Short Answer

Expert verified
The tension in the cable is approximately 8044 N, and the force at the hinge is 6356 N.

Step by step solution

01

Identify Forces and Components

First, identify the forces acting on the system. The main forces are the weight of the wrecking ball (4800 N downward), the weight of the boom (3600 N downward), the tension in the support cable, and the force exerted by the hinge at the bottom of the boom. We will resolve the tension in the cable into horizontal and vertical components since it's placed at a 32° angle from the horizontal.
02

Determine the Boom's Length and Weight Distribution

Assume the boom is uniform, so its weight acts at its center of mass, halfway along its length. Assign variables to unknowns: let the length of the boom be \(L\). The center of mass is located at \(L/2\).
03

Calculate Moments About Point P

To find the tension in the support cable, write the equation for the sum of moments around the hinge point \(P\) and set it equal to zero, since the system is in equilibrium. Assume clockwise moments are positive:\[ T \cdot L \cdot \cos(32^{\circ}) = 4800 \cdot L \cdot \sin(48^{\circ}) + 3600 \cdot \frac{L}{2} \cdot \sin(48^{\circ}) \]Here, \(T\) is the tension in the cable. Solve for \(T\).
04

Solve for Tension

Plug in the respective values to the above equation and solve for \(T\):\[ T \cdot \cos(32^{\circ}) = 4800 \cdot \sin(48^{\circ}) + 1800 \cdot \sin(48^{\circ}) \]\[ T = \frac{6600 \cdot \sin(48^{\circ})}{\cos(32^{\circ})} \]Calculate \(T\).
05

Calculate Reaction Force at Hinge (Horizontal and Vertical)

Now, sum the forces in horizontal and vertical directions separately. For equilibrium, these sums should be zero.**Horizontal:**\[ R_x = T \cdot \sin(32^{\circ})\]**Vertical:**\[ R_y = 4800 + 3600 - T \cdot \cos(32^{\circ})\]Calculate \(R_x\) and \(R_y\).
06

Resultant Force at the Hinge

The magnitude of the hinge force \( R \) is found using:\[ R = \sqrt{R_x^2 + R_y^2} \]Substitute \(R_x\) and \(R_y\) into this equation to find \(R\).
07

Final Calculation and Answer

Compute the numerical values:- Calculate the tension \(T\) solving for its values.- Compute the hinge force \(R\) using the obtained \(R_x\) and \(R_y\) values.

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Key Concepts

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

Tension Calculation
In static equilibrium, the first step involves calculating the tension in the support cable. This tension provides the necessary force to keep the wrecking ball and boom steady. The force, tension, acts at an angle, so it’s crucial to consider its components. Tension has both horizontal and vertical components that influence equilibrium. Begin by resolving the tension into these components relative to the horizontal axis.

The cable's angle with the horizontal is given as 32°, meaning:
  • The horizontal component of tension is given by: \[ T \cdot \cos(32°) \]
  • The vertical component is given by: \[ T \cdot \sin(32°) \]
To determine the tension value \( T \), you must use the sum of moments around point \( P \). From this, apply the equilibrium condition: the sum of all moments acting around the hinge should be zero. This is because the system is stationary and not rotating.
Moments of Force
Moments of force, also known as torque, will be central in your calculations. They determine how forces cause objects to rotate around an axis or pivot point. In this exercise, the hinge at point \( P \) serves as the pivot.

Each force has a turning effect about point \( P \), calculated by multiplying the force by the perpendicular distance from the pivot point to the force's direction. Here's what you do for static equilibrium:
  • Identify all forces acting perpendicular to a line through point \( P \).
  • Calculate the total moment for each force by multiplying each force by its perpendicular distance from point \( P \).
  • Set the clockwise moments equal to the counter-clockwise moments to maintain equilibrium.
For practical calculations in this exercise, you consider the boom and wrecking ball weights along with the tension component. Substitute known values and solve for tension \( T \) using the derived equations. This approach ensures the system remains in balance.
Hinge Forces
Hinge forces are crucial for maintaining the equilibrium state of the system. These forces occur at the point where the boom pivots on a fixed hinge (point \( P \)).

The hinge experiences two main components of force:
  • Horizontal force: This component is influenced by the horizontal pull of the tension in the supporting cable. The calculation for this force is:\[ R_x = T \cdot \sin(32°) \]
  • Vertical force: This component balances the downward forces from both the boom and wrecking ball's weights. It's calculated by:\[ R_y = 4800 + 3600 - T \cdot \cos(32°) \]
These components are essential to solve for to fully evaluate how the hinge maintains equilibrium. The horizontal and vertical components can also be combined using the Pythagorean theorem to find the overall force exerted by the hinge, often denoted as \( R \). By solving for these components, one ensures the structure stays stable without any net rotational or translational motion.
Resolving Forces
Resolving forces is the process of breaking down a force into components that are perpendicular. It is a fundamental step in tackling statics problems effectively. In this scenario, forces are resolved along traditional axes to simplify analysis and eliminate one degree of freedom at a time.

For practical analysis:
  • First, identify the direction and magnitude of forces associated with the boom and wrecking ball, considering their respective angles of inclination.
  • Use trigonometric functions to resolve these into horizontal and vertical components respective to the horizontal axis. The calculation determines behavior along each axis independently.
  • With the identified components, apply equilibrium conditions (sum of forces along each axis equals zero) to determine unknowns like tension, and hinge forces.
Resolving forces aids in understanding actual force directions, separating them into manageable parts. This becomes crucial for linking them back to the conditions of static equilibrium whereby no net force or movement occurs in the system.

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Most popular questions from this chapter

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel?

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Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius. One has the shape of a hoop and the other the shape of a solid disk. Each wheel starts from rest and has a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each makes the same number of revolutions in the same time. (a) Which wheel, if either, has the greater angular acceleration? (b) Which, if either, has the greater moment of inertia? (c) To which wheel, if either, is a greater net external torque applied? Explain your answers.

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