/*! 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 20 A 4.2 -m-long beam is supported ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 20

A 4.2 -m-long beam is supported by a cable at its center. A \(65-\mathrm{kg}\) steelworker stands at one end of the beam. Where should a \(190-\mathrm{kg}\) bucket of concrete be suspended for the beam to be in static equilibrium?

Short Answer

Expert verified
The bucket of concrete should be suspended at a position where \(d = \frac{W1 * L/2 - Wbeam * L/2}{W2}\) from the left end of the beam (where the steelworker stands). However, since the problem does not provide the beam's weight, we cannot calculate the exact numerical position.

Step by step solution

01

Identify Known and Unknown values

We know the length of the beam (L = 4.2 m), weight of the steelworker (W1 = 65 kg) at one end of the beam, and weight of the bucket of concrete (W2 = 190 kg), whose position we need to figure out for static equilibrium. Assume that the worker stands at the left end of the beam and denote the distance from the bucket to the worker (the left end of the beam) as d.
02

Set the Equations for Static Equilibrium

To satisfy the static equilibrium, we set up two equations. The sum of the vertical forces must equal to zero, and the sum of the moments about any point must be zero. The sum of downward forces (-W1 - W2 - Wbeam) equal to the upward force by the cable (T), where Wbeam is weight of the beam and T is the tension in the cable: \(-W1 - W2 - Wbeam + T = 0\). Then, we select the point where the worker stands to calculate the moment (it could be any point, but this selection will simplify the calculation). We obtain that the clockwise moments (caused by Wbeam and W2) equal to the counterclockwise moments (caused by W1): \(W1 * L/2 = (W2 * d) + (Wbeam * L/2)\).
03

Solve for the unknown

Since we are interested in the position of the bucket (d), we can rearrange our second equation from step 2 to: \(d = \frac{W1 * L/2 - Wbeam * L/2}{W2}\). The problem does not provide the weight of the beam. However, our arrangement of the problem does not require incorporating the beam's weight. Thus, the lack of this information does not prevent us from finding the solution. We can now substitute the given values into the equation to find d.

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Most popular questions from this chapter

You're investigating ladder safety for the Consumer Product Safety Commission. Your test case is a uniform ladder of mass \(m\) leaning against a frictionless vertical wall with which it makes an angle \(\theta .\) The coefficient of static friction at the floor is \(\mu .\) Your job is to find an expression for the maximum mass of a person who can climb to the top of the ladder without its slipping. With that result, you're to show that anyone can climb to the top if \(\mu \geq \tan \theta\) but that no one can if \(\mu<\frac{1}{2} \tan \theta.\)

Pregnant women often assume a posture with their shoulders held far back from their normal position. Why?

A uniform solid cone of height \(h\) and base diameter \(\frac{1}{3} h\) sits on the board of Fig. \(12.27 .\) The coefficient of static friction between the cone and incline is \(0.63 .\) As the slope of the board is increased, will the cone first tip over or first begin sliding? (Hint: Start with an integration to find the center of mass.)

A uniform, solid cube of mass \(m\) and side \(s\) is in stable equilibrium when sitting on a level tabletop. How much energy is required to bring it to an unstable equilibrium where it's resting on its corner?

Give an example of an object on which the net force is zero, but that isn't in static equilibrium.
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