Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
9
Static Equilibrium; Elasticity and Fracture
Change chapter

9-1 and 9-2: Equilibrium
9-3: Muscles and Joints
9-4: Stability and Balance
9-5: Elasticity; Stress and Strain
9-6: Fracture
9-7: Arches and Domes

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 3
Q

A tower crane (Fig. 9–48a) must always be carefully balanced so that there is no net torque tending to tip it. A particular crane at a building site is about to lift a 2800-kg air-conditioning unit. The crane’s dimensions are shown in Fig. 9–48b.

  1. Where must the crane’s 9500-kg counterweight be placed when the load is lifted from the ground? (The counterweight is usually moved automatically via sensors and motors to precisely compensate for the load.)
  2. Determine the maximum load that can be lifted with this counterweight when it is placed at its full extent. Ignore the mass of the beam.
Problem 3. (a)
Figure 9-48 (a).
Problem 3. (b)
Figure 9-48 (b).
A
  1. 2.3 m2.3\textrm{ m}
  2. 4200 kg4200\textrm{ kg}
Giancoli 7th Edition, Chapter 9, Problem 3 solution video poster
Padlock

In order to watch this solution you need to have a subscription.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. A counterweight which is mass M is positioned a distance l 2 from the tower so we'll take the tower to be the pivot point. So l 2 is the lever arm of the weight of this counterweight. And the weight is m times g— acceleration due to gravity or gravitational field strength— and that's the counter-clockwise torque and the counter-clockwise torque has to equal the clockwise torque; the clockwise torque is due to the load that's being lifted which has weight mg and lever arm of l 1. So we'll substitute in for the torque's here: Mgl 2 for the counter-clockwise torque and mgl 1 for the clockwise torque; the g's cancel and divide both sides by M—mass of the counterweight— and you get the position of the counterweight needs to be at, the lever arm that it needs in order to have same torque as the torque due to the load. So l 2 is—the load mass—2800 kilograms times—its lever arm—7.7 meters divided by 9500 kilograms— mass of the counterweight— which gives about 2.3 meters is the distance required from the tower to the counterweight. And then in this part (b), let's assume that the counterweight is extended to its full possible length which according to our drawing is 3.4 meters and assuming that the load is also positioned at the end of the boom, 7.7 meters, what is the maximum mass that the load could have? So the mass m, we'll have to redivide both sides by l 1 is the counterweight mass times the counterweight's maximum lever arm divided by the load's maximum lever arm. So that's 9500 kilograms times 3.4 meters divided by 7.7 meters which is about 4200 kilograms.

Find us on:

Facebook iconTrustpilot icon
Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.