The distance from the center of the breastbone to a man's hand, with the arm outstretched and horizontal to the floor, is \(1.0 \mathrm{m} .\) The man is holding a 10.0 -kg dumbbell, oriented vertically, in his hand, with the arm horizontal. What is the torque due to this weight about a horizontal axis through the breastbone perpendicular to his chest?

Short Answer

Expert verified
Answer: The torque due to the weight of the dumbbell is 98.1 Nm.

Step by step solution

01

Calculate the weight of the dumbbell

Weight = mass × acceleration due to gravity We know that the acceleration due to gravity, \(g \approx 9.81\:\mathrm{m/s^2}\). Weight = \(10.0\: kg \times 9.81\: m/s^2 = 98.1\: N\).
02

Calculate the torque

Torque = force × perpendicular distance In this case, the force is the weight of the dumbbell (98.1 N), and the distance is 1.0 m. Torque = \(98.1\: N \times 1.0\: m = 98.1\: Nm\) The torque due to the weight of the dumbbell about a horizontal axis through the breastbone is 98.1 Nm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 46.4 -N force is applied to the outer edge of a door of width $1.26 \mathrm{m}$ in such a way that it acts (a) perpendicular to the door, (b) at an angle of \(43.0^{\circ}\) with respect to the door surface, (c) so that the line of action of the force passes through the axis of the door hinges. Find the torque for these three cases.
Find the force exerted by the biceps muscle in holding a 1-L milk carton (weight \(9.9 \mathrm{N}\) ) with the forearm parallel to the floor. Assume that the hand is \(35.0 \mathrm{cm}\) from the elbow and that the upper arm is $30.0 \mathrm{cm}$ long. The elbow is bent at a right angle and one tendon of the biceps is attached to the forearm at a position \(5.00 \mathrm{cm}\) from the elbow, while the other tendon is attached at \(30.0 \mathrm{cm}\) from the elbow. The weight of the forearm and empty hand is \(18.0 \mathrm{N}\) and the center of gravity of the forearm is at a distance of \(16.5 \mathrm{cm}\) from the elbow.
A mountain climber is rappelling down a vertical wall. The rope attaches to a buckle strapped to the climber's waist \(15 \mathrm{cm}\) to the right of his center of gravity. If the climber weighs \(770 \mathrm{N},\) find (a) the tension in the rope and (b) the magnitude and direction of the contact force exerted by the wall on the climber's feet.
A child of mass \(40.0 \mathrm{kg}\) is sitting on a horizontal seesaw at a distance of \(2.0 \mathrm{m}\) from the supporting axis. What is the magnitude of the torque about the axis due to the weight of the child?
A 124 -g mass is placed on one pan of a balance, at a point \(25 \mathrm{cm}\) from the support of the balance. What is the magnitude of the torque about the support exerted by the mass?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free