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Chapter 8: Problem 85

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.

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Most popular questions from this chapter

A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?
A solid sphere of mass 0.600 kg rolls without slipping along a horizontal surface with a transnational speed of \(5.00 \mathrm{m} / \mathrm{s} .\) It comes to an incline that makes an angle of \(30^{\circ}\) with the horizontal surface. Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline?
The string in a yo-yo is wound around an axle of radius \(0.500 \mathrm{cm} .\) The yo-yo has both rotational and translational motion, like a rolling object, and has mass \(0.200 \mathrm{kg}\) and outer radius \(2.00 \mathrm{cm} .\) Starting from rest, it rotates and falls a distance of \(1.00 \mathrm{m}\) (the length of the string). Assume for simplicity that the yo-yo is a uniform circular disk and that the string is thin compared to the radius of the axle. (a) What is the speed of the yo-yo when it reaches the distance of $1.00 \mathrm{m} ?$ (b) How long does it take to fall? [Hint: The transitional and rotational kinetic energies are related, but the yo-yo is not rolling on its outer radius.]
A person is trying to lift a ladder of mass \(15 \mathrm{kg}\) and length $8.0 \mathrm{m} .$ The person is exerting a vertical force on the ladder at a point of contact \(2.0 \mathrm{m}\) from the center of gravity. The opposite end of the ladder rests on the floor. (a) When the ladder makes an angle of \(60.0^{\circ}\) with the floor, what is this vertical force? (b) A person tries to help by lifting the ladder at the point of contact with the floor. Does this help the person trying to lift the ladder? Explain.
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.
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