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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 脳 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 脳 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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