You are holding a 5-kg dumbbell straight out at arm’s length. Assuming that your arm is 0.70 m long, and that your shoulder acts as a pivot, what torque is the dumbbell exerting?

34.3 N.

Explanation:

Torque: This is a force that tend to cause rotation or twisting in a system. The S.I unit of torque is Nm.

From the question,

Torque = Force×distance from the point of pivot

T = F×R equation 1

Where GT = torque, F = force or weight of the dumbbell, R = length of the arm

But,

F = mg Equation 2

Where m = mass of the dumbbell, g = acceleration due to gravity.

Substitute equation 2 into equation 1

T = mg×R Equation 3

Given: m = 5 kg, g = 9.8 m/s², R = 0.70 m.

Substitute into equation 3

T = 5(9.8)(0.7)

T = 34.3 N.

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

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