## What is the Magnitude of the Torque?

**An athlete at the gym holds a 4.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg.**

**1) What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?**

**Known Variables:**

**Mass of Arm: **4 kg

**Mass of Ball:** 4 kg

**arm length: **70 cm

First you have to find the weight of both the ball and the arm. Multiply them by the force of gravity.

**weight of arm: **4kg x 9.8 = 39.2

**weight of ball**: 4kg x 9.8 = 39.2

We are looking for the Torque

Torque, τ = r_{1} F_{1} sin θ +r_{2} F_{2} sin θ

We need to find r1 and r2

**r1=** 70/2= 35 cm or .35m (arm: to find the center of gravity, divide by 2)

**r2**= 70cm or .70m (ball)

**Since the arms are pointed str****aight out, the angle is 180°**

τ=m1g(L/2) + m2gL.

L is the length of the system.

τ =4 * 9.8 *(0.7/2) + 4 * 9.8 * 0.7

= 40 Nm

**What is the magnitude of the torque about his shoulder if he holds his arm straight, but 55° below horizontal?**

**Now the arms are pointed out at an 55° angle.**

you would be using cosine since its the horizontal distance.

**τ =** .35 x 39.2cos(55) + .70 x 39.2cos(55)

**τ = **7.86 + 15.73

**τ = **24 N*m

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