# Net torque of a person's shoulder, holding a weight

1. Jun 8, 2009

### kathyt.25

1. The problem statement, all variables and given/known data
"An athlete at the gym holds a 5.86 kg steel ball in his hand. His arm is 66.2 cm long and has a mass of 5.56 kg. What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?"

2. Relevant equations
torque = Frsin(theta)

3. The attempt at a solution
I drew the diagram, having the shoulder as the pivot point. There are three forces acting on the arm -
(1) the normal (90 deg to the pivot, pointing UPWARDS)
(2) weight of the arm (90 deg to the arm, pointing DOWN)
(3) weight of steel ball (90 deg to the arm, pointing DOWN)

Since they want to calculate the torque on the shoulder, the normal doesn't produce a torque since it's acting from the pivot point. Where "t" is torque, the equation is:
t(net) = -t(arm) - t(ball)
I made the torques of the arm and ball negative because they would rotate clockwise around the pivot point.

t(net) = -W(arm)*r - W(ball)*r
= -m(arm)g*r - m(ball)g*r
= -(5.56)(9.8)(0.331) - (5.86)(9.8)(0.662)
= -38.0 Nm

I got the value for the distance of the weight of the arm because the centre of mass always acts at the centre of the object... so I divided the full arm's length by two, to get that value of r=0.331m

I'm still getting the wrong answer, and I'm not sure why... I always have difficulty with torque directions. Did I do sometihng wrong?

Last edited: Jun 9, 2009
2. Jun 9, 2009

### djeitnstine

So what is the problem?

3. Jun 9, 2009

### Coto

Try crunching that last calculation one more time. I think you will find your problem. Your thought process is correct.