# A body builder at the gym is lifting a 50kg dumbell. His arm measures 47cm from hand to elbow and 39cm from elbow to shoulder. If he does a front shoulder raise so that his shoulder is in the center of the circle, how much torque will it require?

Nov 23, 2015

Torque = 421 Nm

#### Explanation:

Maximum torque will occur when the arm is horizontal because at that point the line of action of the weight will be perpendicular to the arm. That means that the perpendicular distance between shoulder and weight will be maximum.

First we need to know the weight of the dumbbell:
w=mg = 50× 9.8 = 490 N

Now the total length of the arm is needed, as this is the perpendicular distance between weight and shoulder:
$L = 39 + 47 = 86 c m = 0.86 m$

The torque required at the shoulder is equal to the weight multiplied by the perpendicular length:
T = 490 × 0.86 = 421 Nm