An athlete at the gym holds a 1.2 kg steel ball in his hand. His arm is 64 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?
1. straight but 30 degrees below horizonal?
Torque = force*length*sinTheta
= 1.2*9.8*.64*sin60
the angle theta is the angle between the force and the arm.
To calculate the magnitude of the torque about the shoulder, we need to use the formula:
Torque = force × perpendicular distance
First, we need to find the force exerted by the steel ball. The force can be calculated using Newton's second law:
Force = mass × acceleration
Since the athlete is holding the steel ball, the acceleration in this case is the acceleration due to gravity, which is approximately 9.8 m/s^2. Therefore, the force can be found by:
Force = mass × acceleration
= 1.2 kg × 9.8 m/s^2
= 11.76 N
Next, we need to find the perpendicular distance from the shoulder to the line of action of the force. In this case, the distance is the length of the athlete's arm. Given that the arm length is 64 cm, we need to convert it to meters:
Distance = 64 cm = 64/100 m = 0.64 m
Now that we have both the force and the perpendicular distance, we can calculate the magnitude of the torque using the formula mentioned earlier:
Torque = Force × Distance
= 11.76 N × 0.64 m
= 7.5264 Nm
Therefore, the magnitude of the torque about the shoulder when the athlete holds his arm straight but 30 degrees below horizontal is approximately 7.53 Nm.