Question

The biceps muscle is attached to the radial bone 3.6 centimeters from the elbow joint, the forearm and hand have a combined mass of 2.7 kilograms and a combined length of 36 centimeters, and the ball has a mass of 0.146 kilograms a. What force must be exerted by the biceps muscle to keep the arm in rotational equilibrium? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

Given,

The distance from the elbow joint to the point where the muscle is attached to the radial bone, d=3.6 cm=0.036 m

The mass of the forearm and the hand, M=2.7 kg

The combined length of the forearm and the hand, L=36 cm=0.36 cm

The mass of the ball. m=0.146 kg

From the diagram, the centre of mass of the forearm and the hand is at a distance of L/2.

For the arm to be in the rotational equilibrium, the net torque on the arm must be equal to zero.

That is,

`F_{\text{bicep}}* d-Mg*(L)/(2)-mg* L=0`

Where g is the acceleration due to gravity and F_bicep is the force applied by the bicep muscle.

On rearranging the above equation,

`F_{\text{bicep}}=(Mg*(L)/(2)+mg* L)/(d)`

On substituting the known values,

`\begin{gathered} F_{\text{bicep}}=(2.7*9.8*(0.36)/(2)+0.146*9.8*0.36)/(0.036) \\ =(4.76+0.52)/(0.036) \\ =146.67\text{ N} \end{gathered}`

Thus the force that the bicep muscle must exert in order to keep the arm in rotational equilibrium is 146.67 N