Question

When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.30 kg . If the biceps is connected to the forearm a distance dbiceps = 3.00 cm from the elbow, how much force Fbiceps must the biceps exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor? How much force Felbow must the elbow exert?

Answer 1

Answer:

The answers to the questions are;

The force the biceps must exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor is 166.77 N Upwards.

The force the elbow must  exert is 145.6785 N downwards.

Explanation:

We are required to analyze the system using the principle of moments as follows.

Mass of forearm = 1.30 kg

Weight of forearm = mass × acceleration due to gravity = 1.30 kg × 9.81  m/s²= 12.753 N

Location of point of action of weight of forearm = midpoint  17 cm

Forearm to biceps distance = 3.00 cm from the elbow.

Mass of the ball = 850 g

Weight of ball = mass × acceleration due  to gravity = 0.850 kg × 9.81 = 8.3385 N

Position of the ball = end of the forearm

Therefore, taking moment about the elbow, we have

∑Mₓ = 0, we take clockwise moment to be positive, therefore

8.3385 N× 34.0 cm + 12.753 N× 17 cm - $F_{biceps}$×3.00 cm = 0

Therefore $F_{biceps}$×3.00 cm = 500.31 N cm

Therefore $F_{biceps}$ = (500.31 N·cm)/(3.00 cm) = 166.77 N Upwards

The force exerted by the elbow is given by,

Taking moment about the point of contact between the biceps and the forearm, we get

$F_{elbow}$ × 3.00 cm = 8.3385 N× 31.0 cm + 12.753 N× 14 cm

$F_{elbow}$ = 145.6785 N downwards.