man with a mass of 70 kg balances on one foot while standing on a curb. The only

a. 1715 N b. 2401 N

Explanation:

Let F = force due to calf muscle, F' = force due to tibia and N = force due to ground = weight of man = mg where m = mass of man = 70 kg and g = acceleration due to gravity = 9.8 m/s².

a. Magnitude of the forces exerted on the foot by the muscle

Since the force due to the calf muscle is 6.0 cm behind the ankle joint and the normal force due to the ground is 15.0 cm in front of the ankle joint and the force due to the tibia is at the ankle joint, taking moments about the ankle joint,

F × 6 cm + F' × 0 cm = N × 15 cm

6F = 15N = 15mg

F = 15mg/6

= 15 × 70 kg × 9.8 m/s²/6

= 1715 N  

b. Magnitude of the forces exerted on the foot by the tibia

Taking moments about the calf muscle force, we have

F × 0 cm + F' × 6 cm = N × (15 cm + 6 cm)

6F' = 21N = 21mg

F' = 21mg/6

= 21 × 70 kg × 9.8 m/s²/6

= 2401 N