Question

A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750 N. Ignore the weight of the foot itself. The other significant forces acting on the foot are the tension in the Achilles tendon pulling up and the force of the tibia pushing down on the ankle joint. If the tension in the Achilles tendon is 2225 N, what is the force exerted on the foot by the tibia

Answer 1

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force $N^>$ (750 N)  and the tension in the Achilles tendon $F^>_{Achilles}$ (2225 N) must be equal to the force exerted on the foot by the tibia;

so

| $N^>$ | + |$F^>_{Achilles}$ | = | $F^>_{Tibia}$ |

so force exerted on the foot by the tibia will be;

| $F^>_{Tibia}$ | = |$N^>$ | + |$F^>_{Achilles}$ |

so we substitute IN OUR VALUES

| $F^>_{Tibia}$ | = 750 N + 2225 N

| $F^>_{Tibia}$ |  = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N