A 840N athlete standing on one foot on a very smooth gym floor lifts his body by

Question

A 840N athlete standing on one foot on a very smooth gym floor lifts his body by pivoting his foot upward through a 30.0degrees angle, balancing all of his weight on the ball of the foot. The forces on the foot bones from the rest of his body are due to the Achilles tendon and the ankle joint. The Achilles tendon acts perpendicular to a line through the heel and toes; it is this line that has rotated upward through 30.0degrees. Assume that the weight of the foot is negligible compared with that of the rest of the body.

a) What is the magnitude of the force that the floor exerts on the athlete's foot?

b) What is the direction of the force that the floor exerts on the athlete's foot? (upward or downward)

c) What is the tension in the Achilles tendon?

d) Find the horizontal component of the force that the ankle joint exerts on the foot.

e) Find the vertical component of the force that the ankle joint exerts on the foot.

f) Use these components to find the magnitude of this force

Explanation / Answer

a) Normal force on the foot is equal to magnitude of Weight.

N = mg = 840 N

b) direction of force is upward.

c) Taking moment about ankle joint

(Fgcs30) * 0.16 - T x 0.045 = 0

T = 840cos30 x 0.16 / 0.045 = 2586.53 N

d)
Horizontal force in ankle joint ,

Fx = Tsin30 = 2586.53 sin30 = 1293.26 N

e) In verticl ,

Fy = N + Tcos30 = 840 + 2586.53 cos30 = 3080 N

f) Net force = sqrt( Fx^2 + Fy^2 )

         = 3340.50 N

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