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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure P8.24a. The total gravitational force on the body, F g , is supported by the force n exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in Figure P8.24b, where T is the force exerted by the Achilles tendon on the foot and R is the force exerted by the tibia on the foot. Find the values of T, R, and θ when Fg = n = 700. N.

images

Figure P8.24

To determine
The values of T, R and θ .

Explanation

Given Info: The total gravitational force on the body is Fg , force exerted by the floor on the toes is n, force exerted by the Achilles tendon on the foot is T and force exerted by the tibia on the foot is R.

Explanation:

The following free body diagram shows the force acts on the foot.

From the free body diagram, employing equilibrium condition at the point O (pivot point) in the horizontal position is,

Fx=0

  • Fx is the net force acting on the foot

Write the formula to calculate Fx .

Fx=Rsin15οTsinθ

Rewrite the equilibrium condition by substituting the above relation for Fx .

Rsin15οTsinθ=0

The above relation gives the following result.

Rsin15ο=Tsinθ (1)

  • Fx is the force acts in the horizontal direction.
  • R is the force exerted by the tibia on the foot.
  • T is the force exerted by the Achilles tendon on the foot.

Rewrite the above relation in terms of R.

R=Tsinθsin15° (2)

From the free body diagram, employing equilibrium condition at the point O (pivot point) in the vertical position is,

Fy=0700NRcos15°+Tcosθ=0

Rewrite the above relation.

(700N)+Tcosθ=Rcos15ο (3)

At the point O, the net torque is zero. Therefore,

τO=0(700N)(18.0cm)cosθ+T(25.0cm18.0cm)=0 . (4)

From the above equations,

(1+tan215ο)cos4θ+[(2tan15ο)(0

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