When a person stands on tiptoe (a strenuous position), the position of the foot

Question

When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure (a). The total gravitational force on the body, vector F g, is supported by the force vector n exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in Figure (b), where vector T is the force exerted by the Achilles tendon on the foot and vector R is the force exerted by the tibia on the foot. Find the values of vector T , vector R , and ? when vector F g = 655 N. (Do not assume that vector R is parallel to vector T .)

Explanation / Answer

The weight of the person, F g, must equal the magnitude of vector n, because the toes are the only part of the foot that is touching the floor.
n = 655 N
The sum of the vertical forces must equal 0 N. Let upward forces be positive,    

655– R * cos 15 + T * cos = 0
T * cos – R * cos 15 = -655

The sum of the horizontal forces must equal 0 N.
R * sin 15 – T * sin = 0
R * sin 15 = T * sin
T = R * sin 15 ÷ sin
counter clockwise torque = Clockwise torque
The pivot point is at left end, where n is pushing up on the foot.
Counter clockwise torque = T * 0.25, Clockwise torque = R * 0.18
T * 0.25 = R * 0.18
T = R * 0.18/0.25
T = R * 0.72

R * sin 15 ÷ sin = R * 0.72
Divide both sides by R
sin 15 ÷ sin = 0.72
sin = sin 15 ÷ 0.72
= 21.07

T = R * 0.72 and = 21.07
Substitute the values above into the vertical forces’ equation above.
T * cos – R * cos 15 = -840
R * 0.72 cos 21.07 – R * cos 15 = -655
R = -655 ÷ (0.72 cos 21.07– cos 15)
R = 2227.41 N

T = 2227.41 * 0.72 = 1603.7352 N

Check
T * cos – R * cos 15 = -655

1603.735* cos 21.07 – 2227.41 * cos 15 = -655

1496.5123– 2151.5128 = -655

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