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 = 720 N. (Do not assume that vector R is parallel to vector T .)

T = 1564.9 N Incorrect: Your answer is incorrect. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

R = 2250 N Incorrect: Your answer is incorrect. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N

? = 21.8 Incorrect: Your answer is incorrect. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.°

All of my answers are wrong can someone please help me!!!

Explanation / Answer

For rotational equilibrium, the sum of the moments about any point must be zero. Let's use the point where the normal force "n" acts.
M = 0 = T*25.0cm - R*cos15º*18.0cm
T = 0.6955*R

For vertical equilibrium, the sum of the vertical forces must be zero:
Fv = 0 = 720N + T*cos - R*cos15º = 720N + T*cos - 0.966*R

and finally, for horizontal equilibrium,
R*sin15º = T*sin substitute for T
0.2588*R = 0.6955*R*sin R cancels
sin = 0.372
= 21.8º

Substitute for T and in the vertical force equation:
0 = 720N + 0.6955*R*cos21.8º - 0.966*R = 720N - 0.320*R
R = 720N / 0.320 = 2250 N
T = 0.6955*R = 1564.875 N

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