A gymnast weighing 750 N executes the iron cross. The ring exerts an upward force on...
A gymnast weighing 750 N executes the iron cross. The ring exerts an upward force on the arm of 375 N. The force exerted by the shoulder joint is R, while the muscles exert a total force of M on the arm. M acts at an angle of 45 degrees below the horizontal at a distance of 4 cm from the shoulder joint. The distance from the shoulder joint to ring is 70 cm. Ignore the weight of the arm. Note that one ring supports half the weight of the body.
a) calculate the magnitude of force M
b) calculate the magnitude and direction of the reaction force R
Homework Answers
Using the shoulder joint as the point of rotation. Then...
.
torque due to force from ring - torque due to Fm = 0
.
375 * 70 - Fm* 4 * sin45 = 0
.
Fm = 9279.4 Newtons
Fr cos(θ) = 9279.4*sin45 = 6562.5
Fr sin(θ) +375 = 9279.4*cos45 = 6562.5
Fr sin(θ) = 6187.5
θ = direction of R = 43.315 degrees
Fr = 9019.5 N
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