Question

6.10 Problems Problem 6.1. A man is performing dumbbell kickbacks to strengthen his triceps (Fig. P.6.1). He weighs 55 kg. His forearm and his hand together constitute 2.5% of his body weight. The length of his forearm is 26 cm. The lever arm of the triceps with respect to the center of rotation of el- bow is 2.4 cm. The weight he carries is equal to 5 kg. Assuming that the man performs the dumbbell kickbacks slowly, determine the triceps force as a function of angle e his forearm makes with the vertical axis. Answer: FT604 sin ON. Problem 6.2. The man performing the dumbbell kickbacks in Problem 6.1 wants to know how the rate of rotation of his forearm affects the ease of the exercise. So, he begins to raise his forearm from vertical to hor- izontal position as quickly as he can, still using the same weight. When his forearm makes 45° with the vertical axis, the angular velocity and angular acceleration of the forearm becomes -8 rad/s and -28 rad/s?, respectively. A minus sign associated with angular velocity refers to clockwise rotation of the right hand. Determine the force produced by the triceps at the elbow at the instant considered. If = 0.021 kg-m? Answer: F = 478 (N).
Please solve problem 2 showing all steps and free body diagrams.

Answer 1

Problem 6.8 www Solution :- m=5 kg, r = 26 cm Ic -0.021 kg.m?, si = 24cm Ther= Ic toner T = rf Sino = rng sino = 9 ( 205 x 55 x 0.26 sin 450)+90540.261 - f x0-044 -Tk s -0.588 - IX = 0.021x(-28) = e = -0.588 = [10•241 - F (0.024) fx 0.024 = (10.02419 +0.588 0.024 F = 45IN 451 N

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