HARD PHYSICS PROBLEM HELP PLEASE!!?
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The figure (a) shows an outstretched arm with a mass of 4.1 kg. The arm is 56 cm long, and its center of gravity is 21 cm from the shoulder. The hand at the end of the arm holds a 6.0-kg mass.
1. What is the torque about the shoulder due to the weights of the arm and the 6.0-kg mass?
τ_0 = 40 N * m
2. If the arm is held in equilibrium by the deltoid muscle, whose force on the arm acts 5.0 degrees below the horizontal at a point 18 cm from the shoulder joint (the figure (b)), what is the force exerted by the muscle?
F = ??? kN
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Verified answer
1. τ = 4.1kg * 9.8m/s² * 0.21m * cos15º + 6kg * 9.8m/s² * 0.56m * cos15º
τ = 40 N·m ← agreed
2. The angle between the perpendicular to the arm and the deltoid is 80º (= 90º - (15º - 5º)).
F*cos80º * 0.18m = 40 N·m
F = 1278 N
utilising v = u + at the place u is preliminary velocity and v very final velocity all of us comprehend that for the period of the two situations v = 0 when you consider that the two automobiles come to sit down down back. preliminary velocity u: u + (-4.2)t1 = 0 or u = 4.2t1 preliminary velocity 2u : utilising same theory 2u = 4.2t2 Dividing 2d equation by using first 2u / u = 4.2t2 / 4.2t1 t2 / t1 = 2 t2 = 2 t1 This shows the 2d vehicle takes two times the time to end. u1 = sixteen m/s so t1 = sixteen / 4.2 = 3.80 one s u2 = 32 m/s so t2 = 32 / 4.2 = 7.sixty two s