Explanation The free body diagram of the arm is shown below. The arm is in equilibrium so that the sum of the torques must be zero. Write the mathematical expression for rotational analogue of Newton’s second law of motion for the arm. Σ τ = 0 (I) Here, Σ τ is the net torque. Refer to the free body diagram and write the expression for Σ τ . Σ τ = F M d sin θ − m g D − M g L Here, F M is the force of the deltoid muscle required, d is the perpendicular distance from the elbow joint to the point where the force of the deltoid muscle acts, θ is the angle is the angle from the horizontal at which the force of the deltoid muscle acts, m is the mass of the arm, g is the acceleration due to gravity, D is the perpendicular distance from the shoulder joint to the point where the weight of the arm acts, M is the mass in hand and L is the distance between the shoulder joint and the mass. Put the above equation in equation (I) and rewrite it for F M . F M d sin θ − m g D − M g L = 0 F M d sin θ = ( m D + M L ) g F M = m D + M L d sin θ g (II) Conclusion: It is given that the mass of the arm is 3.3 kg , the value of D is 24 cm , the mass that has to be hold in the hand is 8.5 kg , the value of L is 52 cm , the value of d is 12 cm and the value of θ is 15 °

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Chapter 12, Problem 34P

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