Explanation Given: Let τ be the torque about O acting on the robotic arm. Let m 1 and m 2 , m 3 be the masses ( in Kg ) Let P is the located at the end point To given: r 1 = 1 2 L 1 ( sin θ 1 i + cos θ 1 j ) r 2 = L 1 ( sin θ 1 i + cos θ 1 j ) + 1 2 L 2 ( sin θ 2 i + cos θ 2 j ) r 3 = L 1 ( sin θ 1 i + cos θ 1 j ) + L 2 ( sin θ 2 i + cos θ 2 j ) Where g = 9.8 m/s 2 Calculation: We denote by O , P and Q the points shown in the figure. The co-ordinate of O and Q are 0 = ( 0 , 0 ) , Q = ( L 1 sin θ 1 , L 2 cos θ 1 ) ∴ The midpoint of the segment O Q is, r 1 = O Q = ( 0 + L 1 sin θ 1 2 , 0 + L 1 cos θ 1 2 ) r 1 = O Q = ( L 1 sin θ 1 2 , L 1 cos θ 1 2 ) Since, the mass m 1 is assumed to lie at the midpoint of the arm, the position vectors of m 1 is r 1 = L 1 2 ( sin θ 1 i + cos θ 1 j ) Now, to find the position vector of r 2 of m 2 , we have r 2 = Q M = L 1 sin θ 1 i + L 1 cos θ 1 j r 2 = L 1 ( sin θ 1 i + cos θ 1 j ) The vector Q M make an angle of − ( 90 o − θ 2 ) . With the x-axis and has length L 2 2 hence Q M = L 2 2 ( cos − ( 90 o − θ 2 ) i + sin ( − 90 o − θ 2 ) j ) Q M = L 2 2 ( sin θ 2 i − cos θ 2 j ) Comparing above, we have, ⇒ r 2 = L 1 ( sin θ 1 i + cos θ 1 j ) + L 2 2 ( sin θ 2 i − cos θ 2 j ) Finally, the position vector of r 3 Substitute above answers

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 13.4, Problem 65E

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To prove that τ=−g(L1(12m1+m2+m3)sinθ1+L2(12m2+m3)sinθ2)k

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Chapter 13.4, Problem 64E

Chapter 13.4, Problem 66E

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Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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To prove that τ = − g ( L 1 ( 1 2 m 1 + m 2 + m 3 ) sin θ 1 + L 2 ( 1 2 m 2 + m 3 ) sin θ 2 ) k

Solution Summary: The author demonstrates that tau =-g(l_12msintheta)p is the located at the end point.

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To prove that τ=−g(L1(12m1+m2+m3)sinθ1+L2(12m2+m3)sinθ2)k

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