In the picture: m 1 is at the 25 cm mark, m ms is at the 50 cm mark, the fulcrum

Question

In the picture: m1 is at the 25 cm mark, mms is at the 50 cm mark, the fulcrum is at 60 cm, and m2 is at the 85 cm mark. The stick is a total of 100 cm.

Given the situation in the figure. The mass m1 is 0.50 kg and it is located at x1 = 25 cm. The pivot point is represented by the solid triangle located at x = 60 cm. The mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.31 kg and it is located at x2 = 85 cm. Calculate the net torque (in N%u22C5m with the proper sign) due to these three weights. Use g = 9.79 m/s2.

Explanation / Answer

Net torque T = m1.g.(60-25)*10^-2 + Mms.g.(60-50).*10^-2-M2*g.(85-60)*10^-2

T = 0.5*(9.79)*(35) + 0.40.(9.79)(10) - 0.31(9.79)*(75)}10^-2

T=0.1375 N.m with direction counterclock wise

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