Given the situation in the figure. The mass m1 is 0.68 kg and it is located at x

Question

Given the situation in the figure. The mass m1 is 0.68 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 m. The mass m2 is 0.26 kg and it is located at x2 = 85 cm. Calculate the net torque about the pivot (in Nm with the proper sign) due to these three weights. Use g = 9.79 m/s2. Do not include units with the answer.

Explanation / Answer

here,

m1 = 0.68 kg

m2 = 0.26 kg

mms = 0.4 kg

the net torque about the pivot , T = m1 * g * 0.35 + mms * 9.81 * 0.1 - m2 * g * 0.25

T = 0.68 * 9.81 * 0.35 + 0.4 * 9.81 * 0.1 - 0.26 * 9.81 * 0.25 N.m

T = 2.1 N.m

the net torque about the pivot 2.1 N.m

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.