1. The center of mass of a 0.2 kg (non-uniform) meter stick is located at its 46

Question

1. The center of mass of a 0.2 kg (non-uniform) meter stick is located at its 46-cm mark. What is the magnitude of the torque (in Nm) due to gravity if it is supported at the 26-cm mark? (Use g = 9.79 m/s2).

2.

Suppose that a meter stick is balanced at its center. A 0.14 kg mass is then positioned at the 6-cm mark. At what cm mark must a 0.20 kg mass be placed to balance the 0.14 kg mass?

3.

Given the situation in the figure. The mass m1 is 0.53 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 0.24 kg and it is located at x2 = 85 cm. Calculate the net torque (in N*m with the proper sign) due to these three weights. Use g = 9.79 m/s2.

4. The apparatus described in the previous question is in equilibrium.

True or False?

Explanation / Answer

1) torque = mg*x = 0.2*9.79*(46-26)/100 =0.3916 Nm

2) 0.14*(50-6) + 0.2*(50-46) = 0.2*(x-50)

=> x = 84.8 cm

3) net torque = [0.53*(60-25) + 0.4*(60-50) - 0.24*(85-60)]*9.8/100 = 1.620245 Nm (taking anti clockwise to be positive)

4) FALSE

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

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