2. A uniforn long, thin rod has a mass of M = 20.0 kg and a length of L = 1.50 m

Question

2. A uniforn long, thin rod has a mass of M = 20.0 kg and a length of L = 1.50 m. It is pinned so that it can pivot in the horizontal plane about an axis that is one-third of the way from the end of the rod as shown in the figure below. Three forces, all with a magnitude of F = 75.0 N, act on the rod. Force F1 points at an angle of ? = 30.0° relative to the rod and acts at a distance from the other end of the rod. Forces F2 and F3 act perpendicular to the rod, with F2 acting at the center of the rod and F3 acting at the end of the rod. What is the angular acceleration of the rod? Fi F3

Explanation / Answer

torque = r x F

torque about the pinned point,

torque = -(F L / 3) - ((L/2-L/3)(F)) + (2L/3 F sin30)

= - FL/3 - F L/6 + FL/3

= - FL /6

and torque = I alpha

I = Icm + m d^2 = M L^2 / 12 + M (L/6)^2 = M L^2 / 9

So, - F L / 6 = (M L^2 / 9) alpha

alpha = -3 F / 2 M L

= - 3 x 75 / (2 x 20 x 1.50)

= - 3.75 rad/s^2

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