(10%) Problem 2: A uniform thin rod of mass m = 3.1 kg and length L = 1.8 m can

Question

(10%) Problem 2: A uniform thin rod of mass m = 3.1 kg and length L = 1.8 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 3.5 N, F2 = 3.5 N, F3 = 13 N and F4 = 15.5 N. F2 acts a distance d = 0.13 m from the center of mass.

Randomized Variablesm = 3.1 kg
L = 1.8 m
F1 = 3.5 N
F2 = 3.5 N
F3 = 13 N
F4 = 15.5 N
d = 0.13 m 20% Part (a) Calculate the magnitude 1 of the torque due to force F1 in Nm.
20% Part (b) Calculate the magnitude 2 of the torque due to force F2 in Nm.
20% Part (c) Calculate the magnitude 3 of the torque due to force F3 in Nm.
20% Part (d) Calculate the magnitude 4 of the torque due to force F4 in Nm. 20% Part (e) Calculate the angular acceleration of the thin rod about its center of mass in rad/s2. Let the counter-clockwise direction be positive.

Explanation / Answer

A) 1 = r*F1*sin(90) = r*F1 = (1.8/2)*3.5 = 3.15 N-m clock wise

So 1 = -3.15 N-m

B) 2 = d*F2*sin(45) = 0.13*3.5*sin(45) = 0.321 N-m

C) 3 = 0 N-m

D) 4 = 0 N-m

E) net torque is Tnet = I*alpha

alpha = Tnet /I = (0.321-3.15)/[(1/12)*M*L^2)]

alpha = -2.829/0.837 = -3.37 rad/s^2

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