A uniform thin rod of mass m = 3.3 kg and length L = 1.1 m can rotate about an a

Question

A uniform thin rod of mass m = 3.3 kg and length L = 1.1 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F_1 = 8.5 N. F_2 = 2.5 N. F_3 = 11 N and F_4 = 18.5 N. F_2 acts a distance d = 0.13 m from the center of mass. Randomized Variables m = 3.3 kg L = 1.1 m F_1 = 8.5N F_2 = 2.5 N F_3 = 11N F_4 = 18.5 N d= 0.13 m Calculate the magnitude tau_1 of the torque due to force F_1 in N-m. Calculate the magnitude tau_2 of the torque due to force F_2 ui N-m. Calculate the magnitude tau_3 of the torque due to force F_3 in N-m. Calculate the magnitude tau_4 of the torque due to force F_4 in N-m. Calculate the angular acceleration alpha of the thin rod about its center of mass in rad/s^2. Let the counter-clockwise direction be positive

Explanation / Answer

t1 =F1*(L/2)= 8.5*(1.1/2) = 4.675 N.m

t2 =d*F2*sin45= 0.13*2.5*sin45 = 0.21 N.m

t3 = 0

t4 = 0

Tnet = 4.885

tnet = I*alpha

4.885 = (1/12*3.13*1.12)*alpha = 15.48 clockwise

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