Please help with all five parts A uniform thin rod of mass m = 0.55 kg and lengt

Question

Please help with all five parts A uniform thin rod of mass m = 0.55 kg and length L = 1.7 m can rotate about an axle through is center. Four forces are acting on it as shown in the figure. Their magnitudes are F_1 = 8.5 N, F_2 = 3.5 N, F_3 = 1.5 N and F_4 = 17.5 N. F_2 acts a distance d = 0.11 m from the center of mass. Randomized Variables m = 0.55 kg L = 1.7 m F_1 = 8.5 N F_2 = 3.5 N F_3 = 1.5 N F_4 = 17.5 N d = 0.11 m (a) Calculate the magnitude r_1 of the torque due to force F_1 in N.m. Part (b) Calculate the magnitude r_2 of the torque due to force F_2 in N.m. Part (c) Calculate the magnitude r_3 of the torque due to force F_3 in N.m. Part (d) Calculate the magnitude r_3 of the torque due to force F_3 in N.m. Part (e) Calculate the angular acceleration a of the thin rod about its center of mass in rod/s^3. Let the counter-clockwise direction be positive.

Explanation / Answer

(a)
Torque, t = F*r
t1 = F1 * L/2
t1 = 8.5 * 1.7/2
t1 = 7.23 Nm

(b)
t2 = F2 * sin(45) (L/2 - d)
t2 = 3.5 *sin(45) * (0.11)
t2 = 0.272 Nm

(c)
t3 = F3 * sin(60) * 0
t3 = 0

(d)
t4 = F4 * sin(0) * L/2
t4 = 0

(e)
Moment of Inertia of rod, I = 1/12 * M*L^2
Torque = I *
- 7.23 + 0.272 = 1/12 * 0.55 * 1.7^2 *
= - 52.5 rad/s^2

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