A uniform beam of mass m = 0.6 kg and length L = 0.3 m can rotate about an axle

Question

A uniform beam of mass m = 0.6 kg and length L = 0.3 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 1.5 N, F2 = 1.5 N, F3 = 1.5 N and F4 = 1.5 N. F2 acts a distance d = 0.12 m from the center of mass.

a) calculate the magnitude t_1 of the torque due to force F_1 in N*m

b) calculate the magnitude t_2 of the torque due to force F_2 in N*m

c) calculate the magnitude t_3 of the torque due to force F_3 in N*m

d) calculate the magnitude t_4 of the torque due to force F_4 in N*m

e) calculate the angular acceleration a of the beam about its center of the mass in radians/s^2, let the counter clockwise direction be positive

A uniform beam of mass m = 0.6 kg and length L = 0.3 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 1.5 N, F2 = 1.5 N, F3 = 1.5 N and F4 = 1.5 N. F2 acts a distance d = 0.12 m from the center of mass. calculate the magnitude t_1 of the torque due to force F_1 in N*mcalculate the magnitude t_2 of the torque due to force F_2 in N*m calculate the magnitude t_3 of the torque due to force F_3 in N*m calculate the magnitude t_4 of the torque due to force F_4 in N*m calculate the angular acceleration a of the beam about its center of the mass in radians/s^2, let the counter clockwise direction be positive

Explanation / Answer

A)T1=d*F1SIN(90)=0.15*1.5=-0.225 N.m..
B)T2=d*F2*sin(45)=0.127 N.m

C)T3=d*F3 sin(60)=0
D)T4=d*F4* sin(0)= 0 N.m

E)alpha=Tnet/I


here I is moment of inertia=m*l^2/12=0.0045 kg.m^2


alpha =(T1+T2+)/I=-21.77 rad/s^2

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