What is the net torque about an axis through point A? What is the net torque abo

Question

What is the net torque about an axis through point A? What is the net torque about an axis through point C? A long, uniform beam with mass m and length L is attached by means of a pivot, located at L/4, to a vertical support as shown above. The beam is connected to a support line oriented at an angle 0 relative to the horizontal. The beam is in equilibrium in the horizontal position when a weight of mass 2m is placed at the left end of the beam and a mass of 3m is placed at the right end of the beam. Calculate the tension Tin the support line. Calculate the x and y components of force acting on the pivot point. Given the moment of inertia of the beam is 7/48 mL^2. Calculate the angular acceleration of the beam if the support line is now cut.

Explanation / Answer

1) Net Torque about pont A, TA = 30*1.5*sin(45) - 10*3*sin(30)

= 16.82 N.m

2) Net Torque about pont C, TC = 20*1.5*sin(30) - 10*1.5*sin(30)

= 7.5 N.m

3)

a) As the beam is in equilibrium, net force ant net torque actong on the beam must be zero.

Apply net torque about point P = 0

T*(3*L/4)*sin(theta) +2*m*g*(L/4) - m*g*(L/4) - 3*m*g*(3*L/4) = 0

T*(3/4)*sin(theta) +2*m*g*(1/4) - m*g*(1/4) - 3*m*g*(3/4) = 0

T*(3/4)*sin(theta) = 2*m*g

T = 8*m*g/(3*sin(theta)) <<<<-----Answer

b)
Fx = T*cos(theta)

= 8*m*g*cos(theta)/(3*sin(theta))

= 8*m*g*cos(theta)/3 <<<<-----Answer

b)

Fy = 6*m*g - T*sin(theta)

= 6*m*g - 8*m*g/3

= 3.333*m*g <<<<-----Answer

c) Net to rque acting abou pivot = m*g*(L/4) + 3*m*g*(3*L/4) - 2*m*g*(L/4)

= 2*m*g*L

so, algular acceleration, alfa = T/I

= 2*m*g*L/(7*m*L^2/48)

= 96*m*g*L/(7*m*L)

= 96*g/7

= 96*9.8/7

= 134.4 rad/s^2

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