- #1
klandestine
- 16
- 0
This problem is regarding the attached jpeg.
I am supposed to find the tension in the cable. I know that since the setup is in equilibrium, all net forces and torques equal zero. So,
Since net F(x) = 0, Tcos@ = F
and net F(y) = 0, Tsin@ = g(m1+m2)
@=theta
F=the normal force exerted by the wall
Also, the center of mass is (m1(L/2)+m2(d))/( m1+m2)
I know that the net torque is zero, otherwise the beam would be rotating. If I chose the cm as the point around which i calculate this torque, do I count the weight of the beam and the man? I want to say no, because if this is pointing straight down on the axis, it won't cause rotation. But I don't have a really good grasp of torque.
I am supposed to find the tension in the cable. I know that since the setup is in equilibrium, all net forces and torques equal zero. So,
Since net F(x) = 0, Tcos@ = F
and net F(y) = 0, Tsin@ = g(m1+m2)
@=theta
F=the normal force exerted by the wall
Also, the center of mass is (m1(L/2)+m2(d))/( m1+m2)
I know that the net torque is zero, otherwise the beam would be rotating. If I chose the cm as the point around which i calculate this torque, do I count the weight of the beam and the man? I want to say no, because if this is pointing straight down on the axis, it won't cause rotation. But I don't have a really good grasp of torque.