- #1
agusb1
- 7
- 1
- Homework Statement
- An 8m bar receives a 60 N upwards force 3 meters far from its pivot point, and a F2 force down at its end (as shown in image). Find force F2 so that the bar is in equilibrium.
- Relevant Equations
- T = F * r * sin(θ) = F * d
60 N + F2 = 0
F2 = -60N
However, in order to have rotational equilibrium:
60 N * 3m + F2 *8 m = 0
60 N * 3m - 60 N * 8
180 - 480 = 0
The equation doesn't make sense. This makes me think that the body can't be in translational and rotational equilibrium at the same time. But I found this exercise in a video of a teacher, and when he solved it there, he did it in a different way. He wrote:
net force = force F1 + force F2 + force of reaction (F1 and F2 were forces acting on the body)
net torque = torque1 + torque2 + torque of the reaction forceAnd finds that F2 is 22,5 N.
torque1 + torque2 + torque of the reaction force = 0
60 * 3 - F2 * 8 + 0 =
180 = F2*8
180/8 = F2
22.5 N = F2
Then, force of reaction= -37.5 (60-22.5)
force F1 + force F2 + force of reaction = 0
60 N - 22.5 N - 37.5 N = 0
In that way it does work. But I don't understand why he used the reaction force. I read that translational equilibrium happened only when the net force ACTING ON a body is zero. But the reaction force is not a force acting on the body, it's a force being exerted by the body. In the same way, rotational equilibrium is when the net torque acting on the body is zero. So why is the reaction force relevant? Is the person in the video wrong, or should I use the reaction force to calculate the net force and net torque?