Torque: find the force necessary for a body to be in equilibrium

[agusb1]

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

The question doesn't specify whether we're talking about translation or rotational equilibrium, so I suppose it's both: In order for the body to have translational equilibrium:
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?

 

[scottdave]

The picture looks like a hinge. Think about what reaction forces the wall can apply to the bar through the hinge. Any lateral force? Up or down? What about torque. Also, I'm guessing the bar is massless?

 

[kuruman]

60 N + F2 = 0
F2 = -60N

This is not correct. There is a vertical reaction force at the hinge that you did not take into account. Not doing so gave you a false value for $F_2$. Calculate the sum of torques about the hinge to find the correct value for $F_2$.

Also, I'm guessing the bar is massless?

That's my hunch, too.

 

[PhanthomJay]

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.

do you happen to recall Newton’s 3rd Law??

 

[agusb1]

do you happen to recall Newton’s 3rd Law??

The way I thought about it is: if it's a reaction force, it means it's reacting to an original force ("action") exerted by body one. The "action" force is the one that is applied by body one to body two. Then body two generates a reaction force that acts on body one. This is what I mean with "it's a force being exerted by the body" I interpreted the bar as being body two.

I'm guessing I was just misinterpreting the phrase "reaction force". I also thought that they meant the reaction of the bar after being applied those two forces. But now I think I understand it: there's another force ("reaction force") acting on the bar, because there is an object holding the bar by giving it a force near the pivot point, right?

 

[kuruman]

The way I thought about it is: if it's a reaction force, it means it's reacting to an original force ("action") exerted by body one. The "action" force is the one that is applied by body one to body two. Then body two generates a reaction force that acts on body one. This is what I mean with "it's a force being exerted by the body" I interpreted the bar as being body two.

I'm guessing I was just misinterpreting the phrase "reaction force". I also thought that they meant the reaction of the bar after being applied those two forces. But now I think I understand it: there's another force ("reaction force") acting on the bar, because there is an object holding the bar by giving it a force near the pivot point, right?

Right. If the hinge did not exert a force on the bar, then the hinge might as well not be there because it's not doing anything. However you know what would happen if the hinge were removed: the bar would become unhinged.

 

[PhanthomJay]

The term “reaction force” is in common use when referring to the force exerted on the beam or system by the supports. It is really technically not a reaction force , which is why it probably confused you. When analyzing a beam or other system in equilibrium, the applied loads, the gravity loads(weight) , and the loads at the supports, are all external to the system, acting on the system, and must all be included in applying the equilibrium equations (and that applies to systems not in equilibrium also).