I'm working on an answered physics question in the book but I'm having trouble understanding why the problem could be solved by setting net torque equal to 0 and not net force. diagram 1 You are asked to find the magnitude of the tension in the cable (the magnitude of the force from the beam on the cable). With an axis of rotation at the hinge so as to eliminate any forces at that point, the solution is completed by balancing torques: image 2 I understand why that works, but I'm not sure why you can't instead balance forces as was done in previous problems. In other words:

(Fh=force of the hinge on the beam; Ft=tension force in the cable; m=mass of the beam; M=mass of the block)

x-horizontal, y-vertical



these equations produce: Fhcos(theta)=Ft & Fhsin(theta)=g(m+M)

If you divide the equations, you can eliminate Fh and solve for Ft... why is this not appropriate?

  • 1
    $\begingroup$ I can't actually view the problem statement and solution, so I can't provide a concrete answer, but when you are balancing a net force, you are making sure that the CoM of the object does not accelerate. When you are balancing net torque, you are making sure that the object does not rotate. Just because an object's CoM stays stationary does not mean it doesn't rotate and vice versa. In general, you would have to maintain both net force = 0 and net torque = 0 to specify static equilibrium. However, most problems are set up so that one of those equalities is trivially true. $\endgroup$
    – enumaris
    Commented Apr 2, 2018 at 16:31
  • $\begingroup$ Does doing that get you a different solution? It seems like you're still using the fact that net torque = 0 to solve the problem; just you're solving for a different variable. I'm actually not even sure if they are solving for "torque" in this question. It looks a lot like their "$T_C$" is actually a force as well. $\endgroup$
    – JMac
    Commented Apr 2, 2018 at 17:47
  • $\begingroup$ @enumaris I understand the mechanics of using torques and forces, but yes, I'm getting a different solution when I use forces as I laid out in my question. I think I get about 6400N when the answer is 6100N. $\endgroup$
    – khajiit
    Commented Apr 2, 2018 at 18:38
  • $\begingroup$ @JMac I'm getting a different solution as I explain in my comment above. I'm not using torque at all--I'm simply using the fact that Fnet=0. $\endgroup$
    – khajiit
    Commented Apr 2, 2018 at 18:39
  • $\begingroup$ I'm having trouble understanding where you got your equations from. For example, what is Ft? I'd suggest using Mathjax formatting so that it is more clear what each variable is supposed to be. Things like Ft and Fhy are hard to differentiate the variables and subscripts. $\endgroup$
    – JMac
    Commented Apr 2, 2018 at 18:47

1 Answer 1


Balancing forces is useful, but it is not enough. There aren't enough constraints on the problem to find all the forces just by setting the net force to zero. In your diagram, you could add any value to $\vec{F_h}$. As long as you add the same to $\vec{T}_c$, the net force doesn't change. So find some sum of forces that comes to zero, then add 1 Newton to both horizontal forces on the problem, and you have a new zero-net-force solution.

On the other hand, as long as the net force and net torque are both zero on a stationary rigid object, it will stay stationary. We need both because net force only tells us about the acceleration of the center of mass. If a beam starts spinning around its center, it has zero net force on it (because the center of mass isn't moving), but there was some net torque that got it spinning. So setting $F_{net} = 0$ is not enough.

As for what happened in your work, it's hard to tell. First, try to stick with well-defined notation. This will make it much easier for people to understand what you mean. Your question defines the variables $\vec{T}_c$, $\vec{F}_v$, and $\vec{F}_h$, which you didn't use. Your proposed solution uses the undefined variables "Fhx" "Fhy" and "Ft". Then you write about "theta" without defining it, either.

My best guess is that you assumed that the net force from the wall on the beam acts along that beam. That is, in the notation of the problem, you assumed

$$\frac{\vec{F}_v}{\vec{F}_h} = \frac{a}{b}$$

but there is no reason to assume this. That sort of assumption works for strings or ropes, which can support any shear, but for beams, the wall can exert forces in the horizontal and vertical directions independently. You won't know how big those forces are unless you impose both the condition that the net force on the beam is zero, and the condition that the net torque on the beam is zero.

  • $\begingroup$ Okay, thanks. I think it was the last part that cleared it up for me. I'm just not entirely seeing the difference between the tension force in a rope, which can be broken down into its horizontal and vertical components using the angle of incline, and a beam, which can't... $\endgroup$
    – khajiit
    Commented Apr 2, 2018 at 19:29
  • 2
    $\begingroup$ Both can be broken down into components. However, in the beam the force is not guaranteed to be in the same direction as the beam. Imagine a water bottle sitting on your desk as the beam. Put a small sideways force on it with your finger. It doesn't go anywhere, but the net force from the table is not vertical any more; it's partially vertical (from the normal force from the table) and partially horizontal (from the friction force from the table), and we can't assume that the force is along the axis of the water bottle. $\endgroup$ Commented Apr 2, 2018 at 19:33
  • $\begingroup$ For a string hanging from the ceiling suspending the water bottle, it's different. If you apply a horizontal force to hanging water bottle, it will move and be at an angle. The net force the string can exert is always long its own length. This is just basically what we mean by a "string" or "rope" in elementary physics problems. $\endgroup$ Commented Apr 2, 2018 at 19:34
  • $\begingroup$ @MarkEichenlaub Sometimes in statics problems, they may also define that beam as a two force member. In that case, the force also would have to be applied at the same angle as the beam. That may have been OP's confusion, especially if they have recently learned about two force members. $\endgroup$
    – JMac
    Commented Apr 3, 2018 at 13:30
  • $\begingroup$ @MarkEichenlaub The distinction between a string/rope and a beam definitely cleared it up for me--thank you. $\endgroup$
    – khajiit
    Commented Apr 3, 2018 at 17:16

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