# When calculating a force, why do I get different result when I try to calculate via torque vs via sum of forces at an axis? If I just do it by writing that the sum on y axis is 0 I get

$$100+x=200$$

$$x=100$$

If I try torque around the top:

$$x*800=200*500$$

$$x=125$$

So what am I doing wrong?

• In addition to what is said below about how this system won't be in static equilibrium, why are you calculating the torque from the top? If the top is fixed then the force becomes negligible Typically, a rod such as this would have it's pivot point be calculated around the center if it's a free body, or possibly around the 200N point if you're balancing something. – Mars Aug 26 '19 at 1:39

Essentially, the problem is poorly conditioned. As drawn, it cannot be in static equilibrium, which is why the equations of equilibrium are returning inconsistent solutions. We can confirm this by summing the moments about the point where force "X" is applied. Let's take counter-clockwise moments as positive.

$$\Sigma M = (-200*300) + (100*800) = 20,000 \neq 0$$

The possibilities are:

1. One of the forces you have drawn is incorrect. For example, if the 100N load were also considered an unknown, we could solve for both reactions and ensure static equilibrium.
2. The forces may be drawn correctly but there are additional reactions (namely, moment reactions) that have not yet been considered. If a moment reaction also existed at the point where force "X" is applied, we could use two equations of equilibrium and solve for both "X" and the moment, again ensuring static equilibrium.
3. The element is not in static equilibrium, and the equations of statics don't apply. Additionally, while it is somewhat unclear from the sketch what the actual support/restraint conditions are, it's worth noting that at least 3 restraints are needed for a structure to be stable. Sometimes one of these support reactions will be zero for a particular load configuration, but in principle, all three are still necessary for stability. If Force "X" is the only support reaction, then the structure is unstable.
• Or it's a multi-part question, and the last part is "is the beam at equilibrium in rotation?" – TimWescott Aug 25 '19 at 16:07

Let's accept the 100N as the top force or reaction and go from there.

in order to maintain equilibrium in this beam there are set of prescribed positions and forces, or pairs of (Y, F)that will work. Any other position with wrong conjugate force will break the balance.

The curve containing all the conjugate pairs of (Y,F) has this equation which is just a statement of $$\Sigma M_{(Y_0) } =0$$

$$Y*F-100N*800=0 \\ F=\frac{80000}{Y}$$

This is of the family of Y=1/X curves which appears in many physical concepts.

So if we plug 300 for y in this equation we get $$F=80000/300=266.6N$$

Which is obviously not what you have in your diagram.

Just by inspection we should be able to tell a force of 200N would work only at Y=400 and would not work at any other position.

The answer is you have more variables than the equations, look at the general form of the problem: We have in total five unknowns here and two equations of equilibrium:

$$y-z-x=0$$ $$-z(d+e)+yd=0$$

The degree of freedom is equal to the difference between the number of variables and the number of equations here the degree of freedom is equal to three, however in your figure you assumed four known values, so at least one of the values in your problem is definitely incorrect.

If the number of equations are more than the number of variables you don't have any freedom and the system of equations are not solvable.

The above answers are correct, but I want to add something more. If you treat the structure as a point mass then the force equilibrium should give you the right answer, but the structure here has a finite dimensions. According to the second law of Newton (true for rigid bodies):

$$\sum_iF_i=Ma_G$$

The sum of all external forces on a rigid body is equal to the mass of the body times the acceleration of its centroid.

The static equilibrium implies that the right hand side should be zero. However because of the finite dimensions of the body, not all the forces going through the centre of the mass so you have net non-zero torque around the instantaneous centre of rotation, so the body can rotates but doesn't translate.

The question seems to be incomplete. You haven't mentioned whether the system is in a static equilibrium or not. If I suppose it were the question then the conditions are incomplete as their net torque is not zero at each and every point. In such a case there will always be an angular velocity and an angular acceleration for the given which will be too much calculation involving when you take the angular acceleration into account.

There is a never a case where you will be getting two different answers for a same thing in classical mechanics.