Something simple is tripping me up in a design, and the problem can be reduced to a simple 'seesaw' setup: a rigid beam with pin joint and a weight hanging from each end. I have attached an image with dimensions and values of interest:

Beam and Fulcrum.

The pin joint is not centered in the beam so that one end is further from the joint than the other. The beam has been rotated to an angle, and the length of each side's projection onto the horizontal is known. My question is this: In analyzing this problem I end up showing that the torque induced by each weight is equal but the reaction induced at $F_2$ by $F_1$ does not equal $F_2$, and vice versa. Have I made a mistake in my analysis, and if so where?

Known: $A=30\ deg;\ R_1=10\ inch;\ R_2=5\ inch$

Given: $F_1=10\ lbs;\ F_2=20\ lbs$

Determine the vertical reaction force $F_{1,2}$ induced at $F_2$ by $F_1$:

1) $F_{1,tangent}=F_1\cos(A)$

2) $L_1=R_1/\cos(A)$

3) $\tau_1=F_{1,tangent}L_1=F_1R_1=100\ in.lbs$

4) $L_2=R_2/\cos(A)$

5) $F_{1,2,tangent}=\tau_1/L_2=100\cos(A)/R_2$

6) $F_{1,2}=F_{1,2,tangent}\cos(A)=15\ lbs$

Using this result of $F_{1,2}=15$ and going backwards yields $\tau_2=75\ in.lbs\neq\tau_1$. So it seems like I made an error, but I can't find it.

To look at it another way, wouldn't this result suggest $F_2$ cannot be greater than $15\ lbs$ without breaking equilibrium. Yet when $F_2=20$ the torque induced is $\tau_2=100=\tau_1$.

And so I keep going in circles. I could really use some help! Thank you for your time.


2 Answers 2


We do not need to bother with the angle A, or you would have to consider the projection back and forth of the force F1 into its component orthogonal to the beam too. The reaction of the F1 at the other end of the beam is

$$ 10lbs*10in/5in= 20lbs \ up $$

This reaction is of course being canceled by F2, 20lbs pointing down.

As a rule of thumb vectors such as force can be applied at any point along their trace. So when you want to calculated their torque about a point you multiply the force by its distance from the fulcrum.

  • $\begingroup$ Thank you, that was my intuitive understanding but once I started digging in I started to second guess myself. I see now that I was stuck between the thoughts 'ignore the axial force going through the fulcrum' and 'account for all component forces.' I did not consider the axial force and therefore did not add it's vertical component. Thanks again, really helped me out! $\endgroup$
    – daDib
    Jul 28, 2019 at 18:37
  • $\begingroup$ @daDib, right. happens to all. $\endgroup$
    – kamran
    Jul 28, 2019 at 19:00

As @Kamran stated:

F1xL1=F2xL2 Therefore: F2=[F1XL1] ÷L2

Radial calculations are a mechanical assumption when gravity is acting upon all equally and therefore discounted too. This means discounting angle A also from the equilibrium equation.


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