Okay so I've been doing a math assignment on torque which has to have real-world applications so I've gone off on a massive tangent trying to incorporate physics principles and trying to get them as correct as possible (even though whether the actual physics is correct or not is irrelevant)... But now I've become so obsessed with the physics that I am just really curious to understand how it all really works.

In the image below, what you can see is a snippet of a diagram I've been working on. The mass has been lifted slightly (on some angle, theta) by a sign post due to a force being applied (out of view) at the top of the post (net torque in anticlockwise direction) and it's pivoting off an adjacent object's corner.

The mass being lifted is, obviously, also exerting some kind of force. What I'm trying to figure out is how big of a force. I know that to cause torque, I must find the component acting orthogonal to the post and all the like - but what I'm first trying to understand is this: How is the weight of the mass distributed? At first I saw there were two contact points so, assuming uniform mass, I thought it would probably be accurate enough to say that the weight is distributed evenly over the two contact points. And of course, this would probably be true if the mass was horizontal with two contact points at equal distances from the centre of mass. But what I'm worried about is the angle, theta, to which the mass is being lifted. Surely it has some effect on the centre of mass and thus weight distribution. It looks as though now the top right corner of the mass would be exerting some kind of force as well (perhaps I could conceptualise this situation in terms of torque? i.e. with right bottom corner of the mass as pivot point... not sure how I would go about it exactly though. I've just seen it done before with simple situations where torque is used theoretically to find unknown forces)

Anyways if anyone could shed some light on how to calculate the force exerted by the lower left corner of the mass on the post in terms of M and theta, that would be amazing, thank you!

enter image description here

  • $\begingroup$ Is the box being lifted slowly and is in static equilibrium the entire time, or is it thrown up fast and its vertical acceleration needs to be accounted for? $\endgroup$ Apr 12, 2016 at 20:07
  • $\begingroup$ For the purpose of this question lets say the system has achieved equilibrium and the clockwise torque from the mass is cancelling the anticlockwise torque from the other applied force. $\endgroup$ Apr 12, 2016 at 20:12
  • $\begingroup$ You know gravity always acts through the center of mass. That is how you find the distribution. See my answer. $\endgroup$ Apr 12, 2016 at 20:20

1 Answer 1


Here is the beginning of a free body diagram.


Each contact force $A$ and $B_y$ acts on the contact normal direction. For $B_y$ this is vertical and for $A$ this is perpendicular to the lever. The weight $W=m g$ acts on center of mass. Now you have to add friction to the contacts as shown with force $B_x$. Here friction is shown with a positive value acting to the right, although we expect for the actual friction to oppose $A$ somewhat and so the result should be a negative value for $B_x$. It doesn't matter as the equations end up being the same.

You find the solution as if there is infinite friction available by doing sum of forces and sum of moments equals zero. The point you add up the torques does not matter. You have three equations (two forces + one torque) and three unknowns, $A$, $B_x$ and $B_y$. The details are just trigonometry in order to get the force directions and moment arms correctly.

If you have a finite friction coefficient $\mu$ then you cap $B_x$ such that

$$| B_x | \leq \mu | B_y | $$

The result would be a horizontal acceleration since the forces would not add up to zero anymore.


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