2
$\begingroup$

Suppose you have a shape consisting of two perpendicular rods (the whole shape is a rigid body) which stands upright so the plane of the rods is perpendicular to the plane of the ground, and the hypotenuse of the resulting triangle is on the ground. Each leg is a massive rod, so the forces acting on this right triangle are the normal force at each vertex (pointing upwards) and the weight of each leg (pointing downwards). Also assume the legs are rods of uniform mass, so the weight acts on the center of gravity of each leg. We wish to find the normal force on each leg. Considering net force, we can find the sum of the normal forces as the sum of the weights of each leg. We can find more information by taking torques about the point of contact between the two rods.

My question is how do we take the torque about this point? Can I simply choose one of the rods as a lever arm and say that the torques along this level arm must cancel, or must find the torques from each rod and then say those torques cancel? If the first statement is true then so is the second, but the first statement is stronger than the second so it may not be true.

Also, how would things change if the the rods were not glued together so as to form a rigid body - for example, how would it change if the rods were hinged at the point of contact so as to allow movement at this point?

I know how to solve this problem. My wording was not clear but what I intended to find out was made apparent by the below answers anyways.

Improve this question
$\endgroup$
4
  • $\begingroup$ Small nitpick: "Also assume the legs are rods of uniform mass, so the weight acts on the center of gravity of each leg. " By definition, weight acts on the center of gravity, even if the rods are not of uniform mass. $\endgroup$
    – Brionius
    Jan 7, 2015 at 2:01
  • $\begingroup$ So these rods are "hinged" at their intersection and have frictionless contact points with the "ground"? $\endgroup$
    – DWin
    Jan 7, 2015 at 3:12
  • $\begingroup$ Brionius: you are right, I meant to say geometrical centers. @DWin: in the scenario i presented, they are not hinged at their intersection. The entire system is a rigid body (imagine a triangle with no hypotenuse). In the second scenario the two rods do not together form a rigid body - rotation can occur at the hinge. $\endgroup$
    – Joshua Benabou
    Jan 7, 2015 at 5:09
  • $\begingroup$ Try it with forceeffect.autodesk.com/index.html $\endgroup$
    – John Alexiou
    Jan 7, 2015 at 5:48

3 Answers 3

Reset to default
1
$\begingroup$

In a static situation, torques must balance at every point. You are therefore free to pick whatever "hinge" you choose, and you will be able to solve the problem. However, depending on your assumptions some choices will make your life easier.

It is always helpful to draw a diagram of the situation. Here, I would draw:

enter image description here

The four vertical forces represent the weight of the legs, and the normal force from the ground. I am assuming no horizontal friction. Now the two forces on the left leg give rise to a clockwise torque of $\frac12 w d$, while the forces on the right leg give rise to an equal and opposite torque. The calculation of these torques is actually independent of the pivoting point you choose, since for any point with horizontal offset $x$ you find that

$$\Gamma = wx - w(x-\frac{d}{2})$$

from which you can see that $x$ cancels...

Now if you assume that there is a hinge at $P$ and yet that the legs don't spread, then there must be friction. In fact, the force of friction along the surface and the horizontal reaction force at $P$ must be such that they exactly cancel the torques due to the legs, so if the force is $F$ then

$$F h = \frac{wd}{2}$$

Of course for a right equilateral triangle, $h = \sqrt{2}d$, so the force of friction must be

$$F = \frac{wd}{2\sqrt{2}d}=\frac{w}{\sqrt{8}}$$

This is the horizontal force you would experience at $P$ if there was a hinge, and sufficient friction that the legs don't slide.

You can now also calculate the minimum coefficient of friction needed as $\mu > \frac{1}{\sqrt{8}}$

Improve this answer
$\endgroup$
1
$\begingroup$

I agree with Floris' method, but I believe he used the wrong moment arm for the horizontal reaction force. The Horizontal force should be w/2, as Fd = wd/2. It is incorrect to use the hypotenuse as the moment arm for the horizontal reaction force, as the moment arm should be perpendicular to the applied force.

Improve this answer
$\endgroup$
0
$\begingroup$

To calculate the torque at the point where the rods meet, cut the shape in two and consider each of the rods by itself. Draw a free-body diagram of each rod and at the joining point include the x-y forces and torque that would be applied by the other rod. The force balance equations between the two rods can then be used to find the moment where the rods meet. I believe you should have 7 equations and 7 unknown variables, which is solvable.

If the rods were hinged then there would be no torque at the joining point, since a perfect (non-frictional) hinge can't transmit torque. Although, there may be a moment present at other points along the span of the beams. The loading situation is certainly different and the structure will only be stable due to friction between the legs and the surface they rest on. If there is no friction then the legs will slide apart.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.