0
$\begingroup$

You have one horizontal beam. Its left point is attached to a free pivot at the vertical wall. Its right point is suspended by a rope from the ceiling. Nothing is moved. So $\sum F=0$ and $\sum M=0$. To analyze this, we write $F_a-F_b=0$, where $F_a$ is the force from the rope and $F_b$ is the force of the weight of the beam. At the same time, $Μ_a-M_b=0$, where $M_a$ is the moment from $F_a$ and $M_b$ is the moment from $M_b$. Am I right?

Now we have the same system, but the beam's right point is dropped a bit, so it's not horizontal, but it has an angle. What are the equilibrium equations? Are they exactly the same?

Improve this question
$\endgroup$
6
  • $\begingroup$ There will be an extra term in the reaction from the pivot. Just resolve all the forces along your choice of axes, and include the term for hinge reaction. $\endgroup$
    – GRrocks
    Commented Nov 8, 2015 at 12:00
  • $\begingroup$ can you tell me the equations please? I don't get what you mean. What will be the difference exactly? $\endgroup$
    – ergon
    Commented Nov 8, 2015 at 23:46
  • $\begingroup$ Homework and 'check-my-work' type questions are off-topic for Physics.SE. Please see this post about how to ask homework problems. Additionally, responding to your own post in that way is not a constructive way to convince people to help you. $\endgroup$
    – DilithiumMatrix
    Commented Nov 10, 2015 at 0:13
  • $\begingroup$ The modification you describe doesn't change the underlying principles at all. The only difference in the calculation is that you should describe the equilibrium in both the x and y directions, instead of just in the y-direction like you describe initially. $\endgroup$
    – DilithiumMatrix
    Commented Nov 10, 2015 at 0:14
  • $\begingroup$ did I say it changes the underlying principles? I said it MAY change the equations. Can you tell me please the exact equations? Consider x,y as vertical and horizontal axes. thanks! $\endgroup$
    – ergon
    Commented Nov 10, 2015 at 20:43

2 Answers 2

Reset to default
1
$\begingroup$

The Equations for the $\sum F=0$ will be the same as the forces will maintain the same orientation. But the equations of the $\sum M=0$ will not be the same as the forces will be at an angle to the point of reference. You may have to use a little bit of sines and cosines to find the perpendicular distances to the forces from the point of reference.

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

Am I right?

About the force balance: A force is missing. $F_a$ is the tension in the rope, $F_b$ is the weight, but what about the hinge at the wall? This point also helps to hold up the beam, so your force balance must include this: $$\sum F=F_a+F_b+F_c=0$$

About the force-moment balance: It is correct, if you turn about the pivot/hinge at the wall (which you of course do), because then the missing force causes no moment. Remember to always state, which point such "rotation" equation is set up for.

Now we have the same system, but the beam's right point is dropped a bit, so it's not horizontal, but it has an angle. What are the equilibrium equations? Are they exactly the same?

The force and force-moment balance equations have the same terms, yes, but just remember that the values of the moments have changed, since they are calculated from the perpendicular force component only, which is not the whole force anymore.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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