0
$\begingroup$

Take a look at this question.

enter image description here

The solution by the author is

enter image description here

My question is why the weight force is ignored in the free-body diagram? Is it because the roller(A) and the pint(B) which cancel the effect of the gravitational force?

$\endgroup$
  • $\begingroup$ It must be that the weight of the member is assumed to be negligible. $\endgroup$ – Bob D Sep 19 '19 at 21:22
  • $\begingroup$ @BobD, that actually possible. Unfortunately, nothing stated in the question. Also, this chapter talks about free-body diagram so I've assumed I have to include all forces acting on the body. $\endgroup$ – CroCo Sep 19 '19 at 21:25
  • $\begingroup$ Since it wasn't stated you need to assume so. You assume correctly. A FBD must include all forces, including weight. You can't determine reactions at the pin and roller otherwise. The roller and pin vertical reaction must equal the weight as well as the vertical component of the 390 lb force. $\endgroup$ – Bob D Sep 19 '19 at 21:31
  • $\begingroup$ Where did the 390 lb come from? I would take the y-component of that as the weight. And, therefore, the weight is used. $\endgroup$ – jmh Sep 19 '19 at 21:36
  • $\begingroup$ @jmh, well, may be an external force who knows except the author. But it seems the question is not very well written. Also, if this is the weight vector, why it is not drawn as an vertical downward vector?! I will go with BobD since the mass of the body is not mentioned. $\endgroup$ – CroCo Sep 19 '19 at 21:37
2
$\begingroup$

It must be that the weight of the member is assumed to be negligible in comparison to the vertical component of the 390 lb force. With no information to the contrary you have to assume this to draw the FBD.

Also, you assume correctly that a FBD must include all forces, including weight. You can't determine reactions at the pin and roller otherwise. The sum of the roller and pin vertical reactions must equal the sum of the weight and the vertical component of the 390 lb force.

Hope this helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think the fact that the drawing shows an extended body for all aspects of the system it seems wrong to conclude the object as massless! $\endgroup$ – jmh Sep 19 '19 at 22:08
  • $\begingroup$ @jmh I never said it was "massless". I said it must be assumed to be negligible. And what I mean by negligible is in comparison to the vertical component of the 390 lb force. In any case, if the member is uniform the CG would not be the point of application of the 390 lb force since the vertical member would only be 1/4 the weight of the horizontal member, assuming the members are uniform. I will edit accordingly. $\endgroup$ – Bob D Sep 19 '19 at 22:24
  • $\begingroup$ @jmh I would add it is not unusual to neglect the weight of the beam in elementary statics problems. What is unusual is for the problem not to state ignore the weight of the beam. That’s where this problem falls short $\endgroup$ – Bob D Sep 19 '19 at 22:52

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.