0
$\begingroup$

A block is pushed to the right at the pivot point in such a way that it balances like a wheelie.

FBD:

Free body diagram of block

If torques are taken around the pivot point however, it appears that the torque from gravity is unbalanced even if there is no angular acceleration.

What is the explanation for the net torque not being zero at that specific pivot point? Thanks!

$\endgroup$
0
$\begingroup$

This is not an answer but a note too long for comments
There's a small problem with your FBD. It's not wrong, but it's not right. The diagram is indeed correct, but it is not the right one to draw when you need to balance the torques. Check out this one.
FBD
When the body has to balance the torques about the pivot point (in order to attain zero angular acceleration) this is pretty much possible that the body has turned to such an angle that all the forces might have their line of action passing through the pivot point yielding zero individual torques themselves. In such a case as you've described, this happens when the diagonal passing through the pivot point becomes perpendicular to the ground. Then, the centre of gravity (or the centre of mass) turns out to be vertically over the given point.

$\endgroup$
0
$\begingroup$

You cannot use an arbitrary point to evaluates torques in a dynamic system as you do in statics because the sum of forces is not zero.

Evaluate the torque about the center of mass to see that when balanced the torque is zero and hence angular acceleration is also zero.

fig1

Remember the equations of motion are evaluated from an inertial frame at the center of mass.

  • Net force is mass times acceleration of the center of mass.
  • Net torque about center of mass is mass moment of inertia (about CM) times angular acceleration
$\endgroup$

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.