1
$\begingroup$

I have the standard solution of this problem, however I solved it by an unusual and somehow much easier method than the standard solution, but I'm far from sure, if the logic I applied makes any sense.

Here is the problem:

enter image description here

If we assume the culvert doesn't move or rotate on the truck, then its acceleration would be the same as the acceleration of the truck. But this isn't the case and the culvert moves to the left or the other end of the truck, so there should be a force $F_2$ acts on it, this force acts right at the contact point of the culvert with the surface of the carrier, and since the culvert moves in a straight line the acceleration of its center mass is equal to $\alpha r$, with $\alpha$ the angular velocity and r the radius.

Applying the second law of Newton (horizontal direction):

$$ma - F_2 = mr\alpha$$

with $a = 3 \frac{m}{s^2}$.

Moment equilibrium around the center of mass:

$$\Sigma M_G = I_G\alpha = F_2r$$

By solving the two equation simultaneously, I can have the angular acceleration which is numerically leads to the correct answer.

I was wondering if my method is correct? May I assume the fictitious force acts at the contact point? Regardless of the answer I want to know why or why not?

$\endgroup$
1
$\begingroup$

$F_2$ is not a fictitious force. It is the reaction force of the truck on the culvert. You have (in essence) solved the problem in the (rotating and accelerating) centre of mass frame of the culvert. This means you have two fictitious forces in your solution.

  1. The linear acceleration of this frame is $a-r\alpha$, so the corresponding fictitious force (acting at the centre of mass) is $m(a -r\alpha)$. Then your first equation should really have been $$F_2 = m(a -r\alpha)$$ since the centre of mass does not accelerate in this frame.
  2. The angular acceleration of this frame is $\alpha$, giving a fictitious torque $I_G\alpha$ which equals the active torque $F_2r$. The second equation should be $$\sum M_G = I_G\alpha - F_2r = 0 $$ for equilibrium.

Actually it is easier to think about in an accelerating but non-rotating frame, in which case the second equation would simply be $$ I_G\alpha = F_2r $$ (alternatively you could write this and leave out $\sum M_G$).

$\endgroup$
2
  • $\begingroup$ Thank you for the clarification, but still one thing remains vague, if the force acts on the center of mass, then my second equation should be equal to zero, however this can't' be true because the $\alpha$ is not zero, we see the culvert accelerates $\endgroup$
    – Sam B
    Apr 19 '20 at 11:11
  • $\begingroup$ On rereading, I realised you were thinking a little differently and I modified the answer. $\endgroup$ Apr 19 '20 at 11:30

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.