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Determine the maximum ratio $h/b$ for which the homogenous block will slide without toppling under the action of force F.The coefficient of static friction between the block and the incline is $\mu_s$.

I have a doubt.About which point should the rotational equilibrium be applied?Should it be applied about centre of mass?Or should it be applied about the vertex opposite to the vertex where F is applied?Why?

MY ATTEMPT:

Translational Equations $F+mg\sin(\theta) \geq \mu N$ and $N=mgcos(\theta)$

Rotational Equations This is where I'm facing a problem.Depending upon which point the equilibrium is applied the required ratio will be obtained.

MY VIEWS:

Rotational equilibrium should hold at all points if no toppling/rotation happens.However the answer varies depending on the point of application of equilibrium.Strange.

I hope this is a conceptual doubt and will not be closed as off-topic or homework.If it needs to be closed please inform me if the post can be improved somehow.

  • @CountTo10 Added.Anything more? –  Sep 12 '16 at 19:56
  • Nope, it's just the best way of improving your chances if the community see an attempt/ reasoning, rather an almost certain vote to close otherwise. Best of luck with it (as it's not my area, sorry) , I have up voted it. –  Sep 12 '16 at 20:04
  • About which edge will the object rotate? – Farcher Sep 12 '16 at 20:08
  • @Farcher The vertex opposite to F.... –  Sep 12 '16 at 20:15

2 Answers2

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You can apply it to either location, but there are some considerations:

  • If you consider rotation about COM, then you need to understand the torque from the normal force. As you push the box, the normal force will move toward the front to counteract. At the tipping point, all the normal force will be there. See also: When does the shifting of normal force occur?

  • If you consider rotation about the front vertex, you can ignore the forces that act through it. But it is likely that the box as a whole is accelerating down the ramp. If so, that axis is also accelerating. When that happens, fictional forces appear that act on the center of mass. You can't ignore those.

But both considerations should yield the same answer.

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BowlOfRed
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The block needs to slide without tipping.So basically we should ensure two things.

1) FOR SLIDING :The combined effect of gravity and $F$ forcing the block down the slope must exceed the force of friction.

$F+mg\sin(\theta)=\mu_s N \quad$

[where $N=mg\cos(\theta)$]

2) FOR NO TOPPLING There must be rotational equilibrium about any point in the block.Say we write it for the vertex opposite to the vertex where $F$ acts.

$Fh + mg\sin(\theta)\frac{h}{2} \geq mg\cos(\theta)\frac{b}{2}$

On solving we get

$$\frac{h}{b} \leq \frac{1}{2\mu_s - \tan (\theta)}$$

I hope this will be helpful to future viewers!

  • @BowlOfRed I know.I did'nt add the pseudo force due to acceleration.Instead I used the $\geq$ symbol. –  Sep 20 '16 at 01:23