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Consider a square block kept on a rough horizontal surface (friction is sufficient enough to prevent slipping/sliding). You apply a horizontal force F on the topmost point ,constant in magnitude and direction. The magnitude is adjusted so that the block doesn't gain significant kinetic energy. Due to rotation about the axis passing through the line of contact of the ground and block, the potential energy of the block increases as it loses contact with the surface. How does it gain potential energy? I'm confused which force is responsible for this gain. Thanks in advance.

  • Please provide more details of your confusion. Why are you confused? – sammy gerbil Jun 11 '17 at 09:09
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    The normal and weight balance each other. Also there is no displacement of the point of contact.. hence the work done by all the forces in vertical direction is zero. But still there is a gain in potential energy. Also the force F applied is horizontal hence it doesn't do any work in vertical direction. Then what is responsible for gain of potential energy? –  Jun 11 '17 at 09:18
  • In the presence of friction, the normal force needs not be perpendicular to the surface. – valerio Jun 11 '17 at 09:45
  • @valerio92 : By definition, the normal force is perpendicular to the surface. As with a hinge, the reaction from the surface has 2 components : the tangential component is called friction, the perpendicular (normal) component is called normal reaction. Of course friction (or some other force to oppose horizontal sliding) must be present, but this has no vertical component. If F is applied at the base, there is friction but the COM of the block does not rise. – sammy gerbil Jun 11 '17 at 09:49
  • @sammygerbil In the previous comment I meant to write "reaction force", but I ended up writing "normal"...But anyway, I think that your argument of decomposing the reaction force into the "usual" normal force and an horizontal friction force is misleading, because one would like to assign to the normal component a value $mg$ and things wouldn't really work. I prefer to think of it like this: if there is friction, or a constraint, the reaction force needs not be perpendicular to the surface. But the result should be the same, I agree on that. – valerio Jun 11 '17 at 10:01
  • @valerio92 : The normal component is mg only when the COM is stationary, not accelerating. It must be greater than mg to make the COM rise upwards. You are welcome to post your own explanation. – sammy gerbil Jun 11 '17 at 10:07
  • Is it right to picture this block as a particle? –  Jun 11 '17 at 10:08
  • No. You are right, the block is not a point particle. But we should be able to see it as a Free Body; the resultant of external forces on it explains its motion. – sammy gerbil Jun 11 '17 at 10:20
  • Are we getting into a paradox where newtons laws hold normal force responsible. But the work energy theorem straightway contradicts it? –  Jun 11 '17 at 10:26
  • @sammygerbil "The normal component is mg only when the COM is stationary, not accelerating" block sliding on a frictionless plane, accelerating: is the normal component not mg? – valerio Jun 11 '17 at 10:33
  • @valerio92 I should have said, accelerating in the perpendicular direction. The block accelerating on a frictionless horizontal plane is not accelerating perpendicular to the plane, so the normal reaction is the same as when the block is stationary. – sammy gerbil Jun 11 '17 at 10:42
  • @sammygerbil But you see, this is a circular argument: you want to explain why the COM accelerates upwards, and to do that you say that the normal component is not mg because the COM is accelerating upwards. I personally just prefer to drop straight away the assumption that the reaction must be perpendicular when I have friction. In a certain way this is the same, but my point is that friction is essential here: it is because of that that in the end the total reaction can be different from $mg$. But we are going into the philosophical here, so I will just stop here. – valerio Jun 11 '17 at 10:56
  • The horizontal component of reaction force ( friction) opposes translation but increases rotation. If the plane is frictionless there is nothing to stop translation but rotation is limited by a torque due to normal force which opposes the torque due to force F. Maybe this should answer your query if I'm correct. –  Jun 11 '17 at 11:02
  • I think Archimedes explained it. – Hot Licks Jun 11 '17 at 12:27
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    I really appreciate the efforts you are putting in but maybe I'm not getting the answer I want. –  Jun 12 '17 at 05:36

2 Answers2

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If I understood your question correctly, what gets the Center of Mass to rise (and hence gain potential energy) is the couple (torque) that F creates:

Just like what happens in an imperfectly balanced wheel (let's say that more mass is in the lower part), where you get enough torque to get the wheel to start rotating on its axis. In this case, similarly, the Center of Mass rises, you just need enough Torque to overcome the torque from the gravitational force (which creates a torque as soon as the wheel rotates because the CM isn't exactly on the axis of rotation).

image of what I meant

In a rigid-body, staticity is not only from $\sum\limits_{i=1}^{n} F_i = 0$

but also from the sum of all present torques $\sum\limits_{i=1}^{n} R_i x F_i = 0$ which, in this case, is not happening

$\int\limits_{0}^{\theta_f} M_F(\theta) d\theta = W_{nonconservative} = \Delta E_{mec} = mgh + 1/2 I {\omega}^2$

with $h = lsqrt2 /2 - l/2$ and I from H.S. theorem

The integral actually solves out to constants, as $l\sqrt2 F$ applied along the arc of circumference

Community
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Massimo Pesavento
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The normal and weight balance each other.

You are making an assumption here. While the block is stationary the weight and normal reaction balance. But to make the COM start moving upwards there must be a resultant force upwards. The only upward force is the normal reaction.

Also there is no displacement of the point of contact. Hence the work done by all the forces in vertical direction is zero.

If you squat on the ground then stand up, you have raised your COM. Work must be done by some force external to your body. That force can only be the normal reaction.

But I still have not explained how this force can do work if the point of application does not move...

sammy gerbil
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  • If I look it from work energy perspective.. there is no displacement of point of contact therefore work done is zero.. why does this happen then? –  Jun 11 '17 at 09:34
  • But see, the normal reaction force is upwards and it causes the CENTER of mass to be displaced ( upwards) hence it's the normal reaction force that does work to cause the gain of potential energy. – Tausif Hossain Jun 11 '17 at 09:37
  • What about friction? Friction is very important here: without it the block could only move horizontally. So the normal reaction is not the only force responsible for the block lifting, unless you incorporate friction into it. – valerio Jun 11 '17 at 09:39
  • If you allow, I can give you an example. Consider 2 blocks each of mass m attached together with the help of the spring. Both blocks rest on a frictionless plane. One block touches a vertical wall. Now you compress the spring and let the motion start. Even though there is a normal force but it doesn't do any work because displacement of point of contact is zero. But the potential energy of the spring results in kinetic energy of the system even when the work done by normal force is zero. –  Jun 11 '17 at 09:42
  • So it's the internal energy = rise in kinetic energy , in your example, if I'm correct –  Jun 11 '17 at 09:45
  • Yes, internal energy (elastic energy in the spring, chemical energy in the human) ultimately causes the increase in PE. In the case of the block, it is of course ultimately the work done by force F which is converted into PE. The problem is : if F is the explanation, how does a horizontal external force cause a vertical rise in the COM? – sammy gerbil Jun 11 '17 at 10:23