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.
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$\begingroup$ Please provide more details of your confusion. Why are you confused? $\endgroup$– sammy gerbilCommented Jun 11, 2017 at 9:09
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1$\begingroup$ 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? $\endgroup$– user150098Commented Jun 11, 2017 at 9:18
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$\begingroup$ In the presence of friction, the normal force needs not be perpendicular to the surface. $\endgroup$– valerioCommented Jun 11, 2017 at 9:45
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$\begingroup$ @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. $\endgroup$– sammy gerbilCommented Jun 11, 2017 at 9:49
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1$\begingroup$ I really appreciate the efforts you are putting in but maybe I'm not getting the answer I want. $\endgroup$– user150098Commented Jun 12, 2017 at 5:36
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2 Answers
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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).
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
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$\begingroup$ Can you please elaborate a bit more.? Thanks $\endgroup$– user150098Commented Jun 11, 2017 at 9:49
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$\begingroup$ Can please write equation with respect to work and energy? $\endgroup$– user150098Commented Jun 11, 2017 at 9:52
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$\begingroup$ Adding work and energy in a minute $\endgroup$ Commented Jun 11, 2017 at 9:57
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1$\begingroup$ Can you please explain how a torque causes increase in potential energy, thanks in advance. $\endgroup$ Commented Jun 11, 2017 at 10:12
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$\begingroup$ @TausifHossain Take a look at the analogy with the imperfetly-balanced wheel. You see that torque is actually making all the work that not only gets an angular acceleration, but also the CM to rise. $\endgroup$ Commented Jun 11, 2017 at 10:16
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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...
answered Jun 11, 2017 at 9:33
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$\begingroup$ 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? $\endgroup$– user150098Commented Jun 11, 2017 at 9:34
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$\begingroup$ 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. $\endgroup$ Commented Jun 11, 2017 at 9:37
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$\begingroup$ 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. $\endgroup$– valerioCommented Jun 11, 2017 at 9:39
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$\begingroup$ 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. $\endgroup$– user150098Commented Jun 11, 2017 at 9:42
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$\begingroup$ So it's the internal energy = rise in kinetic energy , in your example, if I'm correct $\endgroup$– user150098Commented Jun 11, 2017 at 9:45