How do you calculate the energy required to hold a weight against gravity?

Question

What is the energy required to hold a mass in place? To lift 1kg 1m at sea level I must use 9.8 J. but if I simply grab onto a 1 kg mass that is on a table and the table is removed. I am now using energy to hold it in place. But since the mass is not moving there is no work and therefore supposedly no energy, but obviously if I don't use energy it will fall to the ground. And so if I put it back on the table is the table now imparting some kind of energy to keep it in place against the gravitational potential energy (is the table effectively accelerating it up at 9.8 m/ss?). How would one calculate the energy required to hold the mass in place?

Answer 1

From a physics perspective, zero energy is required to hold a mass in place. Work is defined as the change in energy of an object, and can be calculated as the net force times the distance over which the force is applied. If the object does not move, that distance is zero, so zero work is done on the object.

The biomechanics of the human body make it so that you do get tired from holding a heavy weight, so a biological system typically does consume energy holding an object in place. But this is because muscle fibers are constantly relaxing and contracting by very small amounts, even when attempting to hold an object still. When you hold a weight, you are constantly doing small, nearly imperceptible amounts of work which will eventually tire you out.

A table, on the other hand, does not consume any energy or perform any work at all to hold an object in place. A weight can sit on a table forever, and the table does not require any source of energy to keep the weight in place. What the table does provide is a normal force which exactly balances the force of gravity, meaning the weight experiences zero net force and does not move, which indicates that zero work is being done.

The key difference here is that the table relies entirely on intermolecular forces to hold the book in place. This is what gives rise to the normal force, which occurs "automatically" when two surfaces come in contact. Your arm, on the other hand, relies on muscle power to hold a book in place. If you relax your muscle, your arm will fall, because your muscles do not naturally "lock in" to place - constant tiny adjustments are needed to keep a muscle in place. You could pile books on a dead body and they would be held up by the normal force from intermolecular interactions just like a table does, but to hold them up by muscle power requires some input energy to hold them in place.

As an interesting biological counterexample, we can look at the three-toed sloth's grip. The connective tissue of their claws is structured such that the natural grip position is closed, and requires energy to open. Because of this, sloths can hang from branches even after they're dead, clearly requiring no energy and performing no work to maintain their position.

Answer 2

If you hold a weight in place, your physical effort does not produce physics work. The energy expended is internal. Richard Feynman in his physics lectures explains it this way:

The fact that we have to generate effort to hold up a weight is simply due to to the design of striated muscle. What happens is when a nerve impulse reaches a muscle fiber, the fiber gives a little twitch and then relaxes, so that when we hold something up , enormous volleys of nerve impulses are coming in to the muscle, large numbers of twitches are maintaining the weight, while other fibers relax. When we hold a heavy weight we get tired, begin to shake, ...because the muscle is tired and not reacting fast enough.

When you are simply holding the weight, you do no physics work. But you expend chemical potential energy of your body when your muscle fibers alternatively "twitch and relax" as described by Feynman. Think of it as internal physiological work as opposed to external physics work. If you lift the weight (even at constant velocity) you are doing positive work separating the dumbbell from the earth resulting in the Earth-dumbbell system gaining gravitational potential energy. Some of your chemical potential energy will still be converted to thermal energy simply because your body is not 100% efficient performing the work.

Hope this helps.

Answer 3

I think there are good answers already, but OP seems to be still dissatisfied, so I will try to put the answer into different perspective.

The work is simply force times distance traveled. If the net force is zero or the object does not move, the work is always zero.

In case of a table holding some mass no macroscopic component moves. There is heat movement of molecules, but this is not macroscopic. In fact, when molecules jiggle, they sometimes push on the mass and then they do (very very small) work on the mass sitting on the table, but at another times the molecules of mass push on the table and they do work on the table. Because the system is in equilibrium the amount of work done by table on mass is same as amount of work done by mass on the table so no work on average is being down. But this is all microscopic, at macroscopic level, there is simply no movement and thus no work.

Now, I will not pretend to understand biomechanics very much, but I think what I am about to write should be good enough picture of what is happening.

Your muscles work similar to this simple picture. Have your mass be connected to bunch of strings. If you would hang the mass on the strings, nothing moves and no work is being down to keep the mass in air. This is analogical to holding the mass exactly in line of your bones. Notice, that holding your body in straight position is quite easy, while holding it on bended legs is difficult (the similar situation happens while holding heavy bag with relaxed arm so that you are supporting it with your bones instead of muscles). In fact, if you would be perfectly straight and all you mass is distributed in such a way that it is completely supported by your bones, you would never get tired.

The situation when you bend your legs is similar to raising the mass little higher (so that it is no longer supported by strings - the strings get relaxed) and letting it go. Of course, naturally it would fall down, the strings stretch and start supporting the mass. But you do not want to do this. You want the mass to be kept higher in the air, and you will start pulling randomly on the strings attached to the mass, giving the mass upward impulses. But as you pull, you immediately relax, so only little amount of upward impulse gets transferred to the mass and it starts falling again. To keep it from falling you thus start giving huge amounts of this very short pulls, constantly pulling upward this string or that one with such a frequency as to make mass basically static. This basically how muscles with their twitching works.

So where is the movement which makes work? it is in the fact, that the mass is constantly falling down and every pull you make you rise it back to its original position. The mass constantly jumps up and down, everytime only in a little amount, but it does this tiny motion huge amounts of time. With every pull, you are rising the mass and you are doing work. With every relaxation, the gravitation is doing work on the mass accelerating it downwards, but then you pull again, doing again the work.

Now, in a perfect scenario, you could let go of one string in exactly the same moment as you pull another string, with just right amount of force, so that the mass will remain constantly at the same position - you will give it no time to fall down. In this case you will do no work, because the mass is no longer jumping up and down. But this requires very precise coordination of which muscle is (I guess) not capable.

Answer 4

Interestingly, from the point of view an inertial free falling observer, work is being done on the object that appears to be stationary from the point of view of a non inertial accelerating observer that is at rest with the surface of the Earth. From the inertial observer's point of view, the object is moving upwards with an acceleration of g and covers a distance of $x = u t +\frac12 gt^2$ in time t where u is the initial velocity and g is the acceleration of gravity. The acceleration is proper acceleration, that can be measured by an accelerometer. Substituting $F= mg$ into $F x$ we get work done in time t is $ mg (ut +1/2 g t^2)$ as measured by the free falling inertial observer.