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How much force do I do apply when I lift my leg above the ground? The same amount as gravity does on my leg (mg)? Or MORE than it (greater than mg)? If the displacement from the ground to my lifted leg is h meters, then what's the work done? mgh or more than mgh? Basically I wanna know if I have to apply more force or do more work than gravity to lift my leg up.

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    $\begingroup$ Actually since the leg is pivoted about the hip joint the force to move it is less than $m g$ (assuming the total mass of the leg is $m$). $\endgroup$
    – JAlex
    Sep 22 at 16:10
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    $\begingroup$ But the actual force exerted by your muscle might be 5 ... 10 times larger than the weight you bear, due to the mechanical disadvantage inherent in your skeletal system. See courses.lumenlearning.com/physics/chapter/… for example. $\endgroup$
    – g.kertesz
    Sep 22 at 18:31
  • $\begingroup$ @g.kertesz - I was talking about the equipollent force upwards on the foot. Essentially the force required to produce the same torque about the joint as the muscles do. $\endgroup$
    – JAlex
    Sep 22 at 18:54
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    $\begingroup$ Obviously more than what simple Newtonian physics would suggest, else you would reach absurd conclusions that you could hold a heavy weight above your head indefinitely since there is no change in energy or net work being done. $\endgroup$
    – eps
    Sep 23 at 18:51
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    $\begingroup$ You are mixing a question which is clearly about an ideal situation, and a very messy real-world scenario. Which are you asking about? One important factor is, that the ground, the shoe and the foot are all compressed when at rest in the initial situation. Even a minuscule amount of upward lift will allow these "springs" to uncompres a bit, and allow your leg to raise a bit: micrometers, millimeters, even centimeters if you have soft ground and thick soles. Does that fit the scope of your question, or is this more of a "spherical cow in vacuum" kind of scenario? $\endgroup$
    – hyde
    Sep 24 at 12:54
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Because your leg began at rest, moved for a time $\Delta t$ and ended at rest, the average force it felt was $$ \langle F \rangle = \frac{\Delta p}{\Delta t} = 0 $$ meaning on average your force was equal and opposite to gravity.

However, when you accelerated upward you acted with more force than gravity, and when you decelerated it, you acted with less force.

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    $\begingroup$ You're neglecting air resistance, etc..; in fact, more average force will have to be applied because of this. (However, it's not very much extra force.) $\endgroup$
    – wizzwizz4
    Sep 22 at 19:39
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    $\begingroup$ @wizzwizz4 I suspect internal resistance in the leg (muscles, joints and tendons) is a lot larger than air resistance. Not that that changes the validity of your comment. $\endgroup$
    – Arthur
    Sep 23 at 8:40
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    $\begingroup$ Also something not to forget about in computations is gravitational pull of distant galaxies on the knee that assist knee in overcoming the gravitational pull of the Earth by attracting it in its direction. $\endgroup$
    – wha7ever
    Sep 23 at 14:36
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    $\begingroup$ @wha7ever - The galaxies are also pulling the Earth. What matters is the tidal effect. That's the reason so that lifting your leg is easier when the Moon is high on the sky. $\endgroup$
    – Pere
    Sep 24 at 16:44
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    $\begingroup$ Guys, the art of physics is to know to which accuracy your model holds. Wanting to be more precise using galactic tidal forces is OK, but this information won't add anything because of the errors coming from way stronger forces like radiation pressure from the sun at day time and thermal fluctuations in air pressure (quite ironic they're "strong"). When adding really small magnitude effects, you should check if you forget any other affects with the same magnitude or bigger. Lets agree that Arthur's comment is on spot and wizzwizz4 also adds a good first approximation and settle it there? :) $\endgroup$ Sep 24 at 17:21
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Remember Newton's laws. If you only exert exactly the same force as gravity, then you are balancing out gravity, resulting in no initiation of motion (this is Newton's 1st law):

$$\sum F=F_\text{lift}-mg=0\qquad \text{ when }F_\text{lift}=mg.$$

In order to accelerate your leg upwards you must thus exert a larger force than gravity (this is Newton's 2nd law):

$$\sum F=F_\text{lift}-mg=ma\qquad \text{ when }F_\text{lift}>mg.$$

The work you have to do to lift the leg to a height $h$ is the gravitational potential energy needed $U_g$ and the kinetic energy supplied $K$:

$$W=U_g+K=mgh+\frac12 mv^2.$$

If you move your leg up and then keep it there, then its kinetic energy is zero (you keep it there at rest). Only then is the work equal to just $mgh$. Meaning, if this is the end situation, then you only have to do work equal to the work done by gravity when the leg falls back down later on.

Now, with all this being said, keep in mind that this is all an idealized description.

  • Firstly, lifting your leg is not a clear-cut process such as lifting a box because you have parts of your limbs such as the thigh that are only lifted partly.
  • Secondly, be aware that the work $W$ mentioned above is only the work done on the lifted object. It is not the work done by the body. Your body is generally inefficient and might spend a lot of energy just to produce the force that lifts your leg. All this behind-the-scenes energy usage and waste is not included in the above calculation of work.
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    $\begingroup$ "...resulting in no motion" Did you mean no acceleration? Doesn't N1 say an object at rest or in moving at constant velocity remains at rest or moving at constant velocity unless acted upon by net force? $\endgroup$
    – Bob D
    Sep 22 at 10:53
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    $\begingroup$ @BobD Sure, I was merely assuming that the leg starts from rest - and thus stays at rest if the upwards force equals the downwards force. $\endgroup$
    – Steeven
    Sep 22 at 14:05
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    $\begingroup$ Understand. Don't mean to make a big deal out of it. I was just concerned with associating N1 with no motion without the qualification. Especially since once motion is initiated, the forces can balance and constant motion occur. May I suggest replacing "no motion" with "no initiation of motion". Bob $\endgroup$
    – Bob D
    Sep 22 at 15:00
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    $\begingroup$ @BobD Good suggestion, done. $\endgroup$
    – Steeven
    Sep 22 at 20:02
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    $\begingroup$ Your welcome 👍 +1 $\endgroup$
    – Bob D
    Sep 22 at 20:34
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Yes, if you apply same force as gravity then your leg would be at rest as the two forces will be cancelling eachother. Now if you have to move the leg upward you need to give your leg a motion from resting position. Now to give a body at rest any motion you need to apply force. Hence, yes you need to apply more force. M×a=F-Mg[F is force applied by you, M is mass of your leg, a is acceleration of your leg as a result of extra force applied by you].

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TL; DR The question is poorly defined: it is based on a simplified model of a leg, which is not really applicable.

Indeed, Newtonian physics deals with point-like objects or, at best, a collection of such objects connected by links (aka solid body in the case of rigid links).

Foot as a point-like object
In the first case we can consider foot as a point-like object, and then the forces need to balance themselves to raise the foot with a constant speed. There are some caveats, related to the fact that the foot, if initially at rest, needs to be first accelerated to make it moving and then need to be stopped, so that the force is not constant in time. But, once it is raised, the forces balance each other (from the point of view of Newtonian mechanics).

Leg as series of sticks and joints
The OP speaks about a leg, which cannot be considered as a point-like object, but rather a series of rigid parts connected by joints. One can develop mechanical model of different level of complexity: starting with simple foot-ankle-hip model and moving on to model them mechanically, as, e.g., in the figure below (taken from the first result that came up in Google). enter image description here

In this case we will have many forces and torques, as well as several parts of the leg with different masses, so the question needs to be refined.

Leg as elastic objects
Now, the parts of the leg are not really rigid, but rather elastic - that is they get deformed when the leg moves. This is particularly the case of the muscles, which play a prominent role in the movement. We thus need to involve elasticity theory, which goes somewhat beyond the Newtonian mechanics (although largely based on it), and deals with continuous stresses rather than forces.

Contraction of muscles
Energy needs to be supplied to make the muscles contract, and needs to be continuously supplied to keep them contracted. This implies largely thermodynamic treatment, far away from simple Newtonian balance of forces. Yet, in terms of work it tells us that we need to spend energy even when the leg is not moving (aka "do more work than gravity").

Chemistry and biology
The above levels of description have not gone beyond physics, which however would be necessary, if we were to describe the motion on the cellular and molecular levels.

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Firstly laying out my physics/math assumptions

  • no friction

  • only moving the leg vertically (as if you are lifting a bag)

  • constant gravity, $g$

  • force on leg, $F$, only varies with time

  • for $t<0$ the ground provides the force to stop movement

therefore we can limit movement to 1 dimension, $x$ and from newtons second equation we get $a(t)=-mg+F(t)$

Now we want to have the leg to start from rest and finish at rest, if it doesn't finish at rest then you haven't finish lifting it.

If $F(t)$ is a constant (either greater than $mg$ or less than $mg$) then we won't get this behaviour, it will either not start moving ($F=mg$) or not stop ($F\neq mg$).

But if we have a slightly more complicated expression for the force, for example where $t_0$ is chosen so that the leg stops at the right height (which i'm going to call $L$). $$F(t)=\begin{cases} 2mg & 0\leq t<\frac{t_0}{2}\\ 0& \frac{t_0}{2}\leq t\leq t_0 \\ mg & t_0\leq t \end{cases}$$ ie lifting the leg with twice the force of gravity for the first half then letting gravity slow it to a stop for the last half, then providing a constant force to match gravity so that it stays there. (a point of note is that the average force is $mg$)

hence the acceleration is $$a(t)=\begin{cases} +mg & 0\leq t<\frac{t_0}{2}\\ -mg & \frac{t_0}{2}\leq t\leq t_0 \\ 0 & t_0\leq t \end{cases}$$

from this we can find the velocity, by either integrating or recognising that these are the same conditions as newtons equations of motion so would follow the same behaviour $$v(t)=\begin{cases} +mgt & 0\leq t<\frac{t_0}{2}\\ -mg(t- \frac{t_0}{2})& \frac{t_0}{2}\leq t\leq t_0 \\ 0 & t_0\leq t \end{cases}$$ and then the height $$x(t)=\begin{cases} +mgt^2 & 0\leq t<\frac{t_0}{2}\\ -mg(t- \frac{t_0}{2})t+L& \frac{t_0}{2}\leq t\leq t_0 \\ L & t_0\leq t \end{cases}$$

A point to mention is that this isn't the only function that would fit the conditions, there would be an infinite number of possible force functions. But they would still have an average force of $mg$.

hopefully that helps

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The general, "ideal" answer in Newtonian physics is as follows. I'll consider downward forces (and Earth's acceleration g) as negative quantities, and upward forces as positives.

The first thing to repeat and emphasize:

No net forces act on unaccelerated objects.

An object suspended in gravity experiences the gravitational force $F_{grav} = m*g$ (m being their mass and g being Earth's acceleration, $9.81 m/s^2$). Because it is suspended and is not accelerated, no net force is acting on it: There must be something holding it up (the ground, a muscle etc.) by exerting on it the exact opposite of the gravitational force. We can call this suspending force $F_{susp}$. Mathematically, the statement "no force is acting on them" means "when we add all the forces up, the result is 0: $F_{susp} + F_{grav} = 0$. The forces acting on the leg cancel each other out, they have the same magnitude but opposite signs. Another way to express this equilibrium is $F_{susp} = -F_{grav}$. Same force, opposite direction.

$F_{susp}$ is the "effective force" the body exerts on the leg, ignoring levers, joints etc. which may yield a higher or lower force to be "applied" by muscles, tendons etc., as seen from the point of view of the actor. Think of a pulley. But the force on the object opposing the gravitational pull is always $-F_{grav}$ because no net forces are acting on an unaccelerated object, including one at rest. Here you can think of a pulley again, this time from the perspective of the hanging object.

Now let's consider the phases involved in lifting the leg.

  • Acceleration phase: In order to accelerate something against gravity — and accleration is obviously necessary if you want to put it into motion — you must not only apply $F_{susp}$ but a bit more, some additional $F_{up}$. Because $F_{susp}$ is exactly canceled out by gravitation, $F_{up}$ is a net force which accelerates the object upward; as an equation we would add up all acting forces and find that the net residue is exactly $F_{up}$: $F_{susp} + F_{grav} + F_{up} = F_{up}$ (instead of 0).
  • Coast phase: This delta only has to be applied for a short moment until the leg is moving. After that moment, to keep it moving at a steady pace, we have to make sure no net forces act on it, so it is again simply $F_{susp}$ again which is exactly canceled out. No acceleration, no net force.
  • Deceleration phase: Versus the end of its relocation the object slows down in order to come to rest (to rest, at least, relative to your position on Earth's surface which is following a literally convoluted trajectory through space — but I digress ;-) ). During this slowing-down phase, which is physically an acceleration like any other, a net force $F_{down}$ must act on it in the same direction as gravity. This is achieved by lowering the suspending force by this small amount so that gravity can get the upper hand. $F_{down}$ is similar to $F_{up}$, just in the opposite direction. Because the object is accelerated, adding all acting forces up results in a residual net force: $F_{grav} + (F_{susp} + F_{down}) = F_{down}$ (instead of 0). (Because $F_{down}$ is facing downward it is negative; adding it to the upward facing $F_{susp}$ decreases the resulting upward force.) After that slowing-down period the leg comes to rest again, and at rest the acting forces are simply $m*g$ in both directions, again canceling each other out.

This discussion of the idealized physics misses a few issues that play a large role in real life. Obviously, the body works with significant friction: Tendons run through sheets, tissue is deformed, fluids are pressed through vessels when the thigh bends and muscles contract etc. etc. Most of the time we move we are not lifting or net-accelerating anything (when we move an object from A to B we accelerate it and then brake it, but we are not able to recuperate the kinetic energy but instead need to exert more energy to stop the movement until it is at rest again): We are simply creating heat. The mechanical work you perform to lift the leg is much higher than the gain in potential energy of the lifted mass.

More energy spent while moving something the same distance means that the forces applied were larger, which is one real-life answer to your question.

Another issue is that holding limbs in a position that is not the equilibrium is actually quite energy-consuming, even if none of that is transformed into kinetic or potential energy: It is all going to waste, i.e., heat. Keeping muscles contracted requires cell activity which accelerates their metabolism: Sugar and oxygen must be transported there, breathing and heart rate accelerate etc. It is exhausting.

Forces and energy to overcome friction and to fuel the peculiarities of our body are not considered in the ideal Newtonian mechanics considerations above. This is by design: One of the biggest modern-era physics achievements was to divide the chaotic muddle we see when we take in the real world into superimposed subsystems which could be described separately. In a way Newtonian physics is a regression from the Aristotelian concept that every movement needs a mover, a continuing cause. Rejecting this is the core of Newtonian physics: Motion change needs a mover; steady motion as such doesn't. This is an astonishing abstraction, the achievement of a genius thinker, because this literally never happens in macroscopic reality. On Earth nothing ever continues moving without some mover. Unpropelled, everything comes to a stop rather quickly. In cosmological time even celestial objects moving through vacuum experience friction through tidal forces, become tidally locked etc.

Newton enabled us to think of friction as an effect separate from the basic mechanical laws. It could be seen as an impurity, a contamination: In machines and transportation, let alone school room physics, we try to minimize it because it is annoying, less deterministic and comparatively hard to compute.

But friction is not a side effect, an impurity best ignored: Heat losses are systematically described by thermodynamics which became a central part of cosmology in the 20th century. Aristotle was right, after all: Nothing, literally nothing macroscopic moves or changes without friction, without increasing entropy, without being propelled, without a mover. The concept of ever-increasing entropy is a fundamental and far-reaching concept with the same rank as evolution in biology. The simple principle is that unlikely random events don't occur that often.

Newtonian physics is only a subsystem, but a brilliantly carved out, beautiful one.

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Here's a general way to look at it:

For starters assume you have all your weight on one leg, so the other leg is hovering very slightly above the ground. If we didn't assume this we would have to account for the weight shifting from one leg to the other and this would complicate the story ever so slightly. At this point you are still stationary so the force on your standing leg is exactly $mg$. During the lifting of your leg your center of mass moves a little bit upward. Unless rotations are involved you can treat objects like they are point masses located at their center of mass. So for example to calculate the work that was done you can calculate the height of the center of mass before you lift your leg and the height after and use the difference to calculate $\Delta W=mg\Delta h$. The energy expended will generally be a bit higher because your muscles lose energy due to friction.

To look at the force that is exerted on the ground let's look at the height of your center of mass and we will assume there is a period of constant acceleration, a period of constant velocity and a period of constant deceleration. The velocity and acceleration graphs will look something like this:

enter image description here

The acceleration graph is the result of the total force on your body, which is given by $$F_\text{total}=F_\text{leg}+F_\text{gravity}$$ If you ponder a bit about how the signs should be you can conclude that

  1. during the initial constant acceleration $F_\text{leg}>mg$,
  2. during the phase of constant velocity $F_\text{leg}=mg$,
  3. during the last phase of constant deceleration $F_\text{leg}<mg$.

The graph that I showed above is ofcourse not how a real world graph will look like but it should give you a vague idea of what is happening.

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Well ,in theory, yes. Your leg weighs a certain amount, let's say 30 pounds, but when you lift your leg, there is some friction that will always resist, such as the air that you displace being relocated, and the slight amount of friction in your leg joint and sinews rubbing together and such. So you do apply more force, but it is a such a small amount, that it is hardly noticeable. It is so small, that it does not matter for any practical reason. I hope this answers your question.

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To suspend 1kg of mass at the earth's surface requires 96 watts of energy. This can be done by building an electronic circuit that suspends a mass, and computing the power requirements. Many physicists believe that no work is required to hold a mass in a gravitational field, which is false. Try holding a 100 lb mass over your head for an hour, and then tell me...that...lol ed

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  • $\begingroup$ work, in physics, is defined as force times distance (or for variable forces, it is the path integral of the force). So if an object doesn't move then no work is required. Also there is a $1kg$ mass that has be suspended in Paris for over 100 years and it required no power to suspend it. $\endgroup$
    – Nyra
    Sep 23 at 16:01
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    $\begingroup$ Read this: physics.stackexchange.com/q/1984/305718 $\endgroup$
    – ACB
    Sep 24 at 13:16

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