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Homework Help: Moment and forces for a person standing on tiptoe

  1. Jul 9, 2010 #1
    Hi guys! I need some help.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I have tried solving this by assuming the vector sum of three forces to be zero.


    But evidently I got the wrong answer. I am going to try using the fact that the resultant of the moments about any point is zero to solve this but I'd like to know why the working above got me the wrong answer.

    Another thing I'm confused is that why do we not need to take into account the weight of the person? Is the weight included in the compression (force R) of the tibia?
  2. jcsd
  3. Jul 9, 2010 #2
    This seems pretty hard, I'll think about it. Here are some things I find strange:

    1)I'm not very familiar with biology, but I think that the sum of R and T are supposed to give how the body's weight is pushing down on the ankle. N is -mg, holding the whole body's weight. However, when I said the "body's weight" just above, that doesn't include the weight of the ankle, while N does! So here's something to think about: if we change the distribution of weight inside the body so that the overall weight remains the same, but we put like 65kg in the ankle and 5kg to the rest, will R and T change? I think they should...

    2) The ankle can't rotate as well, so summaM = 0 as well. So is it true then that 0.18*R = 0.25*Tcos(11)? Also, the vertical forces have to be 0 as well, which means that R*sin(10)=T*sin(21). If you use this and the above equation, you get nonsense. I only have one explanation for this: for the first equation, the weight of the ankle itself also has an Fk rotational force or whatever you call it (leverage?).

    So yeah I really don't know, maybe what I said made you realize something I didn't and can lead you forward:)
  4. Jul 9, 2010 #3
    You got wrong answer because forces exerted different points. Thats why you can't add them as a vectors.

    Imagine foot as a bar. Better as a lever. I've rotated picture for better point of view:

    [PLAIN]http://img85.imageshack.us/img85/8510/footf.jpg [Broken]

    Foot is fixed! So torques added to zero. First use R point as a pivot - you can calculate T. Then back to first point of view and use N point as a pivot and calculate R.

    T (and R) depends from N. N is equal to weight.

    Last edited by a moderator: May 4, 2017
  5. Jul 9, 2010 #4


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    Although it is true that the sum of all the forces is zero, the closed triangle that you drew assumes that R, T and N are all the forces acting on the foot. That is not necessarily true, because there might be friction acting perpendicular to N. Would this person be able to stand on tiptoe in the absence of friction such as on a skateboard?

    I suggest that you solve the problem by saying that the sum of all the vertical forces is zero and that the sum of all the torques is zero. Two equations, two unknowns and what happens in the horizontal direction is irrelevant.
  6. Jul 9, 2010 #5
    @Bartek: With you're way, I got T=1771.52 and R=2415N.
    @kuruman: the above results don't give the sum of vertical forces to be zero.

    And by the way, I think there can't be any friction. I mean yeah sure there is WHILE you're standing to tip-toe, but after that, there can't be, because if there was friction, since there are no other vertical forces applied to the body as a whole, it would be accelerating... So what I got with Bartek's method can't be right either, since they don't give the sum of horizontal forces to be equal either.
  7. Jul 9, 2010 #6
    @Tusike and Bartek: Yea I use the same method and got T=1770 and R=2420N.

    @Bartek: "You got wrong answer because forces exerted different points. Thats why you can't add them as a vectors." Why not? Let's say a table on the ground. The normal force by the ground on the table exerts on the four legs, which balances the weight of the table down from the centre of gravity. You mean we can't add the vectors up in this case too?

    @kuruman: I think there's no friction like what Tusike said. I'm suspecting that the R doesn't include the weight of the ankle, and there's no way to determine the centre of gravity for the ankle because it doesn't have uniform density, so that's the force unaccounted for and therefore I can't use the vector sum method. EDIT: Oh and besides if the friction is perpendicular to N then it would constitute a moment as well. But then what I said above about the centre of gravity for the ankle can't apply either because it'd again generate a moment which we did not account for :s
    Last edited: Jul 9, 2010
  8. Jul 9, 2010 #7
    That's why you have the horizontal components of the tension and the reaction force. The friction force can only act in reaction to some other force on the mass, in this case, it balances out the horizontal components of T and R.

    You cannot have the equilibrium there without the friction, or else you would end up rotating.
    If you take the torque about the pivot point, with your calculated values of R and T, you'll see that you end up with a net torque, that means that you are not in equilibrium, therefore a mechanical equilibrium without friction is impossible in this system.

    That is not true. A force acting at the point of contact, at the pivot point, cannot provide a torque, by the very definition of torque: [tex]\vec \tau = \vec r \times \vec F[/tex]
    If [tex]\vec r[/tex], the distance from the point of application to the pivot point, is 0, then the torque too, is 0.

    Knowledge of the body's distribution of weight and the inner structure of the ankle and foot is irrelevant. Just as you can look at the body as a whole and conclude that N=mg, you can look at the ankle and make conclusions about its stresses (T and R), the normal force N and the torque each of them provides about the toes.
  9. Jul 9, 2010 #8
    But you're working backwards! I can also assumed that the horizontal component of R and T cancel out each other and therefore there is no need for friction to balance the forces? Why can't I assume that? In fact that's what I did when I use the vector-sum method (which is wrong I know but is friction really the reason?) Is it NECESSARY for such a system to have friction (since the question didn't specify)? I can totally imagine a ice ballerina tiptoeing on ice without accelerating forward.

    But what if I take R as the pivot point, which is what I did in one of the steps to calculate the values above. The friction needs to be factored into the calculation and I didn't and still arrive at the supposedly right answer!
  10. Jul 9, 2010 #9
    Of course we can not!

    Table in your example is a rigid body - not a point. You have to use torques with center of mass as a pivot. Naturally if table is symmetric all angles are equal so all forces are equal.

    But imagine, that one of leg is closer center of table than other are. Why tense (force) of this leg increase? There is no reason if we can add force. Tense will be increased because of decreasing of displacement vector. Torques for all legs are still equal.


    @tusike: I had the same. I was amazed because of high values. Than i tried to stand on tiptoe. WITH SO SMALL ANGLE. And now I believe. :smile:
  11. Jul 9, 2010 #10
    So how, may I ask, will the friction provide the equilibrium? :) And btw, R and T are INNER forces, which means that if you look at the body as a whole, they'd be canceled out. So all you'd be left with is mg, and (if you say there is), the friction's force. That'd mean you're accelerating some way (which actually happens when you stand to tip-toe; as I said, there's friction then, but only then).

    I think the answer lies in the fact that that really isn't a good model for standing on your toes. Just try it! There are actually two points where you're feet touches the ground, the tip of you're toes and where they leave your foot.
    The model IS an accurate representation for when you are standing on your heels. And yes, just stand on your heels, and notice that unless you stand in a specific way (meaning your center of mass is just above your heels), you will fall! And the angles given in this problem aren't one of those ways..

    OP: I'm not sure where you got this problem from, but it seems to me it's not looking for a real in-depth analysis of the situation:) I think it's just trying to make you practice all the summaM = 0, torque things or whatever you call them, so I say just use that calculation, and forget about the sum of the forces not being equal to zero, say there's something compensating.
  12. Jul 9, 2010 #11
    You forget that in that situation, the force of the other legs decrease. The sum of all the forces (4 normal ones and gravity) are still 0:)
  13. Jul 9, 2010 #12
    NOTE: You can only take the place where N is (the toe touches the ground) as the pivot point! For example, if you had a car wheel, the center of it is the only place where you could take the pivot point, you can't just take it wherever you want...:)

    The above is not true if the body isn't attached to anything. For example, if you look at a ruler on a table that has been hit, you can take the pivot point for it's rotation wherever you want to calculate it's spin speed.
  14. Jul 9, 2010 #13

    I have calculated a lot of problems whereby the forces do not exert on a single point but as long as the vertical component and horizontal component are both zero then the system is in translational equilibrium. Moment and torque are different thing obviously.

    This is so vexing because I don't know which is the force unaccounted for and I'm still not convinced that it's friction XD
  15. Jul 9, 2010 #14
    I did exactly what Bartek did...

    ... and I got the correct answer. The moment about ANY point for a system in rotational equilibrium is equal to zero. No?
  16. Jul 9, 2010 #15
    Image a long piece of wood pinned to the wall at the top. Now start pushing it where it was pinned, perpendicular to the wood. Will it start to rotate? no. That's because the only place you can have the pivot point is where the wood is pinned, and there the torque would be zero.
    I think the same applies to the ankle, the point where it touches the ground is where it's pinned. Come to think of it, you can't calculate how Bartek did, you can't use N when talking about R and T's torque, simply because N's torque is zero no matter what.
    All you can say is that R*18 = T*cos(11)*25; each other's torque must be equal. And from that, you can only calculate what R/T must be, but not specific values. And what you get won't lead you to anything, because using it, the some of the forces won't be equal to zero.

    "I think the answer lies in the fact that that really isn't a good model for standing on your toes. Just try it! There are actually two points where you're feet touches the ground, the tip of you're toes and where they leave your foot."

    This would offer us a way to have the forces equal zero, while the torque also does. And this is how it does actually work, stand on you're toes real steep and you'll see.
  17. Jul 9, 2010 #16
    Of course... sum of vertical components have to be equal weight: table is in the rest. What about leaning table? Sum of vertical component are still equal, but forces are different.

    You are right if the table is symmetric and all angles have 90 deg.

    But this is only special conditions. Generally if table is a rigid body we have to use torques.
  18. Jul 9, 2010 #17
    Let's stop arguing about this... If an object is at rest, that can only be possible if the forces acting on it are 0, and if the torque's acting on it are 0. Both need to apply. (and even this way, sometimes an object isn't at rest, for example a spinning wheel may spin at a constant speed if all those are true)
  19. Jul 9, 2010 #18
    It is not necessary, if you don't mind rotating without accelerating linearly.
    Calculate the net torque for your special case with no friction, you will see that it is non-zero, and therefore you have not found the values for equilibrium.

    Your answer is only supposedly right. It is in fact, incorrect, as it does not give 0 torque about any axis, as it must for equilibrium to hold.

    Just because you can find specific values for the force under certain assumptions for translational equilibrium, does not mean that rotational equilibrium entails. Just like with the table example, there has to be some load distribution for it not to tip over, and if you move the legs too much, it won't be able to hold up.

    In your table example, you can add the vectors up, but you cannot assume symmetry. Let's analyze that situation for a second. Let's assume, for simplicity, a square table with a side of length [tex]2b[/tex].
    Let's say we displace one of the legs to the center, the others remaining in their places.



    With the numbers standing in for the position of each of the legs.

    You'll be 100% correct in saying [tex]Mg=N_1+N_2+N_3+N_4[/tex], having taken the vector sum, but that won't give you too much information.
    You have to take the torques about some axis to see what the condition for equilibrium is, and you'll have to take frictions into account as well, since without them the system may be unstable.

    Yes, when you look at the body as whole, there's a whole different condition required, for the external reaction force and the weight to be on the same line of force, but granting that separate equilibrium condition, we are now focusing solely on the foot.

    Hurr, isn't that the other way around? You can take any axis for a static situation, since [tex]\Sigma \vec \tau=0[/tex] about any axis in equilibrium, but for a dynamic situation, unless you want to deal with accelerated reference frames, and adjusting your moment of inertia to account for not taking the torques about the pivot or the center of mass, you end up in a mess, don't you? I mean, I think it's valid to find the angular velocity using any axis, but just way too much work compared to the comfy axes of the system.

    I crunched the numbers with a frictional force at the point of contact, and for mechanical equilibrium it must be non-zero, and pointing to the right. I only had to take torques about the point of contact.

    Looking back at the sheet I worked on, I see that the horizontal component equation for Newton's Second Law is redundant, unless we actually want to know the value of the frictional force. It is sufficient to look at the vertical components and the torque about the pivot point.
    Looking at the torque about T or R would only complicate things by forcing us to consider the nature and value of the frictional force.

    Last edited: Jul 9, 2010
  20. Jul 9, 2010 #19


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    Tell 'em, RoyalCat!!
  21. Jul 9, 2010 #20
    I didn't say anything about whether it's dynamic or static. All I'm saying is that if e.g. you hit a ruler on a table, you can take the axis wherever you want; but hit a ruler that's been nailed down to the table (in such a way that it can still rotate), you can only take the axis to be where the nail is.

    And OK, let's say there is friction. Here's the problem I have using this idea:
    Forces acting on the ankle:
    R, T, N, and friction. (i'm not sure about gravity's pull of the ankle's own weight, but I think it should be included)

    Forces acting on the whole body:
    R, -R, T, -R -these all cancel each other out
    N, mg - these two cancel each other out, since the body isn't moving vertically.
    That leaves us with the friction. We're supposed to be left with nothing, since the body is in equilibrium. Or am I forgetting about a force acting on the whole body that cancels this friction out?
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