Moment and forces for a person standing on tiptoe

arkofnoah

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?

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Tusike

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:)

Bartek

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?
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.

regards

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kuruman

Homework Helper
Gold Member
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.

Tusike

@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.

arkofnoah

@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

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RoyalCat

@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.
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.

@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
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: $$\vec \tau = \vec r \times \vec F$$
If $$\vec r$$, 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.

arkofnoah

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.
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.

A force acting at the point of contact, at the pivot point, cannot provide a torque, by the very definition of torque:
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!

Bartek

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?
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.

regards

@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.

Tusike

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.
A force acting at the point of contact, at the pivot point, cannot provide a torque
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.

Tusike

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.
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:)

Tusike

But what if I take R as the pivot point
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.

arkofnoah

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:)
Exactly!

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

arkofnoah

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.
I did exactly what Bartek did...

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.
... and I got the correct answer. The moment about ANY point for a system in rotational equilibrium is equal to zero. No?

Tusike

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.

Bartek

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:)
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.

Tusike

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)

RoyalCat

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!
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 $$2b$$.
Let's say we displace one of the legs to the center, the others remaining in their places.

Diagram:

1-------
|||||||||
||||4||||
|||||||||
2-------3

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

You'll be 100% correct in saying $$Mg=N_1+N_2+N_3+N_4$$, 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.

Tusike said:
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..
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.

Tusike said:
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.
Hurr, isn't that the other way around? You can take any axis for a static situation, since $$\Sigma \vec \tau=0$$ 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.

Tusike said:
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) [/tex]

I disagree. :) Arguing is the quintessence of science and learning. Unless we pit all the ideas against each-other and pick out only those that survive, we wouldn't be making much progress, just stagnating.

Anyway, the situation you referred to is called dynamical equilibrium. Where the components in a system are moving with constant velocities but not accelerating. Remember Newton's First Law, an object in motion tends to stay in motion unless acted upon by an outside force.

That statement defines force in the way Newton's Second Law defines it, $$\vec F= m\vec a$$, it is what changes the velocity of an object with mass.

If you wish to look at it mathematically, $$F=ma = m\frac{dv}{dt}$$
$$\Delta v=\int_{t_0}^{t'}a \cdot dt = \int_{v_0}^{v'} dv=v'-v_0$$
Obviously, if the acceleration is 0, then the first integral is 0, implying $$v'=v_0$$, that is, the velocity goes unchanged.

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kuruman

Homework Helper
Gold Member
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.
Tell 'em, RoyalCat!!

Tusike

Hurr, isn't that the other way around? You can take any axis for a static situation, since LaTeX Code: \\Sigma \\vec \\tau=0 about any axis in equilibrium, but for a dynamic situation
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?

RoyalCat

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?
You could include the weight of the foot, but that would only complicate the algebra. It is negligible in comparison with the other forces at play. (Let's say a foot weighs one kilogram, http://wiki.answers.com/Q/How_much_does_a_human_foot_weigh I can't believe google had the answer to that.) That means that the force of gravity on it is approximately 10 N, that's two orders of magnitude smaller than everything else at play, except friction, but that's on another axis altogether, so the point still stands.

R and T do NOT cancel each other out. You assumed that without any justification.
You are not allowed to take the vector sum of some of the forces and claim that it is equal to 0 at equilibrium. That fails to consider the other forces at play, and gives you nonsense. :)

The way to state the equilibrium conditions for simple mechanical systems is to apply Newton's Second Law according to how many independent coordinates you have in the system.
In our case, we need two independent mutually perpendicular axes, and an equation for the rotation. In 3D we'd need another equation for the third dimension and so on.

There are two good choices for the coordinate system here. The first is with y along the direction of N, and x perpendicular to it.
The other is with y along the direction of R with x along the direction of the rod. Either way the results are the same. The question is just which trigonometry you end up using.

If you're still getting mixed up results, I'll be glad to post my solution.

EDIT:
I think I see now what you've done. You looked at R, T as forces internal to the body, that thus, do not act on it externally to produce any acceleration, and included mg as a force acting from the center of gravity. This is a legitimate approach, but what it does is ignore the question you were asked.
As you've pointed out, N=mg, and if we assume friction exists, then it must be 0 for equilibrium to be established.
If you now look at the torque about the toes, you'll see that this situation is possible provided one condition is satisfied, and that is that the center of gravity of the body is directly above the toes. That way there is no torque and equilibrium is achieved.

Oh. Now I see something. The analysis of the foot alone, assuming a horizontal force at the point of contact shows that there must be a force there to the right.
So now we've found all the forces at the point of contact, the normal force and the horizontal force there. Now if we say that this force is friction, that is to say, external to the body, then yes, there will be a net acceleration.
What we're left to conclude is that there cannot be a frictional force there as it would cause the body as a whole to accelerate. (Note that having two points of contact does not resolve this problem, consider the case where the center of gravity is directly above one of them, you still need a friction to overcome the internal horizontal stresses, and you're left back at square one.)

I can't find an answer to this problem..

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kuruman

Homework Helper
Gold Member
One more time. You have two unknowns, R and T, therefore you need two equations. One is not sufficient. Three is too many and unless you introduce a third unknown, your system of equations will be overdetermined. Now then, which two equations do you think you should write down?

Tusike

R and T do NOT cancel each other out. You assumed that without any justification.
I didn't say they do. I said that a tendon somewhere in the body is the cause of R, and another is for T (or bone or whatever). The point is, that whatever's causing it, is inside the body itself. Therefore, if you look at the body as a whole, and not just the ankle, R and T appear nowhere because of R' and T', which cancel them out. (Newton's third law). It's the same thing when you try to lift yourself. Suppose you act a force R with your arm on an object in a way that the reaction would make you move up. Now, if that object was an outside object (e.g. pullup-bar), you'd be moving up. But if that other object was also you, say, you pushed your knee, you wouldn't move up, you'd stay in place (well, your feet wouldn't leave the ground that's for sure).

Tusike

@kuruman: here are two equations:
Sum of torques is 0, which would mean:
(1) 18*R = 25 * T*cos(11)

Sum of forces is 0, son vertically:
(2) T*cos(21) - R*cos(10) + N = 0
and horizontally:
(3) T*sin(21) - R*sin(10) = 0

Divide (1)/(3), you get nonsense. If we somehow manage to solve the friction thing, and say there is friction so (3) doesn't apply, from (1) and (2) we get:
T=1678,65N
R=2288.63N

Now I'm very happy with this and I'm hope it's right and everything, I just don't see why there would be friction taking care of the horizontal components... Because with these values, (3) wouldn't be true...

kuruman

Homework Helper
Gold Member
@kuruman: here are two equations:
Sum of torques is 0, which would mean:
(1) 18*R = 25 * T*cos(11)

Sum of forces is 0, son vertically:
(2) T*cos(21) - R*cos(10) + N = 0
and horizontally:
(3) T*sin(21) - R*sin(10) = 0

Divide (1)/(3), you get nonsense. If we somehow manage to solve the friction thing, and say there is friction so (3) doesn't apply, from (1) and (2) we get:
T=1678,65N
R=2288.63N

Now I'm very happy with this and I'm hope it's right and everything, I just don't see why there would be friction taking care of the horizontal components... Because with these values, (3) wouldn't be true...
I agree with your values for R and T. Now for the third equation. If you introduce friction, you get

T*sin(21) - R*sin(10) + f= 0

Knowing the values for R and T, you can find (if you wish) how much f is needed and in what direction to keep this thing in equilibrium in the horizontal direction too. Although the problem does not ask for it, friction is there. Static friction, like the normal force, is a contact force that is whatever is necessary to provide the observed acceleration, in this case zero.

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