Homework Help: One horse starts moving or slips

1. Mar 29, 2014

mcastillo356

1. The problem statement, all variables and given/known data
Hello. I've got three doubts I can't solve by myself: ¿which is the force a horse must make with its hoofs to start a movement forward, and what force is needed to slip?. And the third question is what is the diagram of forces in these two cases?.

2. Relevant equations
Newton's laws, force of friction (fe maxeFn), normal force (Fn), force of gravity (Fg), and horse's applied force(Fh).

3. The attempt at a solution
I Know that two horizontal forces are involved: the force made by the horse, and the friction of the ground. But I really don't know how to apply Newton's third law, and explain whose are these two forces' reaction forces; and I don't know to argue when does the horse start movement, and when does slip.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 30, 2014

tycoon515

In order for the horse to start moving forward, a net force must act on it whose horizontal component acts in the forward direction. The horse applies a push force against the ground surface, $F$$H$, at the angle $\alpha$ its legs make with the ground. This vector has a horizontal component that acts backward, $-F$$h$ $cos$ $\alpha$ as well as a vertical component that acts downward, $-F$$H$ $sin$ $\alpha$. The normal force exerted by the ground on the horse's hooves, $F$$N$$=F$$W$ $+$ $F$$H$ $sin$ $\alpha$ is what determines the amount of static friction the horse has available to start moving. Static friction is always equal to $F$$H$ $cos$ $\alpha$ up to $μ$$s$ $F$$N$. Setting these two expressions equal to each other and simplifying will let you solve for the maximum value of $F$$H$ before the horse starts to slip. Also, provided that $\alpha\neq90$, any value of $F$$H$ up to the result determined in the previous equation will cause forward motion. Furthermore, if it is known that $\mu$$k$ is not zero, then any value of $F$$H$ will cause the horse to move forward since even if it does cause slippage immediately, that slippage will still result in a forward kinetic friction force.

3. Mar 30, 2014

mcastillo356

¡Perfect!. Last question: $F_{H}\mathcal{cos}\alpha$ must be less or equal to $\mu_{s}F_{N}$. And if $F_Hcos\alpha$ is bigger, the horse slips. But I find difficult to match this with Newton's third law: if $F_Hcos\alpha$ is bigger, ¿which is the opposite and equal force?

Thanks

4. Mar 30, 2014

haruspex

In the absence of a diagram or clear description, I have to guess what the set up is.
FH appears to be the compression force in the horse's legs, exerted at angle α to the horizontal. FW is clearly the weight of the horse. Is the horse perhaps pulling a load? Say that's a horizontal force L, and put Ff for the frictional force between hooves and ground.
In equilibrium we have:

L = Ff = FH cos α <= μsFN
FN = FW = FH sin α (not FW + FH sin α ).

It will only get to exceed μsFN briefly. While it does, the horse's hooves accelerate at a rate corresponding to that difference divided by their mass. From then on, the horse simply cannot exert a force greater than μkFN.

5. Apr 1, 2014

mcastillo356

haruspex, a horse is capable to exert on the ground more force than it's own weight, so tycoon515 is right.

tycoon515's words not only have explained this question; they've made me think twice about something: horse's legs are better designed than human being's. Anyhow, our evolutive asset is not to be fast.

Thanks!

6. Apr 1, 2014

haruspex

You have not confirmed or denied my guess at the set-up, so I still assume the horse is pulling horizontally on a load. In that case, the horse cannot exert a vertical force on the ground greater than its own weight (unless it jumps!). The total force is greater since it is the vector sum of the normal force (the horse's weight) and the frictional force). In magnitudes (the signs depend on perspective):
L = Ff = FH cos α <= μsFN
FN = FW = FH sin α
where
FW = weight of horse
FN = normal force
FH = force along leg
α = angle of leg to horizontal
Ff = frictional force (the horizontal force from the ground)
So the total force the horse exerts on the ground is FH = √(FW2+L2) >= FW

If the load force is not horizontal, the horse may be pulling upwards on the load. In that case, yes, the vertical force from the ground will exceed the weight of the horse. But tycoon515's equation would still be wrong. If the load is at angle β to the horizontal then we have:
FN = FH sin α = FW + L sin(β)

7. Apr 2, 2014

mcastillo356

I must confess I am an absolute beginner. And I may have not stated well my doubts. There is no load in the case I wanted to solve.

I didn't understand your previous message. And I've read more carefully your last one. I find perfect your reasoning for the case of a load on the horse.

I hope you agree with tycoon515 for the case there is no load.

8. Apr 2, 2014

haruspex

No, FN = FW + FH sin α is wrong in all cases. FH sin α includes a contribution from FW.
Think of what's happening where the hooves meet the ground. The force the legs exert there is a consequence of the horse's weight, not additional to it.

For the no load case, just set L=0 in the equations I posted. But you wish to consider the horse starting to move, so now we need an acceleration term instead. It just replaces the load:

Ma = Ff = FH cos α <= μsFN
FN = FW = FH sin α

9. Apr 2, 2014

mcastillo356

And how do you express normal force $F_N$ if the horse jumps?.

10. Apr 2, 2014

haruspex

While the horse is accelerating upwards, the normal force will exceed the horse's weight accordingly: Ma = ∑F.

11. Apr 3, 2014

mcastillo356

Excuse the delay answering. ¿Don't you think that, if a horse starts moving, it can exert on the ground a force whose vertical component exceeds its weight?. If you answer is affirmative, then you agreee with tycoon515.

12. Apr 3, 2014

haruspex

If it exerts a vertical force greater than its own weight then it will accelerate upwards: May = ∑Fy. That does not agree with what tycoon515 posted.

13. Apr 4, 2014

mcastillo356

So, which is $\sum{F_y}$?

14. Apr 4, 2014

haruspex

For the horse, taking upwards positive, $\sum{F_y} = F_H \sin \alpha - Mg$

15. Apr 6, 2014

mcastillo356

Fine!!.

tycoon515's approach was not accurate.

First of all I must beg you a pardon: I haven't trust you until now. Moreover, I've read all your posts more carefully, and there is no trace of mistake, there where I found contradictions. My lasts posts' intention was to reveal them. You probably have noticed that intention. It's been unfortunate. I really thought I had understood it before you started posting. I appreciate not only your knowledge, but also your patience.

Hope I find your help next time.

Thank you very much!!

16. Apr 6, 2014

haruspex

17. Apr 12, 2014

tycoon515

I'm sorry; I did not realize the horse was pulling a load. That changes things. But assuming there is no load, I do not see how my original approach is inaccurate. The normal force between the horse's hooves and the ground, $F$$N$, and the coefficient of static friction between its hooves and the ground, $\mu$$s$, determine the maximum amount of friction available for the horse to start moving forward. $F$$N$ is the sum of the horse's weight, $F$$W$, and the vertical component of the force its legs apply at an angle $\alpha$ with the ground, $F$$A$ $sin$ $\alpha$. Now applying Newton's Third Law, any force that the horse exerts on the ground, the ground exerts on the horse but in the opposite direction. The horizontal component of the horse's applied force, $F$$A$ $cos$ $\alpha$, acts backward, but the maximum force the horse can exert on the ground parallel to the ground (i.e. via static friction) is equal to $\mu$$s$$F$$N$. As long as $F$$A$ $cos$ $\alpha$ $\leq$ $\mu$$s$$F$$N$, the ground exerts a forward force of magnitude $F$$A$ $cos$ $\alpha$ on the horse's hooves via static friction, and the horse will accelerate forward. However, if $F$$A$ $cos$ $\alpha$ exceeds $\mu$$s$$F$$N$, then there is force left over that the ground cannot oppose, causing the horse's hooves to accelerate backward (i.e. slipping). Even after slipping, as long as the coefficient of kinetic friction between the horse's hooves and the ground, $\mu$$k$, is greater than $0$, then kinetic friction will enable the ground to exert a forward force of magnitude $\mu$$k$$F$$N$ on the horse's hooves, and the horse will accelerate forward. Concerning the vertical direction, $F$$N$ is the total force the horse's hooves exert on the ground, and the ground pushes back up with a force of magnitude $F$$N$ on the horse's hooves, and if $F$$N$ exceeds $F$$W$, then the horse will accelerate upward. I did not mention this last bit in my original response because I thought the question was about the horse's forward motion only.

18. Apr 12, 2014

haruspex

That's where you're going wrong.
Draw a diagram with the ground, legs and horse's torso as separate items, ignoring the mass of the legs. If the vertical component of the compression in the legs is FL, you have two equations relating it to other forces: one at the horse, one at the ground.
At the horse, FL = FW; at the ground FL=FN.

19. Apr 14, 2014

tycoon515

But the vertical component of the force vector the horse's legs exert on the ground is not equal to the horse's weight. The horse must apply additional force to the ground and therefore additional vertical force to the ground in order to initiate forward motion (besides the case where its hooves push parallel to the ground), so it accelerates upward (briefly). Why does it accelerate upward? Because the ground applies a greater force upward on it than the gravitational force the Earth exerts on it. By Newton's Third Law, the upward force exerted by the ground on the horse is equal in magnitude to the downward force the horse exerts on the ground through its hooves, or as you call it, the vertical component of compression in the legs. I will agree that the normal force is equal to the vertical component of the compression in the horse's legs/hooves, but the vertical component of the compression force is the sum of the horse's weight and any additional force it applies downward.

20. Apr 14, 2014

haruspex

No, it must apply a forward force to get forward motion. Think of a car moving from rest. Does it rise up before it starts going forward?

21. Apr 16, 2014

tycoon515

When a car starts moving forward, it's because the road applies a forward force on it via static friction with the tires. By Newton's 3rd Law, this force is paired with the backward force the tires exert on the road. This force acts parallel to the road surface since the rubber in contact with the road surface pushes in a direction tangent to the tire.

The horse initiates forward motion via static friction as well, but it's not an identical situation. The ground needs to apply a force on it that has a horizontal component acting forward. By Newton's Third Law, the horse needs to exert a force on the ground with a horizontal component acting in the backward direction. One way it could do this is by pushing parallel to the ground. This would require that its lower legs rest on the ground and push backward without imparting a vertical component. This is not the natural way for a horse to start moving. More likely, its legs will be nearly upright, making some angle with the ground. Before it even applies any effort, the vertical component of the compression force in its legs is equal to the normal force, which is equal to its weight, due to the ground pushing up with just enough normal force to counteract the Earth's gravitational pull. Once the horse applies effort through its legs to the ground, the ground pushes back with a reaction force whose vertical component is just great enough to counteract the new, greater value of normal force. We know this because the angle at which the applied force vector is exerted guarantees that the applied force can be resolved into horizontal and vertical components. The vertical component of the applied force adds to the horse's weight to produce a greater normal force, and the horizontal component of the applied force is what may enable the horse to start moving forward. The amount of force the horse can exert parallel to the ground is at most equal to the coefficient of static friction between the horse's hooves and the ground times the normal force. Any horizontally applied force up to this value will be in a Newton's Third Law pair with the horizontal component of the reaction force exerted by the ground on the horse's hooves, which is the net force that causes the horse to accelerate forward. Any horizontally applied force beyond the static threshold is not actually exerted on the ground but rather causes the horse's hooves to accelerate backward (i.e. slipping), at which point kinetic friction takes over, provided the coefficient of kinetic friction between the horse's hooves and the ground is not zero. Regardless of whether the horse slips, the new normal force acting on it is greater than the Earth's gravitational pull on it, so there is momentarily a net upward force as well, causing the horse to accelerate upward. This is why horses gallop instead of sliding along the ground.

22. Apr 16, 2014

haruspex

Which is exactly the same as for a car's wheels.
If it's standing perfectly upright, how do you propose the horse is going to exert an additional vertical force? Stand on tippy-toe?
You have no basis for claiming the horse cannot simply add a horizontal component to the force it exerts on the ground. If I, standing, shift my weight to one leg, I can move the other other foot forward. The first vertical movement of my centre of gravity can be downwards, not upwards.
It can work exactly as with a car tyre. The extra force I add is a torque, making my trailing leg tend to swing backwards, as my leading leg swings forwards.

23. Apr 17, 2014

tycoon515

The normal force between the car tire and the road is equal to the car's weight. The normal force does not change when the car starts moving forward since the tire applies a force parallel to the road only.

The horse's legs are not perfectly upright. As I said, they make an angle with the ground. It is true that that the horse's center of mass would have to drop a little to enable its legs to bend and make an angle with the ground, during which time the normal force is less than its weight. Once its legs do make an angle with the ground, it can apply a force through its legs to the ground that can be resolved into horizontal and vertical components. All I'm saying is that this is how a horse would start moving. Any time a real horse has started moving forward on its own on a level surface, the normal force exerted by its hooves on the ground has exceeded its weight. I suppose it's just a conjecture until I perform an experiment with a horse starting to move while standing on four scales. These scales would measure normal force or apparent weight. You'd probably be right that its center of mass would displace downward initially, and the scales would read less than the horse's weight during this brief interval. But regardless of how it achieves this, the scales will read higher than its weight as it starts moving forward.

24. Apr 17, 2014

haruspex

Are you saying that's how horses stand, or merely how your imaginary horse stand?
I'm sure it is possible for a horse to start moving in such a way that there is an initial rise in centre of gravity, but equally it is possible that they don't, or even that the centre of gravity drops initially.

Look how this dialogue has shifted. It started with your completely wrong equation in which you wrote that the normal force on the ground was the sum of the horse's weight and the thrust in its legs. Then it was your claim that when a horse starts moving the normal force exceeds its weight. Now it's down to, if the horse is standing in a certain way then there will be an increase in force.
I really think this has outlived its value.