# Mass gets cancelled out on downhill slope

1. Jun 8, 2013

### e-zero

I'm looking at a general problem in which a skier is going downhill. I understand 'mathematically' how the mass will cancel out of the final equation when calculating the acceleration, but I don't understand this 'conceptually'. Can anyone describe?

2. Jun 8, 2013

### Staff: Mentor

Start by reading through this FAQ from the "General Physics" forum, see if that helps. If not, come back and ask again

3. Jun 8, 2013

### HallsofIvy

Staff Emeritus
I would not say that mass cancels out in the 'final equation' I would say that it cancels out right at the beginning: The gravitational force is $F=-GmM/R^2$ while acceleration satisfies a= F/m so that a= -GM/R^2. That includes the mass of the earth but not the skier's mass.

If we were to include friction, that might not be true. What equations are you talking about?

4. Jun 8, 2013

### HomogenousCow

The conceptual idea is that all masses accelerate the same in a gravitational field, which is a consequence of the equivalence of inertial mass and gravitational mass.
The whole thing is much more intuitive and elegant in general relativity, where inertial motion is replaced by geodesic motion through a curved background geometry.

5. Jun 8, 2013

### e-zero

Ok, I might as well post the question: A skier has begun descending a 30deg slope. Assuming the coefficient of kinetic friction is 0.10, calculate the skier's acceleration.

The diagram shows that the 30deg angle is at the bottom of the slope where the slope meets the ground.

Forumlas for force of gravity on skier:
Fgx = mg sinθ
Fgy = -mg cosθ

Normal vector gets calculated to:
Fn = mg cosθ

Net Force equations are:
mg sinθ - μk Fn = m ax
Fn - mg cosθ = m ay = 0
(ax and ay are the x and y components of acceleration)

Therefore we can solve for 'm ax':
mg sinθ - μk(mg cosθ) = m ax
(This is the step where 'm' (mass) cancels out)

So I get how this is done mathematically, but I wanted a conceptual understanding of why the mass cancels out in this example. Any input?

6. Jun 8, 2013

### Simon Bridge

What was wrong with the answers so far?

7. Jun 8, 2013

### e-zero

Actually after looking at the examples again, I kinda of get it and I realize the ideas and formulas in the examples is what I will be learning more of in the following chapters.

8. Jun 8, 2013

### Simon Bridge

That's likely - though you've already met F=mg close to the Earth's surface.
All that's happened here is that all the forces are proportional to m.

But imagine the skier is towing a toboggan by a long rope, and the rope is looped around a tree uphill so when the skier travels a distance x downhill, the toboggan goes a distance x uphill. The skier mass is M and the toboggan mass is m < M. Now do the math.

Note:
on your specific problem description - you did well, but you should specify the x and y directions as well.
i.e. "+x is downhill along the slope"

9. Jun 10, 2013

### atyy

HallsofIvy and HomogenousCow have given the answer, but it is so important I want to tell you to read their posts carefully. Imagine what would happen in HallsofIvy's equations if the "m" in the law of universal gravitation were different from the "m" in the second law of motion. The former is what HomogenousCow calls "gravitational mass" and the latter is the "inertial mass". The equality of inertial and gravitational mass is called the "weak equivalence principle".

The one qualification to make is that for the cancelling out, we assume that one of the masses is very much less than the other, so that to excellent approximation the small mass can be thought of as "falling freely" in the gravitational field of the larger mass, but the larger mass is not affected by the gravitational field of the small mass. So things will be more complicated if the two masses are more equal.

Last edited: Jun 10, 2013
10. Jun 11, 2013

All the responses here seem to rely on the math, but the concept here is really simple even if you don't know the equations that govern this system. In fact, the equations come from simple intuitive intuitive concepts, rather than the other way around.

To explain, if you have two skiers the same mass on the same skis skiing next to each other, of course they go the same speed down the slope. Then imagine they hold hands. You can now think of them as one skier with twice the mass — and they're going the same speed.

Now you might complain that they're on four skis instead of two. So imagine just one skier again, and imagine she alternates between skiing on both skis and skiing on only one. Now, this doesn't work perfectly (because the pressure actually changes the coefficient of friction by changing the heat transfer and the way the skis alter the crystals in the snow), but the basic idea is that there's *half* the surface area to slow the skier down, but there's *twice* the normal force. So those cancel out.

So put it together, and have two skiers, each on one ski, holding hands, and their speed is the same as for one skier twice the weight on two skis, or each individual skier on two skis, or each individual skier on one ski.

tl;dr: Imagine two skiers holding hands.

11. Jun 11, 2013

### Simon Bridge

@AlyssaSkier: excellent debut post - welcome to PF.

It is easy to get caught up in the math when you live in the physics world - you can neglect other forms of communication.

The picture you provide shows that it "aught to be" intuitive that the acceleration does not depend on the mass of the skier, but does not show the concept behind how that comes to be the case nor does it necessarily lead the the "m"'s cancelling out in the algebra in the manner asked about in post #1. I suspect that is why everyone opted for "inertial mass is the same as gravitational mass" as the statement of underlying concept.

It's also common in beginning students that they will accept two equal-mass skiers with the same acceleration side by side and even holding hands, but, should one skier jump into the other's arms, their intuition says, the acceleration should be different. Being wrapped in each others arms is different from just holding hands - ask anybody ;)

But at least it shows how to get there without having the result fall out in the algebra.

12. Jun 12, 2013

### atyy

Yes, that's a great way of saying it.

In jargony terms, it's called "universality of free fall", and is a form of the "weak equivalence principle".

It's the observation that heavy and light objects dropped from the same height reach the ground at the same time (assuming air resistance can be ignored, and as long as both falling masses are much smaller than the earth's mass).

13. Jun 12, 2013

### WannabeNewton

The equivalence of inertial and gravitational mass has been verified to great accuracy via experiment. See e.g. the famous Eotvos experiment. I have yet to see a first principles proof of it, be it conceptual or mathematical. Conceptual arguments that pertain to certain earth bound situations can certainly be helpful in understanding those particular situations but they do not prove the equivalence in full generality.

14. Jun 12, 2013

### Simon Bridge

I thought it was empirical?

15. Jun 12, 2013

### WannabeNewton

That's my understanding as well, which is why I was saying conceptual arguments of the above sort are great for specific cases but they don't do much to prove to a person why gravitational and inertial mass are equivalent. If someone wanted a justification for it they would have to turn to experiment.

16. Jun 12, 2013

### Simon Bridge

If it is, fundamentally, a synthetic statement, then it cannot be proved.
Anyone looking for a proof is going to be disappointed aren't they?

Besides, I'm not sure that OP was actually asking for a proof... doesn't post #1 specifically ask for a conceptual understanding?

I guess I don't understand the point you are making: could you clarify please?

17. Jun 12, 2013

### WannabeNewton

I was talking about post #10 (specifically the first paragraph of post #10) in relation to posts #4 and #3 and saying that while post #10 provides a conceptual argument for the specific physical situation it deals with, it does not explain why inertial and gravitational mass are equivalent in general. This is something that must be accepted by virtue of experiment. Perhaps I should have quoted what I was referring to so as to have not creating any ambiguity.

18. Jun 12, 2013

### Simon Bridge

Ah - gotcha... I had noticed that too.
Of course, it doesn't have to.

Yeh - it wasn't clear if you were saying that you thought that the truth of the equivalence principle could be known a-priori ... which, if so, would end a long-standing philosophical argument.

Many students seem to think that everything should be provable from first principles: the post would have worked with a short sentence to the effect that
.

To be fair - post #10 is higher than the usual calibre of debut posts.

19. Jun 12, 2013

### WannabeNewton

Oh heavens no. In fact the Eotvos experiment is probably one of the coolest experiments ever! What's the fun if we know things a-priori :tongue:

Oh for sure, post #10 was nothing short of great. I was just saying that something along the lines of post #9 in particular would suffice in explaining, in general, why mass doesn't matter for free fall in a gravitational field.