Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question already has an answer here:

Is it true that a heavier skier goes faster? If it is, why is that?

My intuition would be that the speed gained by a skier should be independent from its mass, since both its acceleration and the friction are proportional to its mass. Am I overlooking some important effect which would cause a heavier skier to go faster?

share|improve this question

marked as duplicate by Waffle's Crazy Peanut, John Rennie, Qmechanic Apr 2 '13 at 15:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers 2

up vote 0 down vote accepted

I'm quite sure that you've left something. The more their mass, they've got more momentum and of course, Inertia. The heavier the body is, the more their body resists their state of motion (or rest).


Yes. They move faster than skinny guys. If you resolve the forces acting on the skier like his acceleration, gravity, friction (of snow), the normal force and also his air drag, you can find that his velocity does depend on his mass.

Resolving all the forces, we get

$$ma=mg\ cos\theta-\mu\ mg\ sin\theta-\frac{1}{2}\rho Av^2C_d$$

$$a+\frac{1}{2m}\rho Av^2C_d=g\ cos\theta-\mu\ g\ sin\theta$$

As the mass is in the denominator portion, we can clearly see that it supports his acceleration (it should increase in order for balancing other forces) by opposing gravity.

share|improve this answer
1  
I don't seek a reference, only a confirmation or a good motivation for the other opinion. Anyway, I am not convinced by your statement. If we take into account only gravity and friction, then the speed is independent from mass. I am asking if there are other effects making the speed mass-dependent or if the speed is in fact independent from the mass of the skier. –  Daniel Robert-Nicoud Apr 2 '13 at 14:39
    
@DanielRobert-Nicoud: Hi Daniel. Well, you've taken into account of only friction and gravity. But, you didn't take take drag into account. That makes the velocity mass-dependent... –  Waffle's Crazy Peanut Apr 2 '13 at 14:53
    
"If you resolve the forces acting on the skier like his acceleration (ma),..." The product ma is not an individual force acting on the skier. It equals the total force on the skier. –  Ben Crowell Apr 2 '13 at 15:18
    
@BenCrowell: Hi Ben. I've revised my answer for the resolution of forces. You're right that it equals all the other forces. But I'm quite sure that while balancing these forces, we can show that it depends on the initial velocity of the skier ;-) –  Waffle's Crazy Peanut Apr 2 '13 at 16:37
1  
You've got your trig functions messed up. The kinetic frictional force is collinear with the force of air friction, so you can't have a trig factor multiplying one and not the other. Also, if $\theta$ is meant to be the angle of the slope relative to horizontal, then you don't recover the correct result for $\theta=0$ or 90 degrees when friction is absent. And as discussed in my answer, (1) I don't think it's realistic to assume that friction is given by $\mu_kF_N$, and (2) $A$ isn't constant. –  Ben Crowell Apr 2 '13 at 19:54

The skier has the following forces acting on him:

  • a normal force $F_N$
  • a gravitational force $F_W=mg$
  • a force from air friction that is probably well approximated as $F_a=cAv^2$, where $A$ is the skier's cross-sectional area
  • a frictional force from the snow $F_f$

$F_a$ increases with speed, and $F_f$ may also have some velocity dependence as well, and this is why a terminal velocity exists. The terminal velocity is the one at which the vector sum of these forces is zero.

If $F_f$ is negligible, then clearly a heavier skier will have a greater $v$, because we need a bigger $F_a$ to cancel out the component of $F_W$ parallel to the slope.

If $F_f$ obeys the usual freshman-physics model of kinetic friction, then it's velocity-independent, and a simple calculation shows that the same conclusion holds for fixed $A$: a heavier skier goes faster. In reality, a bigger skier has a bigger $A$ as well. However, I don't think the qualitative result is altered, because under geometrical scaling, $m$ grows faster than $A$.

However, I don't think it's necessarily valid to assume that $F_f$ obeys the usual model of kinetic friction, which is usually found to be valid only when a solid, dry surface slides over another solid, dry surface. Here the skier is leaving a track by pushing aside the snow (unless the conditions are very icy). I would guess that something like $F_f\propto F_N v$ holds. If this form of $F_f$ applies and $F_a$ is absent, then all forces are proportional to mass, and the terminal velocity is independent of mass. If this form of $F_f$ exists along with an $F_a$, then I think you'd come back to the same result, which is that heavier skiers would go faster.

But the whole thing really depends on the model of friction, and I don't think any one model of friction will be accurate for a variety of conditions (ice, deep powder, ...).

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.