# Physics sled question?

1. Aug 21, 2014

### fightboy

Two children, Roberto (mass 35.0 kg) and Mary (mass 30.0 kg), go out sledding one winter day. Roberto sits on the sled of mass 5.00 kg, and Mary gives the sled a forward push. The coefficients of friction between Roberto and the upper surface of the sled are μs=0.300 and μk=0.200; the friction force between the sled and the icy ground is so small that we can ignore it. Mary finds that if she pushes too hard on the sled, Roberto slides towards the back of the sled. What is the maximum pushing force she can exert without this happening?

Ok so for this problem I actually got the final answer correct (118 N) but was wondering if the way I attempted the problem is correct since my textbook did it differently. So the way my textbook went through the problem was consider Roberto by himself first, and then Roberto and the sled as one unit, two objects all together. To solve they went through this lengthy process first finding Roberto's maximum acceleration then substituting the acceleration expression into the x component expression of the sled and Roberto as one unit, ending up with the final equation FMary on unit=munitμsg.

The way I attempted the problem is a lot easier imo. I treated Roberto and the sled as one single unit from the get go. I basically only used two equations: ∑Fext,y= normal force -392N= 0
with -392N being the magnitude of the weight in the negative y direction. From this I solved for n (normal force) and plugged the value into the static friction equation, fs,maxsn= (0.300)(392N)=118N.
From this value i concluded that the force cannot be larger than maximum static friction value, or else Roberto would begin to slip back, so 118N is the correct answer.
Both ways of doing it lead to the right answer, but is my way of going about the problem correct or did I just coincidentally get the correct answer? Is it necessary to go back and try the problem using the book's method?

2. Aug 21, 2014

### Nathanael

392N is the normal force between the sled+person (as a single object) and the ground. It doesn't make sense to use this value in the static friction equation, since the coefficient of static friction is measured between the person and the sled (and not between the sled and the ground).

If you look at the person as an isolated object, the force that is accelerating the person is the force of static friction. This static friction has a limit to the maximum force it can apply (depending on the normal force between the person and the sled) and therefore there's a limit to the maximum acceleration of the person (without slipping). If you find that maximum acceleration of the person you can use it to find the maximum force that the the girl can apply (without him slipping).

3. Aug 21, 2014

### Simon Bridge

The text did not treat Roberto and the sled as a single unit because Roberto may slide wrt the sled.
i.e. the problem is set up with Roberto and the sled as separate items.

It's not good enough to get the correct answer in the end, you have to be able to say why it is the correct answer.
i.e. how did you know to apply the friction between Roberto and the sled to the total weight of Roberto and the sled? After all, Roberto only presses on the sled with his own weight. The total weight presses on the ice - which we are told has zero friction.

I bet the text did it like this:

The forces on Roberto are: friction from the sled, and, nothing else.
We know this is the maximum that still lets him sit still so:
$\mu_s Mg = Ma$ where M is Roberto's mass.

The forces on the sled are the applied force F from the push, and friction from Roberto's bottom.
These forces oppose each other, and we know F is in the direction of the resulting acceleration.
$F-\mu_s Mg = ma$ where m is the sled mass.

We know the sled and Roberto must have the same acceleration because Roberto does not slide.
These two equations have to be true simultaneously! There are two unknowns.

It turns out that $F=\mu_s(M+m)g$ ... which is the combined weight on the ice scaled by the friction coefficient between Roberto and the sled. The problem is basically getting you to this result - which you seem to have used accidentally.

Last edited: Aug 21, 2014
4. Aug 22, 2014

### haruspex

The reason that the OP method happened to give the right answer might be more evident by thinking in terms of accelerations. The maximum lateral acceleration that can be imparted to Roberto is μsg, so the maximum force on Roberto+sled = μsg(M+m).
Interesting that Mary's mass is given. Could be just to confuse, or maybe there are more parts to the question.