What Is the Maximum Force Mary Can Apply Without Roberto Sliding Off the Sled?

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In summary, 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. The maximum pushing force that Mary can exert without Roberto sliding towards the back of the sled is 118 N. The textbook presents a more complex solution, considering Roberto and the sled as separate objects, while the easier method treats them as
  • #1
fightboy
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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?
 
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  • #2
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
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 let's 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.
 
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  • #4
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.
 
  • #5


Your approach to the problem is perfectly valid and correct. Both methods, yours and the one presented in the textbook, are just different ways of solving the same problem. It is not necessary to go back and try the problem using the textbook's method, as long as you understand the concepts and equations used in your approach. In fact, sometimes it is beneficial to try different methods to solve a problem as it can help solidify your understanding of the material. Keep up the good work!
 

1. What is a physics sled?

A physics sled is a device used to study the principles of motion and forces in physics. It typically consists of a flat surface with low friction materials, such as ice or smooth plastic, and a pulley system to measure the force applied to the sled.

2. How is a physics sled used in experiments?

A physics sled can be used in various experiments to study concepts such as Newton's laws of motion, friction, and acceleration. The sled can be pulled or pushed with different forces, and the resulting motion and forces can be measured and analyzed.

3. What are the factors that affect the motion of a physics sled?

The motion of a physics sled can be affected by several factors, including the applied force, the mass of the sled, the surface it is on, and the presence of friction. These factors can be controlled and varied in experiments to study their effects on the motion of the sled.

4. How can a physics sled be used to demonstrate friction?

A physics sled can be used to demonstrate friction by conducting an experiment where the sled is pulled with different forces on different surfaces. The distance the sled travels can be measured, and the difference in distances on different surfaces can be attributed to the friction between the sled and the surface.

5. What are some real-life applications of studying physics with a sled?

Studying physics with a sled can have real-life applications, such as in the design and testing of vehicles, roller coasters, and other machines that involve motion and forces. It can also help in understanding and predicting natural phenomena, such as the movement of glaciers or avalanches.

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