# Tug of war Force Conceptual Problems

1. Feb 8, 2009

### zell_D

Since I only got 1 reply on my old post (I think it is because my format is wrong) I am making a new post. Hopefully that is okay

1. The problem statement, all variables and given/known data
a) Person A and Person B are around the same size, they play tug of war. Both pull as hard as they can but none are winning. Compare magnitude, are they same or different? If different, which one is larger? Why?
b) Person A and Person C are different in size, Person C is a child compared to Person C. They play tug of war. Both pull as hard as they can but none are winning. Compare magnitude, are they same or different? If different, which one is larger? Why?
c) Person A and Person C are still playing tug of war. Both pull as hard as they can but Person C (kid) is on a skateboard. Person A is now able to roll Person C across the floor. Compare magnitude, are they same or different? If different, which one is larger? Why?
d) Person A is now dragging Person C across the floor. Compare magnitude of forces exerted, are they same or different? If different, which one is larger? Why?
e) a small car and a big truck collides, both initially moving at 5 mph. compare the force that each exerts on the other, which is greater? which vehicle would you rather be in? why?

2. Relevant equations
Newton's Third Law: F a on b = F b on a
F=ma

3. The attempt at a solution (I am 60% sure about my answers as I just started learning this)
a) System at rest. Net force is zero. This means that the Force exerted by each person should be the same. (Because they are around the same size and neither are winning, frictional forces should also be the same)
b) System at rest. Net force is zero. This means that the Force exerted by Person C (the smaller person) is the same as Person A so that he does not get pulled towards Person A. However, I THINK I can assume that Person C's surface has a greater frictional force than Person A's surface to keep the smaller guy in the game.
c) System is NOT at rest. Net force is NOT zero. This means that the force exerted by Person A is greater than Person C. This is mainly because Person C is on a skateboard and has way less frictional force than Person A.
d) System is NOT at rest. Net force is NOT zero. Force exerted by Person A is greater than Person C. Person C is not digging into the ground for max static friction to take effect? Since kinetic friction is always lower than the max static friction. (I am REALLY lost here)
e) Same velocity means no acceleration (I am assuming the velocity stays constant, NOT SURE) so the force they apply on each other would be the same. (Even though I am really debating myself with this) Now this second part I have no idea of ________. My guess is that i would rather be in the truck instead of the small car, but I have no physical idea as to why other than that it is bigger. (Note that I haven't learned momentum yet, so any mechanics involving that would be invalid as work on my homework.)

This homework is due tuesday, I am asking in advance because I really want to understand the concept and do well in this class. thank you (and PLEASE correct my answers/explanations if they are wrong or insufficient, but remember, this is somewhat of an introductory physics course. any mechanics too advanced I probably won't have learnt before)

2. Feb 8, 2009

### JaWiB

I think you could benefit by drawing free-body diagrams for all of these.
Why do you think there would be a greater frictional force?
When you say "Person C" do you actually mean "the force on the Rope by Person C"? Try drawing the free body diagrams and identify any third law force pairs.
The force given by $${\mu}N$$ is the maximum static friction force between two surfaces. When the friction force needed to keep the object in place is greater than this force, it begins to slide.
The acceleration of either car should not matter for the first part of the question. Newton's third law always applies unless your frame of reference is accelerating (for example, if you decided that the car was accelerating with respect to the ground, but wanted to use the frame in which the car was at rest).

The second part of the question is a bit tricky. If the mass of the car is very small, what does that imply about its acceleration? Or equivalently, if the mass of the truck is very large, what does that imply about its acceleration (relative to the acceleration of the car)? If you were in one of the vehicles, what would your acceleration be? What would that imply about the net force on you?

3. Feb 8, 2009

### zell_D

Since friction = mu*N and N = mg in this case, the bigger person would have a bigger N right? So doesn't that mean his frictional force will be bigger? so Person C would have to be on a surface that allows for a larger mu to compete? I don't know, or both of them are on the surface and neither can generate enough force to pass the max static friction?

I THINK so? or the Tension generated in the rope by the person? I suck at physics...
But okay, I drew the diagrams for the skateboard problem. I think I am even more confused now. Okay, since Person A is winning, does that necessarily means that hes applying more force? I personally DONT think so, since the two can apply the same force but since Person C is on a slippery surface, he is bound to lose...

So is my answer right here? I mean the only way that the kid is sliding is because the max static friction has been bypassed right? and it is now in the lower kinetic friction which allows Person A to dominate.

For the first part, I still don't kind of get it. Are you saying that they will generate the same force? Why though >.<

so for the second part, if the car is small, the acceleration is large while vice versa for the big heavy truck. So the truck would obviously be the safer choice. HOWEVER, what does this imply about the net force on me? I have no idea. If the force generated are the same, isnt the Net still 0? Ahh i feel so frustrated

***I guess what I want to know is that... say a system is at rest. Does the friction force and Tension force have to equal each other in the SAME person? No right? Since as long as the NET force is 0, the system is at rest

4. Feb 10, 2009

### PhanthomJay

For the first part, remember that
1.) Newton 3 applies in every case, whether the object is at rest, moving with constant speed, or accelerating. If A exerts a force on B, then B exerts an equal and opposite force on A, always.
2.) Also note that the friction force is not equal to mu(N). It is less than or equal to mu(N).
3.) And don't confuse the 'system' with parts of the system. When the smaler kids starts moving, there is a net force acting on him (tension on him greater than friction from the ground on him), but there is no net force acting on the bigger kid (tension on him = friction from ground on him). Be sure to identify the objects (persons) on which the forces are acting, with a good free body diagram of each, and apply Newton's laws accordingly.