# Homework Help: 2 motion questions

1. Oct 13, 2008

### tornzaer

1. The problem statement, all variables and given/known data
1. Boat is towing two water skiers, each skier has it's own rope that's at 40 degrees the boats axis and the tension is 450N, how much force does the boat have to have to keep them all moving?

2. The plane is traveling at 200 km/h at the bottom of the loop, and the scale (the pilot happens to be sitting on one) reads 4x normal weight/Fg. What is the radius of the loop.

2. Relevant equations
1. I don't really know...

2. I don't really know...

3. The attempt at a solution
My teacher put these on an assignment that's due tomorrow, so I really need some info fast.

I'm thinking she missed out on some variable... Please correct me if I'm wrong.

Last edited: Oct 13, 2008
2. Oct 13, 2008

### Hootenanny

Staff Emeritus
"I don't really know" isn't an attempted solution. There are no missing variables, the questions are complete and can be answered as set.

Last edited: Oct 13, 2008
3. Oct 13, 2008

### tornzaer

See, the equations I got in my lessons are of circular motion (v=2piR/T) and a pulley equation a=ft/m.

I don't know how to fit them in. Could you please at least direct me in the right direction?

4. Oct 13, 2008

### Hootenanny

Staff Emeritus
Well both those equations are irrelevant for the first question since the skiers aren't undergoing circular motion.

For the second question (which is circular motion), do you have an equation connecting centripetal acceleration/force and velocity?

5. Oct 13, 2008

### tornzaer

I have many equations for circular motion: v=2piR/T a=2piv/T a=4piR/T^2 a=v^2/R

Also centripetal force: Fcent=mv^2/R Fcent=m4pi^2R/T^2 Fcent=m2piv/T

Basically, for circular motion, I don't possess the period and for the centripetal force, I don't have mass and period.

6. Oct 13, 2008

### Hootenanny

Staff Emeritus
This equation looks appropriate for the second question.

7. Oct 13, 2008

### tornzaer

I was looking at that too, but I need a mass variable, which the questions is lacking. Am I overlooking something?

8. Oct 13, 2008

### Hootenanny

Staff Emeritus
What is the net force acting on the pilot at the bottom of the loop?

9. Oct 13, 2008

### tornzaer

Right, it's 4x Fg.

What I'm understanding is this:
Fg=mg
Fg=m(9.8m/s^2)

Then Fg would be multiplied by 4, which will give me the Fcent. Then just plug it into the equation to get R.

10. Oct 13, 2008

### Hootenanny

Staff Emeritus
You're close, but not quite correct. 4mg would be the reading on the scales, which is the normal reaction force of the scales, but this is not the same as the net force acting on the pilot. You need to find the net force acting on the pilot and use this as your centripetal force.

Do you follow?

11. Oct 13, 2008

### tornzaer

I know that at the bottom of a "loop", Fcent is Ft-Fg. Is this what you are getting at?

12. Oct 13, 2008

### Hootenanny

Staff Emeritus
Looks good to me

13. Oct 13, 2008

### tornzaer

Very well. Still, I'm in need of a mass, which the question lacks...

As for my first question, could you steer me as well? My substitute teacher gave us this and I think she messed up on a couple of lessons.

14. Oct 13, 2008

### Hootenanny

Staff Emeritus
Are you sure you need to know the mass?
If you assume that the skiers are travelling at a constant velocity, what can you say about the net force acting on them?

15. Oct 13, 2008

### tornzaer

Mass: not following you there, I'm afraid.

1st question: No acceleration means no net force. Fnet=Ft? So Ft would be 0.

16. Oct 13, 2008

### Hootenanny

Staff Emeritus
Try writing out the equation that I quoted before.
Correct and the same goes for the boat.

17. Oct 13, 2008

### tornzaer

Alright, first is done. :P

Second:
Fcent=mv^2/R
Ft-Fg=m*(55.55m/s)^2/R
Please clarify on the next step.

18. Oct 13, 2008

### Hootenanny

Staff Emeritus
Well what is Ft-Fg?

19. Oct 13, 2008

### tornzaer

Ft-Fg=Fcent=Fnet

20. Oct 13, 2008

### tornzaer

Please respond... this thing is due tomorrow.

21. Oct 14, 2008

### Hootenanny

Staff Emeritus
That is correct, but explicitly in terms of mg?
Please be patient, we all provide help here voluntarily. Plus, when you posted your last reply is was gone 2AM in the UK - we all have to sleep :zzz:

22. Oct 14, 2008

### tornzaer

Right, the g is 9.8m/s^2. What about the mass though? That's where I've been stuck for a while.

And sorry for being impatient. :P Didn't know you were in the UK.

23. Oct 14, 2008

### Hootenanny

Staff Emeritus
If you write out the equation, you'll find that you don't need to know them explicitly.

24. Oct 14, 2008

### tornzaer

I see... so you are suggesting giving the answer with an unsolved variable, like m(whatever)?

25. Oct 14, 2008

### Hootenanny

Staff Emeritus
No, I'm saying that if you write out the equation then you will be able to cancel the masses.

Just try writing out the equation.