# Dynmics, motion along a line

1. Sep 29, 2008

### bluebear19

1. The problem statement, all variables and given/known data
A machine has an 850g steel shuttle that is pulled along a square steel rail by an elastic cord. The shuttle is released when the elastic cord has 23.0N tension at a 45 degree angle.
What is the initial acceleration of the shuttle?

2. Relevant equations

F = ma
v1^2 = v0^2 + at
v1^2 = v0^2 +2a(delta x)

3. The attempt at a solution

F = ma
23 = .85a
a= 27.06 m/s^2
This is the acceleration when the cord is let go, but I have no idea how to find the inital acceleration. I know how to split the force into the x/y components but I'm not sure how that will help. Thanks

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Last edited: Sep 30, 2008
2. Oct 1, 2008

### chansong

perhaps you should use newton's second law to break the force into components:

sum of forces in x-direction = Tcos(theta) = ma_x
sum of forces in y-direction = N + Tsin(theta) - mg = ma_y = 0

Which a are you solving for? (for you to figure out)

You are given enough information from these equations and values to easily solve this I think.

3. Oct 1, 2008

### bluebear19

so Fx = 23cos45 = .85a
so a_x = 19.13m/s^2

and Fy = 0
but how do these equations give you the initial acceleration? Don't they only solve for the acceleration when F = 23N

I am confused by the initial acceleration part and the part describing the string as elastic. I don't know how elasticity affects the problem. And also, should I include friction of steel on steel?

4. Oct 1, 2008

### tiny-tim

Welcome to PF!

Hi bluebear19! Welcome to PF!

(welcome also to chansong )

"initial" means immediately the cord is released …

in other words, while it's at its full length, and it is still at 23N tension.

So you're there!

(and no, you can ignore friction in any exam question that doesn't give you a coefficient of friction )

5. Oct 1, 2008

### bokonon

Is the problem really this simple?

I tried solving it in this way, by finding the Fx component of the tension (23*cos45), and then dividing by the mass (.850 kg), using F=ma. However, this does not yield the correct answer. What is missing?

6. Oct 1, 2008

Do we need to account for steel on steel friction for this problem? I have been having such a headache with this, it is driving me crazy!

7. Oct 1, 2008

### bluebear19

ok I got it, my friends helped me. so you have to find the normal force which is Fy - mg
then to find the friction force you multiply the normal force by the kinetic coefficient of friction for steel on steel. then u take that number and substract it from the Fx. then F = ma and ur done

8. Oct 1, 2008

Cool, thanks!

So acceleration is negative?

EDIT: Ah nvm, I've been at this way too long. Thanks again, everyone!

9. Oct 1, 2008

### chansong

i was assuming there was no friction in your problem (as there was none in the one i had to solve myself)

but it looks like you've solved it so congrats