Box in a Car

Hi, i'm new here. I have a problem.

1. The problem statement, all variables and given/known data
A box has the mass m, held by two horizontal ropes and two vertical ropes. The box is inside a still car. If the car accelerated suddenly with the accekeration of a then the box will remain still inside the car. What is :
a. The acceleration of the box/car? (state it in t1, t2, t3, t4 and g)
b. The separation after the time t? (state it in t1, t2, t3, t4, t and g)

The illustration is here -> xttp://i31.tinypic.com/izvnev.jpg (change x to h)

2. Relevant equations
[tex]\Sigma F = ma[/tex]


3. The attempt at a solution

a.
t3 + t1 = ma
m = (t3 + t1)/a (1)

t4 + t2 = mg
m = (t4 + t2)/g (2)

(t3 + t1)/a = (t4 + t2)/g
a = g(t3 + t1)/ t4 + t2 (3)

b.
S = Vt
S = a * t * t
S = (g(t3 + t1)/ t4 + t2) * t^2
S = g (t^2) ((t3 + t1)/(t4 + t2))


I actually don't really know how the tensions work. I hope someone can explain it to me. And sorry, i'm not used to using latex yet.

 

Oh wow thanks! I actually never knew that tension forces always pull away from the object they act on, duh.... As for the 2 equations of m, i needed them to solve the a(acceleration) in t1, t2, t3 ,t4, and g. My third equation is actually m = m. Sorry, i didn't type it.

I think i have done the correct equations now.
ma + t1 = t3
ma = t3 - t1
m = (t3 - t1)/a (1)

mg + t2 = t4
mg = t4 - t2
m = (t4 - t2)/g (2)

m = m
(t3 - t1)/a = (t4 - t2)/g
a = g(t3 - t1)/(t4- t2 (3)

So the separation would be

s = vt
s = a*t^2
s = (t^2*g(t3-t1)) / t4-t2

Is that correct?

 

You're right, i forgot the velocity is not constant. Then i should use

[tex]s = v0t + \frac{at^2}{2}[/tex]

Because the car was resting then

[tex]v0 = 0[/tex]

[tex]s = \frac{at^2}{2}[/tex]

[tex]s = \frac{gt^2(t3-t1)}{2(t4-t2)}[/tex]

Is that correct?

 

Yeah i think that's what the question meant. Thanks for the help! :)