# Cable connecting the two ships homework

1. Sep 26, 2007

### Naikon

1. The problem statement, all variables and given/known data

A tugboat of mass m pulls a ship of mass M, accelerating it. Assume that water friction on the two vessels' is negligible.

If the force acting on the tug's propellar is F, what is the tension, T, in the cable connecting the two ships

(Variables: F, M, m)

2. Relevant equations

F=ma

3. The attempt at a solution

to get T I thought I would use the equation,

a= (T-mg)/m & a= -(T-Mg)/M

However, I cannot figure out how to derive a forumla for T that only involves F, M, and m.

I can derive:

T= (2gMm)/(M+m)

I'm stuck can anyone help?

2. Sep 26, 2007

### hotvette

Welcome to the forums.

What direction does g apply and what direction are the boats traveling?

3. Sep 27, 2007

### Naikon

I guess g is irrelevant in this question, the boat would be traveling the same direction as the tugboat as it is being pulled so I would only need to consider the horiztonal forces acting on the objects?

Last edited: Sep 27, 2007
4. Sep 27, 2007

### hotvette

OK, if g acts down and the boats are traveling in a horizontal direction, what influence does g have on the motion?

5. Sep 27, 2007

### Naikon

none , i just edited my previous post. sorry

6. Sep 27, 2007

### hotvette

Correct!

7. Sep 27, 2007

### Naikon

Ok so I completely forgot to add up my forces for newton's second law, thank you. I guess now I realize the importance of a force body diagram.

I derived:

(T+F/m) = a

(T/M) = a

Resulting in : -FM/(M+m)

8. Sep 27, 2007

### hotvette

What, no FBD? Shame of shames........... By the way, I got the same answer as you.