Cable connecting the two ships homework

AI Thread Summary
A tugboat pulls a ship, and the problem involves determining the tension in the cable connecting them, given the forces acting on both vessels. The initial attempt to derive the tension, T, involved using the equations of motion but did not yield a formula solely in terms of F, M, and m. Participants clarified that gravitational force, g, is irrelevant since both vessels are moving horizontally. After discussing the importance of a free body diagram (FBD), the correct approach was confirmed, leading to a derived formula for T. The discussion emphasizes the need for proper force analysis in solving such problems.
Naikon
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Homework Statement



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)


Homework Equations



F=ma


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


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:
OK, if g acts down and the boats are traveling in a horizontal direction, what influence does g have on the motion?
 
none , i just edited my previous post. sorry
 
Naikon said:
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?

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