Cable connecting the two ships homework

  • Thread starter Naikon
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
hotvette
Homework Helper
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Welcome to the forums.

What direction does g apply and what direction are the boats traveling?
 
  • #3
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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:
  • #4
hotvette
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OK, if g acts down and the boats are traveling in a horizontal direction, what influence does g have on the motion?
 
  • #5
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none , i just edited my previous post. sorry
 
  • #6
hotvette
Homework Helper
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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!
 
  • #7
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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
hotvette
Homework Helper
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What, no FBD? Shame of shames........... By the way, I got the same answer as you.
 

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