# What is the tension in the rope that connects the tugboat

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1. Nov 7, 2016

### emily081715

1. The problem statement, all variables and given/known data
A tugboat pulls two barges down the river. The barge connected to the tugboat, carrying coal, has an inertia of 2.0 × 105 kg, and the other barge, carrying pig iron, has an inertia of 3.1 × 105 kg.The resistive force between the coal barge and the water is 8.0 × 103 N, and the resistive force between the pig iron barge and the water is 1.0 × 104 N . The common acceleration of all three boats is 0.40 m/s2. Even though the ropes are huge, the gravitational force exerted on them is much smaller than the pulling forces.What is the tension in the rope that connects the tugboat to the coal barge?

2. Relevant equations
F=ma

3. The attempt at a solution
F=( 2.0x10^5)(0.4)- 8.0x10^3
= 7200N

2. Nov 7, 2016

3. Nov 7, 2016

### emily081715

i though my reasoning was pretty straight forward on why i approached it that way.the tension would be on the rope (or whatever is pulling it) in between the tug boat and the coal boat. this would make the mass 2.0x10^5 by the acceleration of the whole system (0.4) yet you'd need to account for water resistance on that boat which is 8.0 × 10^3 N

4. Nov 7, 2016

### TomHart

Does the tug boat connect to the coal barge and the coal barge to the other barge? Or is each barge connected to the tug boat by its own rope?

5. Nov 7, 2016

### emily081715

what step and i missing

6. Nov 7, 2016

### emily081715

i assumed that the coal barge is connected to the tugboat. and other barge connect to the coal. sorta like a train. but the question didn't say

7. Nov 7, 2016

### TomHart

Thank you. That's what I suspected also.

8. Nov 7, 2016

### emily081715

i'm not sure if thats the correct way to interpret it though

9. Nov 7, 2016

### Simon Bridge

You have assumed there are only two forces acting on the barge - is that a sound assumption?
Aren't there two ropes from the coal barge? Doesn't the pig-iron barge pull on it?

Have you sketched the situation described?
Have you drawn a free body diagram for the barge?

10. Nov 8, 2016

### emily081715

yes i've sketched and drawn free body diagrams, would the other rope attached to the coal barge be creating a force of friction too? meaning my solution would need to look something more like
T=( 2.0x10^5)(0.4)- (8.0x10^3+3.1 × 105 )
=-238000N

11. Nov 8, 2016

### I like Serena

Let's take another look.
The forces on the coal boat are:
1. +T - the unknown force that the tug boat exerts.
2. -8.0 × 103 N - the drag on the coal boat.
3. -1.0 × 104 N - the drag on the pig iron boat, which is transmitted through the rope.
4. -3.1 × 105 kg × 0.4 m/s2 - the required force to accelerate the pig iron boat by 0.4 m/s2, also transmitted through the rope.
Summing it up, what will T be?

12. Nov 8, 2016

### emily081715

i already solved it on my own

13. Nov 8, 2016

### I like Serena

Just out of curiosity, how did you? Since it appeared you were not quite on track?

14. Nov 8, 2016

### Simon Bridge

#3 and #4 needs to be justified.
The best-practise is to do the free body diagram for the pig-iron boat as well, then solve the system of simultaneous equations you get.... remembering what I told you (probably you, certainly I've told @emily081715 ) in other threads how you should do all the algebra using symbols first, then plug the numbers into the final equation.

Well done - please share so other people will benefit from the insights you gained.

15. Nov 8, 2016

### emily081715

16. Nov 8, 2016

### Simon Bridge

It goes like this then:

fbd for coal barge: T2<----fc<----[mc]---->T1
(1) $T_1 - T_2-f_c = m_ca$ (fc is the drag on barge c)

fbd for pig-iron barge fp<---[mp]--->T2
(2) $T_2 - f_p = m_pa$

Strategy: solve for $T_2$ in (2) and sub into (1).

$T_2 = m_pa + f_p$ ...
$T_1 - (m_pa + f_p) - f_c = m_ca \implies T_1 = (m_c+m_p)a + (f_c+f_p)$
... which is what you got. Well done.
It's pretty much what you did - you just put the fbd's next to each other where I put them in a list.