# Calculating speed from tension.Watch

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#1
Here’s the question at hand:
A ship is pulled at a constant speed of 2.5ms^-1 by 2 tugs, A and B. Each tug is connected by a cable so that the angle each of the cables make with the direction of travel is 41 degrees. The ship experiences a drag force of 8000v^2 Newtons.
As the tugs attempt to increase the speed of the ship from 2.5ms^-1, tug A breaks down, with its cable becoming slack.
Calculate the speed to which the ship initially decelerates, assuming the tension in the other cable remains constant.

I’m really struggling with this one so any help would be greatly appreciated.
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1 month ago
#2
(Original post by Theo298)
Here’s the question at hand:
A ship is pulled at a constant speed of 2.5ms^-1 by 2 tugs, A and B. Each tug is connected by a cable so that the angle each of the cables make with the direction of travel is 41 degrees. The ship experiences a drag force of 8000v^2 Newtons.
As the tugs attempt to increase the speed of the ship from 2.5ms^-1, tug A breaks down, with its cable becoming slack.
Calculate the speed to which the ship initially decelerates, assuming the tension in the other cable remains constant.

I’m really struggling with this one so any help would be greatly appreciated.
Have you drawn the two diagrams with two tensions and with one tension?
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#3
(Original post by mqb2766)
Have you drawn the two diagrams with two tensions and with one tension?
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1 month ago
#4
(Original post by Theo298)
Draw the boats, lines, ... and put the tensions, angles, .. on there
If you can't draw the diagrams, Id honestly suggest starting with simpler problems.
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#5
(Original post by mqb2766)
Draw the boats, lines, ... and put the tensions, angles, .. on there
If you can't draw the diagrams, Id honestly suggest starting with simpler problems.
I am able to draw diagrams, we were just not supplied with the tensions for the ropes.
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1 month ago
#6
so the key is to realise that with the 2 tugs the net force in the direction of motion is zero newtons. thus the 2 tension components in the forward direction must balance the resistive force in the opposite direction.

you can now work out the individual tensions.

once the cable snaps the only forward force is the tension component from the other tug. this is unchanged, but will not be able to balance the resistive force.
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#7
(Original post by the bear)
so the key is to realise that with the 2 tugs the net force in the direction of motion is zero newtons. thus the 2 tension components in the forward direction must balance the resistive force in the opposite direction.

you can now work out the individual tensions.

once the cable snaps the only forward force is the tension component from the other tug. this is unchanged, but will not be able to balance the resistive force.
Thanks this is very helpful. I now have the drag force, 50,000N. Would the tension in the cable be half of the drag force, for there are 2 cables?
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1 month ago
#8
(Original post by Theo298)
Thanks this is very helpful. I now have the drag force, 50,000N. Would the tension in the cable be half of the drag force, for there are 2 cables?
You'd have to resolve forwards and sideways. The tugs are at an angle.
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#9
(Original post by mqb2766)
You'd have to resolve forwards and sideways. The tugs are at an angle.
So is the force 25000N?
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1 month ago
#10
(Original post by Theo298)
So is the force 25000N?
Can you post your diagram and explain the figure.
The rope tensions are not 25,000
Last edited by mqb2766; 1 month ago
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#11 0
1 month ago
#12
(Original post by Theo298)
So is the force 25000N?
the forward component of each cable is 25000 N.
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1 month ago
#13
(Original post by Theo298) When you put the forces / tensions on there you'll see they are not aligned so need to be balanced forwards and sideways
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#14
(Original post by mqb2766)
When you put the forces / tensions on there you'll see they are not aligned so need to be balanced forwards and sideways
Okay so I think I have that as 33125.32N.

How would I use that to calculate the original question?
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1 month ago
#15
(Original post by Theo298)
Okay so I think I have that as 33125.32N.

How would I use that to calculate the original question?
Sounds right.
So that should balance the new resistive drag (velocity) which will be aligned with the tension.
Last edited by mqb2766; 1 month ago
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#16
(Original post by mqb2766)
Sounds right.
So that should balance the new resistive drag (velocity) which will be aligned with the tension.
Would the new drag be 8000v^2 = 33125.32N, as there is only one tug now?
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1 month ago
#17
(Original post by Theo298)
Would the new drag be 8000v^2 = 33125.32N, as there is only one tug now?
Yes, the forces are in the same (opposite) direction.

Do draw the diagrams and put the tensions on there, even if they're unknown. It makes directions / resolving components clear.
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#18
(Original post by mqb2766)
Yes, the forces are in the same (opposite) direction.

Do draw the diagrams and put the tensions on there, even if they're unknown. It makes directions / resolving components clear.
So that would equal 2.03ms^-1?
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1 month ago
#19
(Original post by Theo298)
So that would equal 2.03ms^-1?
Dunno, I presume you're capable of a division and square root?
Does the answer make rough sense if the initial speed was 2.5?
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#20
(Original post by mqb2766)
Dunno, I presume you're capable of a division and square root?
Does the answer make rough sense if the initial speed was 2.5?
I was expecting the loss of half the tugs to have more of an impact, but I can complete the required process the get the answer.
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