# Calculating speed from tension. Watch

Announcements

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.

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.

0

reply

Report

#2

(Original post by

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.

**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.

0

reply

(Original post by

Have you drawn the two diagrams with two tensions and with one tension?

**mqb2766**)Have you drawn the two diagrams with two tensions and with one tension?

0

reply

Report

#4

(Original post by

I have not, no. I do not know how to got about this.

**Theo298**)I have not, no. I do not know how to got about this.

If you can't draw the diagrams, Id honestly suggest starting with simpler problems.

0

reply

(Original post by

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.

**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.

0

reply

Report

#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.

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.

1

reply

(Original post by

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.

**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.

0

reply

Report

#8

(Original post by

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?

**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?

0

reply

(Original post by

You'd have to resolve forwards and sideways. The tugs are at an angle.

**mqb2766**)You'd have to resolve forwards and sideways. The tugs are at an angle.

0

reply

Report

#10

(Original post by

So is the force 25000N?

**Theo298**)So is the force 25000N?

The rope tensions are not 25,000

Last edited by mqb2766; 1 month ago

0

reply

Report

#12

(Original post by

So is the force 25000N?

**Theo298**)So is the force 25000N?

0

reply

Report

#13

(Original post by

**Theo298**)
0

reply

(Original post by

When you put the forces / tensions on there you'll see they are not aligned so need to be balanced forwards and sideways

**mqb2766**)When you put the forces / tensions on there you'll see they are not aligned so need to be balanced forwards and sideways

How would I use that to calculate the original question?

0

reply

Report

#15

(Original post by

Okay so I think I have that as 33125.32N.

How would I use that to calculate the original question?

**Theo298**)Okay so I think I have that as 33125.32N.

How would I use that to calculate the original question?

So that should balance the new resistive drag (velocity) which will be aligned with the tension.

Last edited by mqb2766; 1 month ago

0

reply

(Original post by

Sounds right.

So that should balance the new resistive drag (velocity) which will be aligned with the tension.

**mqb2766**)Sounds right.

So that should balance the new resistive drag (velocity) which will be aligned with the tension.

0

reply

Report

#17

(Original post by

Would the new drag be 8000v^2 = 33125.32N, as there is only one tug now?

**Theo298**)Would the new drag be 8000v^2 = 33125.32N, as there is only one tug now?

Do draw the diagrams and put the tensions on there, even if they're unknown. It makes directions / resolving components clear.

0

reply

(Original post by

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.

**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.

0

reply

Report

#19

(Original post by

So that would equal 2.03ms^-1?

**Theo298**)So that would equal 2.03ms^-1?

Does the answer make rough sense if the initial speed was 2.5?

0

reply

(Original post by

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?

**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?

0

reply

X

### Quick Reply

Back

to top

to top