How do you calculate the tension in a rope of a water skier? Watch

TiernanW
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Hey guys!

I have this Physics question for AS Physics:
http://i65.tinypic.com/2qlbn9h.jpg

We are currently doing Newton s Law and so I used F = m*a to get the force in the opposite direction of the friction like 70 * 3.4 and 850 * 3.4 (it doesn t say the boat is accelerating but I assumed so as the person is accelerating something has to make him accelerate. Then I said that the forces to the right must equal the forces to the left, but I am not sure if this applies to accelerating objects, but anyway I said that 3128 (sum of the 2 m*a) = x + 325 and calculated x (The tension) from there. But I have a feeling its incorrect.

Anyone any ideas?
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Smithenator5000
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(Original post by TiernanW)
Hey guys!

I have this Physics question for AS Physics:
http://i65.tinypic.com/2qlbn9h.jpg

We are currently doing Newton s Law and so I used F = m*a to get the force in the opposite direction of the friction like 70 * 3.4 and 850 * 3.4 (it doesn t say the boat is accelerating but I assumed so as the person is accelerating something has to make him accelerate. Then I said that the forces to the right must equal the forces to the left, but I am not sure if this applies to accelerating objects, but anyway I said that 3128 (sum of the 2 m*a) = x + 325 and calculated x (The tension) from there. But I have a feeling its incorrect.

Anyone any ideas?
Hello there,

As I see it, you do not need to consider the acceleration of the boat; you only need to focus on the skier. The skier has two forces acting upon them: the tension  T and the resistive force of  75 N . As you know the acceleration to be  3.4 ms^{-2} and the mass of the skier to be  70 Kg , you can now use Newton's Second Law . . .





[T - (75)] = (70)(3.4)



T - 75 = 238



T = 313 N

. . . to find that the tension is  313 N . I hope that this has been helpful.
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TiernanW
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(Original post by Smithenator5000)
Hello there,

As I see it, you do not need to consider the acceleration of the boat; you only need to focus on the skier. The skier has two forces acting upon them: the tension  T and the resistive force of  75 N . As you know the acceleration to be  3.4 ms^{-2} and the mass of the skier to be  70 Kg , you can now use Newton's Second Law . . .





[T - (75)] = (70)(3.4)



T - 75 = 238



T = 313 N

. . . to find that the tension is  313 N . I hope that this has been helpful.
Ah thank you. The tension is the force that is pulling the skier so T - the drag is the resultant and because F in ma is the resultant, it must equal.

So now, if I wanted to calculate the total driving force produced by the propellers of the boat, I would have to add the resultant force of the boat and the resultant 238 from the skier?
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Smithenator5000
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(Original post by TiernanW)
Ah thank you. The tension is the force that is pulling the skier so T - the drag is the resultant and because F in ma is the resultant, it must equal.

So now, if I wanted to calculate the total driving force produced by the propellers of the boat, I would have to add the resultant force of the boat and the resultant 238 from the skier?
Hello again,

If you assume that the string is taught, you can also assume that the skier and boat are travelling at the same velocity and experience the same acceleration of  3.4 ms^{-2} .. Three main horizontal forces act on the boat: the driving force  D , the tension  313 N and the resistive force  250 N . As we know the mass of the boat to be  850 Kg , we can once again use Newton's Second Law . . .





[D - (250) - (313)] = (850)(3.4)





D - 563 = 2890





D = 3453 N

. . . to find the driving force to be  3453 N .

Whenever faced with a question like this, try and isolate one of the objects and consider the forces which act on that object alone. Newton's Second Law should then be applicable, provided you know the mass and acceleration. Once again, I hope that this has been helpful.
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TiernanW
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(Original post by Smithenator5000)
Hello again,

If you assume that the string is taught, you can also assume that the skier and boat are travelling at the same velocity and experience the same acceleration of  3.4 ms^{-2} .. Three main horizontal forces act on the boat: the driving force  D , the tension  313 N and the resistive force  250 N . As we know the mass of the boat to be  850 Kg , we can once again use Newton's Second Law . . .





[D - (250) - (313)] = (850)(3.4)





D - 563 = 2890





D = 3453 N

. . . to find the driving force to be  3453 N .

Whenever faced with a question like this, try and isolate one of the objects and consider the forces which act on that object alone. Newton's Second Law should then be applicable, provided you know the mass and acceleration. Once again, I hope that this has been helpful.
Thank you! You've been extremely helpful.
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Smithenator5000
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(Original post by TiernanW)
Thank you! You've been extremely helpful.
You're welcome.
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