What is the net work done on the boat by the two locomotives?

  • Thread starter leezak
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  • #1
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The drawing shows a boat being pulled by two locomotives through a canal of length 2.39 km. The tension in each cable is 4.17E4 N, and = 16.7°. What is the net work done on the boat by the two locomotives?

first I used W=fd... i multiplied 4.17E4 * 2 = 8.34E4 to get the total tension in each string. then to get the total work i multiplied that by 2390 (m) = 1.99E8... i divided that by two to get the work in each string = 9.97E7... then i realized that would be the tension in each string if the string was perpendicular to the boat so i found the hypotenuse using cos(16.7)=9.97E7/H that gave me 1.04E8. in order to get the net work i multiplied that by two becuase there are two strings and i got 2.08E8 J. that answer is wrong and i can't figure out why... can someone help please?! thanks!
 

Answers and Replies

  • #2
hotvette
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Your approach seems correct. You may have a math mistake somewhere. I got 1.91E8 J.
 
  • #3
Fermat
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leezak said:
... so i found the hypotenuse using cos(16.7)=9.97E7/H that gave me 1.04E8. ...
Looks like you divided by cos(16.7) instead of multiplying by it :smile:

I got the same as Hotvette.
 
  • #4
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but if i am trying to find the hypotenuse of the triangle... shouldn't i divide instead of multiply?
 
  • #5
hotvette
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The tension on the cable represents the hypotenuse. What you want is the component in the direction of travel.
 
  • #6
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oh thank you!!!
 

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