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

[leezak]

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!

 

[hotvette]

Your approach seems correct. You may have a math mistake somewhere. I got 1.91E8 J.

 

[Fermat]

... 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

I got the same as Hotvette.

 

[leezak]

but if i am trying to find the hypotenuse of the triangle... shouldn't i divide instead of multiply?

 

[hotvette]

The tension on the cable represents the hypotenuse. What you want is the component in the direction of travel.

 

[leezak]

oh thank you!