Work Done by Horses & Friction on Canal Boat

[tony873004]

Two horses on either side of a canal pull a canal boat of mass 1.0*10^4 kg at a constant speed for a distance of 10.0 km. (Figure 7.32). One horse exerts a force of 3.0*10^2 N at an angle of 20º to the canal, and the other exerts 5.0 * 102 N. Find the work done by each horse and the work done by friction between the boat and the water.

I believe I would set it up like this:
Call the direction parallel to the canal the x axis
The work of a horse whose angle is given will be the x-component of the force * distance in meters
W_{horse1}=cos(20)*3*10^2N*10000m

The work done by the horse whose angle is not given will be the same as the first horse. Again, it will be the x-component of the force, which will be equal to Horse 1's x-component, and although the question doesn't ask for it, I could now compute Horse 2's angle.

But the work done by the friction between the boat and the water? I don't know how to solve that? What if the question said they were on ice instead of water?

The best I can come up with is
W_{friction}=Force_{friction} * 10000 meters

 

[Locrian]

First, your assumption that the forces in the x-direction does not seem sound to me. On the contrary, I would expect the forces in the y-direction to be the same, but with opposite sign, and then the force in the x-direction of the second horse to be determined after that. I am missing the figure, and also I think there is a mistype, but am not sure (is the second horse actually exerting 5.0 * 10^2N?), so I may not have all necessary information.

Secondly, as to your question about the work that friction does, I hope you don't mind me asking this question: If an object is at a constant velocity for a period of time, what is the net work that has been done on it? Or better yet, if this object is at a constant velocity, what is the net force on it, and what does that mean for the work done?

Hope this helps.

 

[Gale]

To find the force of friction use

\Sigma F= ma

if the velocity is constant, then acceleration=0.

[edit] I'm too slow...

 

[tony873004]

Thanks, that does help. You're right about the typo 5.0*102 = 5.0*10^2.

Thinking about it, it does make more sense that the y-components should be equal or the boat will hit the side of the canal.

So after setting the y's equal, I have to compute both horses' x-components of force.

You're right, the constant velocity is the key. Water friction must equal the horses pulls but in the opposite direction or there'd be acceleration.

Thanks Locrain!

 

[tony873004]

[edit] I'm too slow...

but helpful nonetheless. Thanks, Gale!