How Do You Calculate the Angle in a Towing Scenario Using Work and Force?

[nissanfreak]

Work done by a constant force!

A small plane tows a glider at constant speed and altitude. If the plane does 2.00 X 10^5 J of work to tow the glider 145m and the tension in the tow rope is 2560N, what is the angle between the tow rope and the horizontal?

This is one of my homework problems. I think that I might be using the wrong formula for this. What formula should I use? And If its possible could someone please show me how to solve this step by step I really want to understand this stuff, so any and all help would be greatly appreciated!

 

[Doc Al]

Why don't you show us what you've done so far?

Hint: The glider moves horizontally. So what component of the rope tension is doing the work on the glider?

 

[nissanfreak]

I know that the answer is 57.4 degrees, but I just can't seem to get it? The formula I am using is W=Fd(cos phada) And I am trying to solve for phada. Can someone please work out this problem step by step for me? I want to learn!

 

[Astronuc]

or Force = Work / distance.

Resolve the force applied horizontally and vertically. Which applies to the work being done?

 

[Knavish]

Since the plane is pulling the glider horizontally (and not vertically), the horizontal force corresponds to the work being done. Thus F=2560cos(theta).

I think you can solve from there.

 

[Doc Al]

I know that the answer is 57.4 degrees, but I just can't seem to get it? The formula I am using is W=Fd(cos phada).

Realize that you are given values for W, F (that's the tension in the rope), and d. All you need do is solve for cos(theta). Then use your calculator to find theta.

To solve for cos(theta): Divide both sides of your formula by Fd.

 

[nissanfreak]

Well I've tried solving it like that but the answer I keep getting is 5.38793103e-11. Then to get rid of the e-6 I move the decimal over to the right 11 spaces and I then get 54 degrees as my answer. But the books answer is 57.4 degrees. What am I doing wrong?

This is how I had the equation looking .0000200/(2560)(145)=cos theta

 

[Doc Al]

This is how I had the equation looking .0000200/(2560)(145)=cos theta

The work is given as 2.00 X 10^5 J = 200,000 J. (Not 2.00 X 10^-5 J = 0.0000200 J.) You messed up the exponent.