Pre Calculus Question: Couldn't find anywhere else to ask.

[Bhu_gunner]

I am in physics, so when I saw this problem I thought "this should be fun." NOPE.

The question is as follows; Find the work done by a 10lb force acting in the direction <1,2> to move an object from (0,0) feet to (3,0) feet.

The equation I've been given, but told in class that I won't need, is; |F| cos(x) |AB|

So, I've gotten this far:


However, I don't seem to be able to solve any further. My teacher said that F, which is the force of 10lbs, goesfarther than the distance of the object (3,0). Yet, I still can't find the work done. This in vectors, btw. We were doing vector projections.

 

[MarkFL]

I would use exactly the equation you were told you wouldn't need:

[MATH]W=(10\text{ lb})\frac{1}{\sqrt{5}}(3\text{ ft})=\,?[/MATH]

 

[Steven G]

MarkFL is absolutely correct. Why were you told not to use this equation?

Also your diagram has the wrong force F. Do you see your mistake?

 

[Bhu_gunner]

Why is my force wrong? The question said there was a force of 10lbs in a vector traveling <1,2>. Did I perhaps forget vertical or horizontal components.

Also, though I'm not sure why, my teacher said I could solve such a problem without the equation. Even the using the equation is correct, which makes me mad, seeing that if I have an equation, I should be able to use it, but for some reason, my teacher showed us a way where you didn't need it. Thank you for agreeing with me on using the equation though. Do you perhaps know any other method? I was told that by somehow extending the force vector out longer than vector AB, I could somehow solve.

 

[Dr.Peterson]

The question is as follows; Find the work done by a 10lb force acting in the direction <1,2> to move an object from (0,0) feet to (3,0) feet.

The equation I've been given, but told in class that I won't need, is; |F| cos(x) |AB|

I don't seem to be able to solve any further. My teacher said that F, which is the force of 10lbs, goesfarther than the distance of the object (3,0). Yet, I still can't find the work done. This in vectors, btw. We were doing vector projections.

I hope you realize that it's meaningless to say "the force of 10lbs, goes farther than the distance of the object": a force is not a distance! The vector (which you drew as <2,1> by mistake) only tells you the direction of the force; the length of the vector is irrelevant, as it is measured in pounds, not feet.

As for how to do it without that formula, can you tell us what else you know about work besides the formula? I would guess that you are expected to go back to definitions. For example, perhaps you are to multiply the distance by the component of force in the direction of the movement. (Never mind that the formula is directly based on that definition.) What definition do you have?

 

[MarkFL]

It looks like your force vector terminates at (2,1) instead of (1,2).

We have a force vector whose magnitude is 10 lb. But, only the horizontal component is doing the work, and it has a magnitude of \(2\sqrt{5}\). The vertical component has a magnitude twice that of the horizontal component, or \(4\sqrt{5}\). We could find this via the Pythagorean theorem:

[MATH]x^2+(2x)^2=10^2[/MATH]
[MATH]5x^2=100[/MATH]
[MATH]x^2=20[/MATH]
[MATH]x=2\sqrt{5}[/MATH]
But, we are still using what is defined as the cosine of the angle subtended by the force and displacement vectors.