# Thread: Calculus III & Physics

1. ## Calculus III & Physics

I am working a problem I'm a bit stuck on. I would love to hear your suggestions!

A person pulls a wagon 100 feet up an incline that makes an angle of 30o with the horizontal by exerting a force of 20 pounds on a handle that makes an angle of 30o with the horizontal. Find the the work done.

Reference picture:

Could I just use the following reasoning:
$\displaystyle W = \vec{F}\vec{X} = ||\vec{F}||\cdot ||\vec{X}||\cdot cos\Theta$

$\displaystyle \vec{F} = <20,0>$
$\displaystyle \vec{X} = <86.6,50>$

$\displaystyle \therefore W = \vec{F}\vec{X}$
$\displaystyle = ||\vec{F}||\cdot ||\vec{X}||\cdot cos\Theta$
$\displaystyle = \sqrt{20^{2}+0^{2}}\cdot \sqrt{86.6^{2}+50^{2}}\cdot cos(30^{\circ})$
$\displaystyle = 20\cdot 100\cdot cos(30^{\circ}) = 1732$

This seems a bit of excessive, no? Or maybe not. I'm just unsure if I'm going the right way with this.

2. ## Re: Calculus III & Physics

Yes, you have it right, but I think you are over-complicating the calculation. Keep in mind that work is the dot product of force vector and motion vector. The value for theta is the angle of the force applied relative to the direction of movement of the cart - that angle is 30 degrees (the angle of the handle relative to the incline). So to find the work done on the cart by the force simply use F= 20 lb, D = 100 ft, and theta = 30 degrees. There was no need to find vertical and horizontal components of distance. Also be sure to specifythat the answer is in units of foot-pounds.

BTW - the problem statement says the handle is at 30 degrees to the horizontal, but I assume you meant it's at 30 degrees to the direction of motion of the cart.

3. ## Re: Calculus III & Physics

Originally Posted by ebaines
Yes, you have it right, but I think you are over-complicating the calculation. Keep in mind that work is the dot product of force vector and motion vector. The value for theta is the angle of the force applied relative to the direction of movement of the cart - that angle is 30 degrees (the angle of the handle relative to the incline). So to find the work done on the cart by the force simply use F= 20 lb, D = 100 ft, and theta = 30 degrees. There was no need to find vertical and horizontal components of distance. Also be sure to specifythat the answer is in units of foot-pounds.

BTW - the problem statement says the handle is at 30 degrees to the horizontal, but I assume you meant it's at 30 degrees to the direction of motion of the cart.
Why wouldn't you take into consideration the extra 30 degrees? Because then this would just be the same as a problem in a flat surface, right?

But if I do what you've suggested then it should be the following:

$\displaystyle \therefore W = \vec{F}\vec{X}$
$\displaystyle = ||\vec{F}||\cdot ||\vec{X}||\cdot cos\Theta$
$\displaystyle = 20\cdot 100\cdot cos(30^{\circ}) = 1732$
Which gives me the same answer...
EDIT: Shouldn't it be $\displaystyle cos(60^{\circ})?$

4. ## Re: Calculus III & Physics

Yes, I did say that you already have the right answer - I just was trying to point out that there was no need to calculate the horizontal and vertical components of the distance up the ramp. The angle is 30 degrees because that's the angle between the applied force and the direction of motion (the angle of the handle relative to the incline).

BTW, if you really want to break the problem into horizontal and vertical components you can. It's a bit more work but yields the same result:

The vertical component of force is 20 x sin(60) = 17.3 pounds and the vertical movement is 100 sin(30) = 50 feet, so work in the vertical direction is 17.3 pounds times 50 feet = 866 ft-lbs. The horizontal component of force is 20 x cos(60) = 10 pounds, and the horizontal movement is 100 cos(30) = 86.6 feet, so work in the horizontal direction is 10 pounds times 86.6 feet = 866 ft-lbs. Hence the total work performed is 866 + 866 = 1732 ft-lbs.

5. ## Re: Calculus III & Physics

Originally Posted by DrKittenPaws
I am working a problem I'm a bit stuck on. I would love to hear your suggestions!

A person pulls a wagon 100 feet up an incline that makes an angle of 30o with the horizontal by exerting a force of 20 pounds on a handle that makes an angle of 30o with the horizontal. Find the the work done.

Reference picture:

Could I just use the following reasoning:
$\displaystyle W = \vec{F}\vec{X} = ||\vec{F}||\cdot ||\vec{X}||\cdot cos\Theta$

$\displaystyle \vec{F} = <20,0>$
$\displaystyle \vec{X} = <86.6,50>$

$\displaystyle \therefore W = \vec{F}\vec{X}$
$\displaystyle = ||\vec{F}||\cdot ||\vec{X}||\cdot cos\Theta$
$\displaystyle = \sqrt{20^{2}+0^{2}}\cdot \sqrt{86.6^{2}+50^{2}}\cdot cos(30^{\circ})$
$\displaystyle = 20\cdot 100\cdot cos(30^{\circ}) = 1732$

This seems a bit of excessive, no? Or maybe not. I'm just unsure if I'm going the right way with this.
The statement of the problem means the direction of the applied force is parallel to the incline. the work done would simply means force*distance: W = 20*100 = 2000 lb.ft

That answer is not consistent with the reference figure you provided. Is the statement of the problem is all that is given in your textbook?

6. ## Re: Calculus III & Physics

Originally Posted by DrKittenPaws
Find the the work done.
Find the work done by what? There are three forces on the box: A normal force, a weight, and the applied force. Are you possibly looking for a net work done on the box?

-Dan

7. ## Re: Calculus III & Physics

Originally Posted by topsquark
Find the work done by what? There are three forces on the box: A normal force, a weight, and the applied force. Are you possibly looking for a net work done on the box?

-Dan
I understood the question as since the weight is not given, the only work done is the work of the applied force. The work by the normal force is 0 since it perpendicular to the direction of the motion

8. ## Re: Calculus III & Physics

Originally Posted by votan
I understood the question as since the weight is not given, the only work done is the work of the applied force. The work by the normal force is 0 since it perpendicular to the direction of the motion
Sorry, while I was thinking about this one I chose 20 lbs to be the applied force, then assumed it was a weight.

Thanks for the catch.

As to the Normal force, I know it doesn't do any work. I was essentially asking if the OP knew that. But as my approach is flawed it doesn't have an impact on the problem, so nothing to worry about.

-Dan