# weight increse over incline

• Feb 8th 2010, 08:41 PM
CarriageWhip
weight increse over incline
A wagon with people weighs 2,850 pounds being pulled by a horse weighing 1700 pounds can be moved very easy on a flat surface but what is the weight load being put on the horse when he takes that same load up a 5 degree incline?

10 degree incline the same load would feel like ? pounds

20 degree incline the same load would feel like ? pounds

Thank you
• Feb 9th 2010, 12:44 AM
Hello CarriageWhip

Welcome to Math Help Forum!
Quote:

Originally Posted by CarriageWhip
A wagon with people weighs 2,850 pounds being pulled by a horse weighing 1700 pounds can be moved very easy on a flat surface but what is the weight load being put on the horse when he takes that same load up a 5 degree incline?

10 degree incline the same load would feel like ? pounds

20 degree incline the same load would feel like ? pounds

Thank you

The component of a vertical weight force, $\displaystyle W$, down a plane inclined at an angle $\displaystyle \alpha$ to the horizontal is $\displaystyle W\sin\alpha$.

So, if $\displaystyle \alpha = 0^o$, the component is $\displaystyle W\sin 0^o = 0$. In other words, there is zero component on a horizontal plane.

If $\displaystyle \alpha = 10^o, \sin10^o \approx 0.174$. So just over $\displaystyle 17$% of the weight acts down the plane. If the weight is $\displaystyle 2850$ lbs, this is about $\displaystyle 495$ lbs.

$\displaystyle \sin20^o \approx 0.342$. So just over $\displaystyle 34$% of the weight will act down a $\displaystyle 20^o$ plane. In the case you quote, this is about $\displaystyle 975$ lbs.

• Feb 14th 2010, 09:03 AM
CarriageWhip
Thank you for the time to work with my problem. Next stupid question would be when you say "down the plane" is that mean pulling up hill or force put on a down hill and is there a difference?
• Feb 14th 2010, 09:54 AM
I personally think for problems involving inclined planes, it is more convinient to choose the coordinate system axes with x along the incline and y perpendicular to it, than the "standard" horizontal & vertical axis. With these axes, you reperesent a force of magnitude $\displaystyle mg cos \theta$ along the negative y axis. And so the component of vertical weight force is $\displaystyle mg cos \theta$.