# Vertical Load Reactions On A Ramp

Hello,

I'm working on a small hardware project.

I have two load sensors located at distance $S_1$ ($x=0$) and $S_2$. Assuming we ignore the weight of the ramp ...

• What is the load on $S_1$ and $S_2$ as a function of $x$ where $x$ is an object's horizontal static position on the ramp?

Here's what I have so far:

• As the object moves horizontally away from $S_1$, the load on $S_1$ monotonically decreasing.
• As the object moves horizontally closer to $S_1$, the load on $S_1$ monotonically increasing.
• When the object is directly over $S_1$ ($x=0$), the load on $S_1 = w=mg$.
• When the object is directly over $S_2$ ($x = L = S_2$), the load on $S_1 = 0$.

I have feeling adding an incline does not change the vertical forces acting upon $S_1$ and $S_2$.

My question is, can this problem be reduced to a simply supported beam who's equations are:

$$S1 = (\frac{L - x}{L})w$$

and

$$S2 = (\frac{x}{L})w$$

Any other observations or insights would be great. Thanks!

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A note: you may find our Robotics site useful :) –  Manishearth Jan 2 at 13:49
Oh nice! I didn't see that in the bottom of the page ;) –  Brian Chavez Jan 2 at 13:58
It's not there. It's a beta site -- on graduated sites, only other graduated sites are listed on the bottom. –  Manishearth Jan 2 at 14:03
It is called the lever rule and you are correct in splitting the weight across $S_1$ and $S_2$, but only for a static case. –  ja72 Jan 2 at 21:08
Take S1 as the origin. There is a torque about S1 due to the vertical force which is just $mg$ independent of the incline. This is the torque which the support S2 must counter to stop the ramp from rotating about S1. This gives you the force on S2, and the force on S1 is just $mg - F_{S2}$. So nothing essential in the problem depends on the incline and your simple beam equations should work.