# I Weight distribution on both ends of a ramp?

1. Oct 15, 2016

### xshovelfighter

Hey guys,

I am building a small water ramp that will extend from a dock and into the water with a buoy at the end. I could probably just use trial and error, but thought it would be fun to make some rough calculations to help my design.

This picture (http://imgur.com/a/bHYfG) shows the general layout of the ramp. Ignore the fact that the ramp extends into the water past the buoy and assume that the board ends at the buoy.

So, I am wanting to calculate the vertical force at point A, where the board and buoy extend into the water. I came up with the equation: force at point A is equal to the weight of the board times sin(theta). Based on this equation, if theta is 90 degress, then the force at A is equal to the full weight of the board, which checks out logically. Further, the force at A decreases proportionally as theta approaches 0, which makes sense as well.

Anyway, just wanted to see if this sounds right? Any comments, corrections, suggestions, etc are appreciated!

2. Oct 15, 2016

### CWatters

Your approach is incorrect. For example you have ignored the support provided at the other end of the ramp.

The answer is actually quite simple... if the ramp is uniform then each end caries half the weight.

3. Oct 15, 2016

### CWatters

PS This is true for all angles except absolutely vertical. In that position you cannot determine which end carries what percentage of the weight. It's said to be statically indeterminate.

4. Oct 15, 2016

### xshovelfighter

However, how can this be correct? Take for example, a person in the push-up position. The system is static, no work is being done, and we are simply looking at the pressure and force on the hands and feet of the person. In scenario 1, the person's feet are way below their hands and they are near vertical. The force of the hands is minimal in this instance. On the opposite end of the spectrum, the person's feet are above their head. In this scenario, the weight and force on their hands is significantly greater than at their feet. This is why it's harder to do a push-up at an angle closer to horizontal.

This concept applies to the ramp, and the weight/force at point A should be a function of the angle theta, correct?

5. Oct 15, 2016

### Tom.G

I just tried the experiment with a 1 foot ruler and a kitchen scale.

Ruler weight: 35 grams
One end of ruler on kitchen counter, half inch in from other end resting on scale platform

Scale pan 1.5 inches above counter: scale shows 18 grams
Scale pan 6 inches above counter: scale shows 13 grams

6. Oct 16, 2016

### CWatters

There are some differences between your ruler experiment and assumptions I made for the ramp. In your experiment with the ruler there are friction forces at both ends which mean the reaction forces at each end aren't necessarily vertical. I made the assumption that there was no friction which is ok for the float end but I should have considered it at the jetty end.

I'll be back with a drawing and explanation of the forces but may take me awhile.

7. Oct 16, 2016

### CWatters

Ok here is a diagram...

Fw is the weight of the ramp acting vertically at the mid point.
Fb is the buoyancy force created by the float. There is no friction at that end so it acts vertically.

Fn is the normal reaction at the jetty end
Ff is the friction force at the jetty end
Together these two vector add to give the total reaction force Fj at the jetty end.

Since the ramp has no angular acceleration the net torque on the ramp must be zero. Taking moments about the jetty end and taking clockwise as +ve...

FbLCos(θ) - 0.5FwLCos(θ) = 0

L and Cos(θ) cancels..

Fb - 0.5Fw = 0

So
Fb = 0.5Fw ...............................(1)
eg The float carries half the weight.

Since L and Cos(θ) cancel the force on the float is independent of the ramp angle.

Now for the jetty end...

Since the ramp isn't accelerating vertically the net vertical force on the ramp sums to zero..

Fb + Fj - Fw = 0

Substitute (1)

0.5Fw + Fj - Fw = 0
Rearrange to give
Fj = 0.5Fw
eg the jetty carries the other half of the weight.

Reason for edit: Got my jetty and ramp mixed up so had to correct that in a few places.

Last edited: Oct 16, 2016
8. Oct 16, 2016

### CWatters

PS: It turns out that that the friction force at the jetty end doesn't make a difference. This is because...

The ramp isn't accelerating horizontally. So the net horizontal force must sum to zero. Therefore the horizontal component of the reaction force at the jetty must be equal and opposite to the horizontal component of friction. This makes the total reaction force Fj vertical.

In the case of the ruler there is friction at both ends. I haven't thought it through but I suspect it accounts for the slight difference in the weight carried by each end that you found. For example the total reaction force at both ends would no longer be vertical as they would also have to balance friction forces at the lower end.

Last edited: Oct 16, 2016
9. Oct 16, 2016

### CWatters

I believe friction at the feet would also account for this.

10. Oct 16, 2016

### CWatters

Regarding the vertical case..

Consider a man hanging from a horizontal bar with his feet just touching the ground. How much of his weight is carried by his feet vs his hands cannot be determined. All you can say is that together they add up to his weight.

11. Oct 16, 2016

### CWatters

It occurs to me that as the ramp angle increases the max static friction at the jetty end will be exceeded. For example in the vertical case the normal force and hence friction is zero. In this case the load carried by the float would rise until it's carrying all the load.

12. Oct 16, 2016

### Tom.G

Just repeated the experiment of post #5 with friction material (rubber grippy used to open jars) at the ruler-platform interface, and with glass support at the 'buoy' end of ruler to minimize friction. The scale reading was constant 18 gram regardless of height of the 'buoy' end. This confirms the evaluation by @CWatters.

13. Oct 16, 2016

### xshovelfighter

Thanks for the very in depth and thoughtful response. I'll digest tomorrow morning and will respond by tomorrow evening.

At a first glance, the math checks out, but just not logically yet. Will look over in more detail and see if any questions come up. Thanks again for the response - would have responded sooner but just returning from a weekend trip and lots of travel.

14. Oct 17, 2016

### xshovelfighter

Thanks again for the thorough explanation! Went thought the math myself and everything is correct. As you mentioned, the equations are all independent of theta any which way you look at it. As a double check, looking at the moment about the center of mass quickly gives Fb = Fj, which also yields Fb = Fj = 0.5Fw. Everything checks out.

I also did some thinking in the car about how this makes sense conceptually. I think I have finally figured out why I couldn't wrap my head around this. Consider two people holding the opposite ends of a couch, standing on a steep staircase. One person is higher than the other. It's easy to think that the person at the bottom is holding more weight and has the "harder" job. However, I think my stubbornness in visualizing the weight being equal for both people is a result of associating this scenario with movement. Carrying the couch down the incline incurs momentum in the direction of the person at the bottom which makes it harder for this person. However; we are looking at a static system, so this doesn't apply. Also, even in a static situation, the way you typically hold/grip an object in this staircase situation seems like it has a lot to do with it. The angle the person up top is gripping the couch likely transfers weight forward, thus transferring additional weight to the person at the bottom to compensate for. However, if both people had a "gripping post" extending vertically from the bottom of the couch to hold onto (so that force on the holding hands is strictly vertical, with no horizontal component), I could much more easily imagine that the weight held would be equal in this situation. In the everyday scenario without the vertical gripping post, the net force for each individual person is likely not strictly vertical.. Hopefully this makes sense, I apologize for the tangent. Can draw a diagram if needed...

Last edited: Oct 17, 2016