Weight distribution on both ends of a ramp?

[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!

 

[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.

 

[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.

 

[xshovelfighter]

Thanks for the reply!

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?

 

[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

 

[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.

 

[CWatters]

Ok here is a diagram...

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

F_n is the normal reaction at the jetty end
F_f is the friction force at the jetty end
Together these two vector add to give the total reaction force F_j 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...

F_bLCos(θ) - 0.5F_wLCos(θ) = 0

L and Cos(θ) cancels..

F_b - 0.5F_w = 0

So
F_b = 0.5F_w ......(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..

F_b + F_j - F_w = 0

Substitute (1)

0.5F_w + F_j - F_w = 0
Rearrange to give
F_j = 0.5F_w
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.

 

[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 F_j 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.

 

[CWatters]

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.

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

 

[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.

 

[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.

 

[Tom.G]

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.

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.

 

[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.

 

[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...

 

[dcb]

@xshovelfighter, I believe you are correct and Tom G's initial post is also correct. I'm not sure why he recanted in his second post. I have also carried a couch down the stairs and you definitely want to be the guy on the upper end. I would also much rather do pushups flat to the ground rather than with my feet raised.

If you take a 10 pound dumbbell and put one end on a scale and hold the other end parallel to the ground the scale will read 5 pounds. As you raise your hand the scale will read more than 5 pounds. At 45 degrees it will read 7.5 pounds and of course at 90 degrees it will read 10 pounds. Note the scale is reading weight in the y direction.

@CWatters mistake is that it's not FbLcos(theta) but FbL/cos(theta) in the torque (perpendicular to the ramp) force. Fb is 1/2Fw*cos^2(theta). You must go on and do the force parallel to the ramp which the lower side bears entirely.

 

[dcb]

I should note that the presumption is that the ramp is resting on both ends (which would be the case for couch moving and inclined pushups) and we are trying to understand the forces in the y direction,

 

[jbriggs444]

@xshovelfighter, I believe you are correct and Tom G's initial post is also correct. I'm not sure why he recanted in his second post. I have also carried a couch down the stairs and you definitely want to be the guy on the upper end. I would also much rather do pushups flat to the ground rather than with my feet raised.

First of all, welcome to Physics Forums. Just because I am about to jump all over your content, that does not mean that we do not like you.

Please be aware that you have responded to a seven year old post.

The couch situation is complicated. It has to do as much with human physiology and psychology as with physics. It does not help that the couch has non-negligible height.

If both people supporting the couch are constrained to provide purely vertical forces and if the couch were compressed to negligible height (like a heavy plank instead of a couch) then the weight would be split 50/50 as @CWatters suggests. A simple argument based on a torque balance assures us that this must be so. With the caveat that for a vertical plank, the torque argument dissolves and the situation becomes statically indeterminate -- as @CWatters also noted.

If the people supporting the couch are free to supply their force in a direction other than purely vertical then there is a strong temptation for the top fellow to "cheat" and lift more or less at right angles to the long axis of the couch. He will push away from the stairs rather than lifting vertically upward. This allows him to reduce the magnitude of the force he supplies. The person at the bottom is also free to cheat. He could let the couch hang from the top person's hands and only supply enough diagonal force to tilt it parallel to the stairs. The situation is statically indeterminate.

But the fellow on the bottom has a better angle -- he can more easily provide more force. If both helpers are lifting as hard as they can, it is the bottom fellow who is usually in a position to push harder.

These strategies may or may not apply in the case at hand. It will depend on how the two ends of the beam are pinned. We are told that the support at the bottom end is a float attached to the beam. It will lift in a purely vertical direction. It follows from a simple force balance that the support at the other end must also supply a purely vertical force. So in the case at hand we indeed have a 50/50 split.

If you take a 10 pound dumbbell and put one end on a scale and hold the other end parallel to the ground the scale will read 5 pounds. As you raise your hand the scale will read more than 5 pounds.

This is not correct. It depends on the direction of the force you apply with your hand. If you want to economize on that force, you will apply it diagonally. Like a workman leaning on a shovel, you may not have to lift at all. You can even push downward instead.

That's statically indeterminate.

 

[dcb]

Hi jbriggs444, thanks for responding. Yes I realize how old this post is and I was hoping someone would respond as I'd Iike to discuss this. I happened on to this as I have a dock and was trying to figure out the forces on it as the water level went up and down. I have a bachelor's and master's degree in Electrical Engineer from the University of Michigan (Go Blue). I never had to take a statics course but for some strange reason, dynamics was required. I had a great professor and loved the course.

To be clear, I am talking about the case where the ends of the ramp are not attached to either end (they are resting on the ends). I couldn't tell from the original picture if that was the case, but it's the case I was interested in and I believe it matches the couch discussion.

First I'd like to ignore X & Y and look at the forces perpendicular and parallel to the ramp. Let's also start with the simple case of horizontal and vertical.

When the ramp is horizontal, the perpendicular forces (torque) are Fw*cos(0) in the middle and 1/2 Fw*cos(0) at each end. This is what I see when I put either end of the dumbbell on the scale (1/2 the dumbbell weight) and also the weight I fell carrying one end of the couch on a flat surface.

The forces parallel to the ramp are 0 (or more correctly Fw*sin(0)).

Now let's flip the ramp on it's end (it's vertical). Now the perpendicular (torque) forces are Fw*cos(90) and 1/2 Fw*cos(90) at each end which is 0.

The force parallel to the ramp is now Fw*sin(90) = Fw. This is what I measure when I stand the dumbbell up on the scale. Also if we're on a really steep staircase with a couch and I cheat at the top and push the couch vertical the poor schmuck on the bottom is carrying the full weight.

Do we agree so far?

 

[jbriggs444]

When the ramp is horizontal, the perpendicular forces (torque) are Fw*cos(0) in the middle and 1/2 Fw*cos(0) at each end. This is what I see when I put either end of the dumbbell on the scale (1/2 the dumbbell weight) and also the weight I fell carrying one end of the couch on a flat surface.

To be clear, you are taking ##F_w## as the weight of the ramp i.e. ##mg##. One assumes that you will be taking torques about the center of mass.

Obviously the force of gravity exerts no torque since its moment arm is zero. The horizontal forces, if any, from the two supports also have a zero moment arm. The only two forces providing a torque are the vertical components of the two support forces. Those moment arms are equal and opposite. It follows that each support force must be equal and support half of the ramp's weight.

So for a horizontal ramp, I agree that the vertical load is shared 50/50 on the two supports regardless of the attachment mechanisms.

The forces parallel to the ramp are 0 (or more correctly Fw*sin(0)).

This is not correct. It can depend on how the supports at the two ends are pinned to the beam. The two horizontal components could be equal and opposite, putting the beam in tension. Or they could be equal and opposite, putting the beam in compression. Or they could both be zero. The situation is statically indeterminate.

Nonetheless, as reasoned out above, the vertical load distribution must be 50/50

Now let's flip the ramp on it's end (it's vertical). Now the perpendicular (torque) forces are Fw*cos(90) and 1/2 Fw*cos(90) at each end which is 0.

Not quite.

Again we have the downward force from gravity matched against the supporting forces from the two ends. But the moment arms for all three vertical forces are zero. The total torque from all vertical forces is sure to add to zero.

The horizontal force components have moment arms that are equal and opposite. The torque balance requires that the horizontal force components be equal. The linear momentum balance requires that the horizontal force components be equal and opposite. It follows that both horizontal components must be zero in this case.

However, the vertical support forces can take on any combination that adds to ##F_w##. It might be 50/50 for a vertical beam supported evenly from both ends. It might be 100/0 for a beam hanging from a hook on a rafter. It might be 0/100 for a beam balanced on its bottom end. It might even be 200/-100 for a beam hanging from a hook on the ceiling and being hauled downward by a cable with a turnbuckle. Or -100/200 for a beam balanced on a floor jack and being shoved into the ceiling.

The situation is statically indeterminate.

Do you agree? Can we proceed to the case of a tilted ramp?

 

[dcb]

No, again to be clear, there is no attachment of the ramp to either end. It is not "pinned to the beam". It is simply resting on either side. I'm not attaching the dumbbell to the scale, I'm resting it on the scale. This is the case I'm interested in.

 

[dcb]

Do you agree that the guy on the lower stair holding the vertical couch is carrying the entire weight?

 

[jbriggs444]

No, again to be clear, there is no attachment of the ramp to either end. It is not "pinned to the beam". It is simply resting on either side. I'm not attaching the dumbbell to the scale, I'm resting it on the scale. This is the case I'm interested in.

There is such a thing as friction, you know. Unless you have rollers on your hand or have oiled the teflon scale platform, horizontal forces are something that needs to be controlled for.

 

[jbriggs444]

Do you agree that the guy on the lower stair holding the vertical couch is carrying the entire weight?

Typically more than half, yes. But as I already pointed out, that's down to psychology and physiology, not physics.

 

[dcb]

You clearly do not understand the problem. If the couch is vertical (or horizontal for that matter) the center of mass of the couch is straight down over the guy. There is no way the guy is not carrying 100% of the weight of the couch.

 

[dcb]

Sorry horizonal with only one person.

 

[dcb]

How can the guy at the top of the couch support any weight whatsoever?

 

[dcb]

I suggest you buy a cheap scale. Place a fork on it and it may weight say 20 grams. Now balance the fork vertically on the scale by tapping it with your fingers and it will say 20 grams.

 

[dcb]

When the fork is vertical you fingers are providing zero force.

 

[berkeman]

Necro thread closed for Moderation...

 

[berkeman]

Typically more than half, yes. But as I already pointed out, that's down to psychology and physiology, not physics.

How can the guy at the top of the couch support any weight whatsoever?

By squeezing with their hands and arms and pulling upward. As @jbriggs444 says, it's down to physiology and how much the top person wants to work to try to "pull" their weight.

Anyway, after a PM conversation with @dcb the thread is back open for a while. It's probably best to use a simple beam supported between two scales for working out the math, IMO.

 

[jbriggs444]

I offer this drawing that conforms to a simple beam setup with purely vertical support forces on a beam between two scales at different heights. The beam is at angle ##\theta## from the horizontal.

Case 1: I claim that the beam can be in equilibrium with both ##F_1## and ##F_2## purely vertical. In this case, both support forces will have magnitude ##\frac{mg}{2}##

Case 2: I claim that an alternate possibility is for ##F_2## to be changed to a diagonal direction, up and to the right. If this is done, ##F_1## must also be changed to a non-vertical direction, likely up and to the left. We might then ask:

"If we fix the bottom end in place with a hinge, what is the minimum magnitude for force ##F_2## that can hold the beam in place? At what angle from the horizontal would ##F_2## then be exerted?

Case 3: I claim that is another possibility. ##F_1## could be changed to a diagonal direction, up and to the right. If this is done, ##F_2## must be changed. This time to a direction that is likely up and to the left. We can again ask a question:

"If we fix the top of the beam in place with a hinge, what is the minimum magnitude for force ##F_1## that can hold the beam in place? At what angle from the horizontal would ##F_1## then be exerted?

Roughly speaking, case 1 corresponds to a toy wagon sitting on a flight of stairs.
Roughly speaking, case 2 corresponds to a roofer pushing a ladder away from the eaves.
Roughly speaking, case 3 corresponds to a person holding a pendulum at an angle.

 

[dcb]

As berkeman pointed out to me, I have morphed the discussion into something different than the original post so I apologize but I was interested xshovelfighter's observation of inclined push up's and carrying a couch.

If you assume the couch has zero height everything you say is correct. The height of the couch changes it's center of gravity as the stair angle increases so it is important to the discussion.Couch Problem

Above is a link to a high resolution jpeg if you can't see my image.

The presumption is that the top & bottom guys grasp the bottom of the couch along it's base and let their arms hang down to support it's weight. L is the length of the couch H is it's height and W it's weight.

Setting the bottom guy (B is his upward force) as the reference, W as the weight at the CG of the couch, and T is the upward force supplied by the top guy then
the horizontal moment arms are:

distance BT = L*cos(theta)
distance BW = L/2*cos(theta) - H/2*sin(theta)
Note this cause the CG of the couch to move back more quickly than the base of the couch. Setting H to 0 causes the B and T forces to be W/2 for all angles.

sum of horizontal moment arms around B=0=T*BT-W*BW thus T=W*BW/BT
sum of the Y forces = 0 = W-T-B thus B=W-T

setting L=7, H=2, W=200# and theta=45 degrees

T=72# , B=128#

Somewhere around theta=74 degrees T=0 and B=W=200# so the top guy is doing nothing.

At 75 degrees the CG goes over the top of the bottom guy and flips over unless the top guy has arms long enough to pull the couch back.

The bottom guy has to adjust his grip to the center of the side of the couch (H/2) to keep the CG from flipping the couch.

Does this make sense?

 

[dcb]

This approximates what I see on a scale when I put one end of dumbbell on it and slowly lift the other end.

 

[jbriggs444]

Somewhere around theta=74 degrees T=0 and B=W=200# so the top guy is doing nothing.

I agree with pretty much everything you are pointing out about a couch with non-zero height. However, I want to explore the situation here with ##\theta = 74## degrees. This is a couch with length 7 and height 2.

At 74 degrees, I expect that it can balance on its lower left corner. But let us check that assertion...$$74 \text{ degrees} = arctan(7/2)$$Yes. It is balanced on the lower left corner.

The bottom guy is supporting the couch at its lower left corner. This is at the balance point. The upper guy is supporting the couch at its upper right corner. This is not at the balance point.

Yes, I agree that in this scenario, if both guys are supporting the couch with purely vertical forces, the bottom guy must support the full weight. The top guy can support nothing.

Of course, if the two fellows were holding the couch from the upper left and upper right corners, the situation would be reversed. The top guy would have to supply all of the supporting force and the bottom guy could supply nothing. The couch would be hanging, balanced, from the top guy's hands (in an incredibly difficult posture).

This approximates what I see on a scale when I put one end of dumbbell on it and slowly lift the other end.

Edit: Some reprasing below as I had lost track of which end of the dumbbell was on the scale.

I claim that this is not a well defined experimental procedure. What horizontal force are you exerting on the scale with your hand as the dumbell pivots about its opposite end the end on the scale?

If the horizontal force is zero, then the vertical support force will be split 50/50 regardless of the tilt angle all the way up to just short of a purely vertical dumbbell.

Of course, you may have difficulty applying a purely vertical force with a platform scale your hand as the dumbbell approaches the vertical. You may have to switch to a cord and a tension based force measurement device and arrange carefully for the cord to remain vertical. If this is done, the situation when the dumbbell is purely vertical is statically indeterminate. This was pointed out in post #3 back in 2016:

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.

 

[dcb]

Thank you for joining in this discussion. Every time I try to figure out the X forces it makes no sense to me.

I think that if both guys could somehow grab their end of the couch at h/2 instead of the base, everything changes.

I'm out of the country in a few days for the month of April so I'd like to pick up the discussion then.