Physics Help Forum Falling object force for starting lever

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 Feb 16th 2018, 05:06 PM #1 Junior Member   Join Date: Feb 2017 Posts: 18 Falling object force for starting lever I need help understanding how to calculate force from a falling object. I have a lever connected to a spring and I want to know the amount of force on the other end of the spring. If I have a model with a force F that pushes on the first end of the lever (the spring is attached to the other end) then I can do this calculation fine. But I don't just have F. I need to calculate F as the result of an object falling on my lever from some height H. My calculations through the lever and spring already consider the movement of the spring of course so that movement I assume wouldn't be part of finding F. But, the calculations I've seen for finding the force of a falling object always have a step dividing by the distance traveled after impact so that can't be 0 and yet an ever smaller number will yield an ever larger force so I don't know how all this fits together. Please help me understand how to find F to enter into my lever and spring calculations as that F is the result of a falling object. For this example, let's say the object weighs 200kg and it falls 3 meters. What's F? Thanks all.
 Feb 16th 2018, 06:55 PM #2 Senior Member   Join Date: Aug 2010 Posts: 372 What, exactly, do you mean by "force for a falling object"? As long as the object is falling, there is NO force. A falling object only applies a force to something that stops it or slows it. If, for example, an object falling at v m/s strikes a surface that stops it in t seconds, then during that contact it decelerates at -v/t m/s^2. Since "F= ma", assuming the object has mass m kg, F= m(-v/t)= -mv/t N. The surface has applied force -mv/t N to the object, ,the object has applied force mv/t N to the surface. Or suppose the sinks x m into the surface before stopping. An object with mass m kg and speed v m/s has kinetic energy $mv^2/2$ Joules. If the surface applies force F to the object, Then it does work Fx to slow it to 0 where it has kinetic energy 0. Conservation of energy gives $Fx= mv^2/2$ and then $F= \frac{mv^2}{2x}$.
 Feb 16th 2018, 09:27 PM #3 Junior Member   Join Date: Feb 2017 Posts: 18 By force of a falling object I mean the force it would apply to the surface it lands on. In this case, the surface is the lever. Sorry, I'm certain to mix up terms or be missing them all together as I'm not a physics student, other than for my project here, so learning as I go. Let's say I have a lever system like in the image attached. If I say F is 200N, then I am able to calculate the force S on the spring and ultimately how far the spring is compressed for a given spring weight. I understand the steps to do this. What I don't understand is how to find F when it is the result of a 200kg weight that has just fallen 3m before hitting the lever at F. I start with: KE = mgh KE = 200 x 9.8 x 3 KE = 5884.2 Now I can use KE to get the Force at impact. Problem is, I don't know what to use for distance of impact. F = 5884.2 / d d = how for the 200 kg mass moves past the impact point. Obviously the mass will move the lever down but at this point I don't know how far. If I knew F, I would be able to calculate how far it would push the lever by knowing how far the spring would compress. Sort of a chicken and egg situation it seems. What do I use for d here so I can get F to put into calculations in the lever and spring forces? Attached Thumbnails
Feb 17th 2018, 01:09 AM   #4
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Hallsofivey has given the compleat answer

 Originally Posted by builder89 If I say F is 200N, then I am able to calculate the force S on the spring and ultimately how far the spring is compressed for a given spring weight. I understand the steps to do this.
F will not be constant , it will vary from the moment of impact , as the spring is compressed till the object comes to rest ...

At the moment of contact F will be zero , increasing to a maximum when the weight comes to rest ...

This maximum is dependent on the stiffness of the spring and hence the distance fallen after first contact.

So in other words the (variable) Force is different for different springs.

 Feb 17th 2018, 08:53 AM #5 Junior Member   Join Date: Feb 2017 Posts: 18 Right... ok, starting to come together.... No F at top of lever. So, I can specify the max distance the spring will travel. Then I can calculate how far the lever at F will travel. Now I have d so I can calculate max F at the point the lever is all the way down. Then I can feed that F through lever calculations and come up with a force S, which will allow me to compute the spring weight that will be fully compressed at that point. Does this method sound right? Seems strange to me I can come up with max F knowing d alone and not knowing how much the lever pushes back yet. But what if I already know the spring weight and I want to know how far a given F will push the lever down? I don't know d yet. I can't get S yet because I don't know how far the spring will get compressed. Will I need to reorganize all the lever math to carry the F through as a set of variables including 'd' and then isolate d=... at the end? Thanks for the input guys.
 Feb 17th 2018, 08:08 PM #6 Senior Member   Join Date: Apr 2017 Posts: 410 Spring weight ??? this will make things more complicated if the spring and levers have mass ... Is this a practical question or just theory ??? Give all the information you can .
Feb 17th 2018, 09:17 PM   #7
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 Originally Posted by builder89 I need help understanding how to calculate force from a falling object.But I don't just have F.
Its not possible to determine F unless the exact details of the object (floor) and the falling object (baseball?) are specified. This is because the force of the object due to the floor depends on the interaction between floor and ball. The force on a steel floor due to a falling steel ball onto a steel ball is much greater than the force of a sponge falling on a rubber floor. That's because all objects have a certain "give" to them which determines the rate of acceleration of the falling object when it hits something. If you know how the accelerates of the falling object then you know the force on that object.

 Feb 17th 2018, 11:53 PM #8 Junior Member   Join Date: Feb 2017 Posts: 18 This is a practical application but the layout of the machine may differ so I'm trying to understand the process of solving the problem. I'm trying to avoid the details of the 2 levers in the model or the real world linkage that sits between the falling mass and the spring to keep the conversation and math more simple. I don't just want the fish. I want to learn to fish. In this case, so I can design the linkage I need and change it as necessary knowing how to do the math in any case. Looking at the image here: On the left side of the machine there is a spring with rate K. On the right side, there is a box with mass m suspended h meters above point L, which is at the end of a lever. When m is dropped, it will gain velocity until it touches the end of the lever at point L having KE of 1/2mv2. It will then begin to push the lever an additional distance d before KE is completely consumed and equal to 0 and the model looks like the second image where the spring is compressed a distance of c where the movement will begin to reverse back to an equilibrium somewhere. So, in this updated model we have a known spring rate in the spring. The question I want to solve is: For a given mass m and height h, how far will the spring be compressed, c? I'm not sure where to start. If I start on the right side, I can calculate KE at L. My limited understanding leads me to think then I should calculate the force that would be applied to the lever at L yielding some force at U and then some force at S which I can then use to calculate how much the force at S compresses the spring given rate K. But I get it... the lever is moving so I wouldn't know the force at L until the model looked like the second image where F = KE/d. But what is d? Furthermore, as the block moves down on the lever, the angles of force are changing so seems I need to break the vertical distance d into increments and do sums of forces or something so I capture more accurately the force at S throughout the different leverage ratios inside the distance d traveled. Calculating from the left side seems easier to understand. Knowing K, I pick a value for C, easily get a value for S, then U required to get S and L required to get U. I can figure the placement of the levers given the value of C and geometry of the system so now I know d and I can calculate the force at L. Then I can calculate a combination of m and h that would work with d to give me L. But this is a guessing game to try and find the right C to ultimately get me the m and h I'm targeting. What is the best way to answer my question and consider all these forces? For a given mass m and height h, how far will the spring be compressed, c? Attached Thumbnails
 Feb 18th 2018, 03:00 AM #9 Senior Member   Join Date: Apr 2017 Posts: 410 It might be best to forget about (for the moment ) all the levers ... They serve no perpose that could not be achieved by adjusting the stifness of the spring , and their mass brings added complications if you want accuracy ... So the mass could just fall onto the spring .. The answers you require are totally dependent on the stiffness of the spring ...first chose one stifness (how much compression in cm for each N force) ...easiest to assume the spring has negligible mass compared to m ....
 Feb 18th 2018, 02:47 PM #10 Junior Member   Join Date: Feb 2017 Posts: 18 OK. I understand the case of the weight dropping directly on the spring. There are plenty of youtube videos about it. :-) It's the linkage levers that make my case complicated for me. Let's make the linkage a black box. We don't care how it works inside but we do care what it accomplishes. At least, it makes the progression from L1 to L4 non-linear. It also allows me to change the relationship between the amount of spring compressed relative to the amount of lever moved. And, in some cases it may be unrealistic to obtain a spring with a rate too high or low but the linkage allows the forces to be increased or reduced to fit available springs. There is an algorithm for the force on the spring given an L force. I understand now (correct me if I'm wrong) that we can obtain the force L2 at the point in the first picture here simply by calculating mgh/d. So, I believe, as has been stated above, it is not relevant at this point what forces are happening inside the black box. Furthermore if we take h out of the model and just want to know L1 without m dropping distance h, but rather just sitting there at L1 in equilibrium, we can just say L1=mgh, which is the PE, without dividing by a d of 0 since we have no KE (since no movement). So my remaining problem, now that I understand the above points. The work done by moving the lever from L1 to L4 is not linear. It may take work w to move the lever from point L1 to L2, 1.5w to move from point L2 to L3 and then .75w to move from L3 to L4... or something like that. Is it still true that the force exerted at L4 is simply mgh/(d1+d2+d3)? Attached Thumbnails

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