Starting force of a falling object on a compound lever

[Builder89]

Thanks everyone for the help. I think I'm good to go now. :-)

 

[haruspex]

Thanks everyone for the help. I think I'm good to go now. :-)

OK, but I'd like to be confident you are calculating the torques correctly: $\tau = Fr\sin(\theta)$

 

[Builder89]

I'd love for you to look over them. :-)

Here's how I calculate the vertical force up (FM) on the swing arm, which would then have the shock angle taken into account as applied to the bottom of the shock.

The free body diagram would look like this:

Let's just say FR = 100N (cause I forgot we already calculated 980N)

I find FM using the torque equilibrium about FL.

FR(cos12)(495) = FM(cos12)(102)
FR(484.14) = FM(99.7)
485.60N = FM

Then, let's say the shock is stood up 50° clockwise from horizontal with the bottom touching the point of FM.

The force into the shock, FS = cos40(458.60) = 371.99N

Look right?

 

[haruspex]

FR(cos12)(495) = FM(cos12)(102)

The force into the shock, FS = cos40(458.60)

No, you have to treat F_R and F_S as two forces acting at different angles on the arm, balancing:
F_Rcos(12)(495) = F_Scos(40)(102)

 

[Builder89]

Hmmm, really? I feel you're wrong. FL is the pivot point, O. FM and FR are torque forces working to twist around FL in opposite directions.

 

[haruspex]

Hmmm, really? I feel you're wrong. FL is the pivot point, O. FM and FR are torque forces working to twist around FL in opposite directions.

Actually I was a bit wrong, but not in the way you think.
You said take the shock as being 50 degrees above the horizontal. That means it is at only 38 degrees to the rear wheel arm. If the compression in the schock absorber is F_S, its component normal to the arm is F_Scos(52), so the torque it exerts about F_L is F_S cos(52) (102). This has to balance the torque from F_R.
F_S cos(52) (102)=F_R cos(12) (495).
So F_S=7.7 F_R.

 

[Builder89]

Yeah, I understand. Your solution is simpler than the way I was trying to do it and it makes sense but I'd still like to know why my method yields a different result.

My first attempt at FS I think was wrong, even for my intended approach. Let me explain.

The shock is 50 degrees above horizontal so it's 40 degrees from vertical.

If FM is a vertical force at that point of 485.60N as calculated above, why can't I get the same FS using the following?
cos40 = 458.6/FS
FS = 458.6/cos40 = 595.58

 

[haruspex]

why can't I get the same FS using the following?

Because part of the compression force on the shock comes from the reaction force at the pivot.

 

[Builder89]

I'd like to believe you but I can't get it to reconcile with other information I think I understand so I must be wrong maybe somewhere else?

FM was calculated above to be the vertical force down on that free body diagram. That makes me think there is mathematically an equal and opposite force pushing up at FM. But maybe that's where I'm wrong, cause intuitively I know the forces are at the ends of the swing arm.

If we CAN treat the diagram as if there is an upward vertical force of FM at that point, then I would think I can feed it into the following (incomplete) free body diagram of the shock where FM is pushing up vertically to a horizontal plane, which transfers into the bottom of the shock as FM/cos40.

But you're saying I can't isolate FM as vertical output from the swing arm to be fed into this model? Seems I should be able to come up with an FM that could then be fed into whatever I want to set on top of that FM, like a shock or a linkage mechanism. And if that was true, then the calculation on this diagram would equal the value found from the direct torque calculation you presented (which I believe is accurate), in this case.

So, why can't FM be isolated as output from the swing arm diagram and fed into some other mechanism after the fact as a vertical force?

 

[JBA]

The torque force vector at the shock connection to the swing arm must be perpendicular to the centerline of the arm because a portion of the vertical FM force is translated along the centerline of the swing arm to its pivot.

The torque load perpendicular to the swing arm centerline from the vertical FM = FM * cos 12 and the angle between the shock centerline and the torque force vector (perpendicular to the center of the swing arm) at the shock connection point is 40 + 12 = 52°, so the force along the shock centerline is:
FS = 495/102 * FM * cos 12 / cos 52° = 7.71 FM (as is presented above).

 

[Builder89]

7.71FM?

FS = 7.7FR above.

 

[haruspex]

why can't FM be isolated as output from the swing arm diagram and fed into some other mechanism after the fact as a vertical force?

Sure, but that is just the vertical component of F_S.
To make it simpler I'll ignore the 12 degree slope for now.
F_S cos 40 = F_M. The horizontal component of F_S is fighting the tension in the arm.
But the proper way is the way I laid out, resolving F_S into components normal to and parallel to the arm.

FS = 495/102 * FM * cos 12 / cos 52° = 7.71 FM

I think you mean F_R, not F_M.

 

[Builder89]

But guys, why is JBAs calculation using 495/102 and cos12 at all? I get that FM is the vertical force but that's the point. The vertical force was calculated so it could be applied to the next step of the mechanism. FM = 485.60N already accounts for the fact the swing arm is slanted 12°. Then FS = FM/cos40 considers the shock is slanted 40° from vertical FM yielding the force vertical FM applies in the direction of FS.

FS cos 40 = FM

But:
cos(theta) = adj/hyp.
cos40 = FM/FS
cos40/FS = FM

 

[JBA]

I apologize, I could not keep both the post #33 diagram and my post on screen at the same time and just got my references mixed in the process of developing the equation.

Obviously, based upon the post #33 diagram, it should have read: FS = 495/102 * FR* cos 12° / cos 52° = 7.71 FR

With regard to the 12° angle used in this calculation, the distribution of the FM force between the torque force vector component perpendicular to the swing arm and its tension/compression component along the swing arm centerline (that transmitted to the pivot) will vary along with the angle of the arm and shock throughout the arm travel arc, while the vertical orientation of the ground contact vector will always be vertical. As a result you must evaluate the shock connection force at each point of interest along the swing arm's travel to determine the point (angle) at which FM will be at its maximum value.

The effect of the FM connection on shock will always be along it centerline; however, when analyzing the stresses on the swing arm it must be resolved into its perpendicular component which will determine the arm's bending stress and its longitudinal component which will determine the tension/compressive stress along the arm's centerline.

 

[haruspex]

cos(theta) = adj/hyp.

In your diagram in post #39, there are three forces acting where the shock meets the arm: F_S, F_M and the compression in the arm. You seem to be resolving in the direction parallel to the shock, but if you do that you must include the compression in the arm. Simpler to resolve normal to the arm.

 

[Builder89]

Guys. Maybe we aren't seeing eye to eye here.

I'm trying to find the FM, which is a vertical force at this point in time where the swing arm is at 12° below the horizon. That FM is a number and that number should be able to be inserted as the up force on the next mechanism to sit on top of this point, such as the shock, or anything else at any other angle.

When calculating FM, we shouldn't consider the shock angle because FM is a vertical force. In all cases vertical force of FM should be the same given the context of post #33. Yes, the next mechanism, like the shock, will have it's own angle(s) but the next force into that shock is calculated starting with a vertical force of FM.

Am I just completely off my rocker here? It feels like you guys are considering too many variables past the point of FM.

What if the shock is replaced by this:

See... I still should be able to have the same FM as input to this mechanism even though the math to apply that FM to the triangle will be totally different than for the shock. Then I can manipulate the shape of the triangle and the length of the rod at FL to take that FM and yield some FR at this point of 12° swing arm. But all those variations will use the same FM. (The shock connects to the FR point of the triangle laying up and left)

Then I'll do it all over again with a different degree swing arm further in the rotation and a new FM for that case, applying that new FM to different triangle configurations.

But... we're just talking about calculating FM on the 12° swing arm in this thread.

 

[haruspex]

I'm trying to find the FM

Why?
Up until this point, I had the impression you wanted the relationship between the compression in the shock, FS, and the normal force from the ground on the rear wheel, FR. FM does not exist as an actual force; it is only a way of representing the torque from FR. Why do you care what its value is?

 

[JBA]

Comment deleted

 

[Builder89]

You're correct. The end goal is to get FS based on FR. But you asked how I would calculate the forces/torques and I told you. Apparently my method is wrong so I'm trying to understand why because the way you suggested, though cleaner in the case of the shock doesn't apply to my triangle case which is my real world case, so although I understand A way to solve my original posted question, the process has told me maybe I have over simplified the problem and ended up understanding incorrectly how to apply this to my real world case.

My thought is that if I know FM as a vertical force, I can then apply it to either a direct shock mount at some angle from the swing arm OR the triangle linkage mechanism.

I don't feel I'm getting a direct answer on why FM vertical can't be used this way because I thought FM = 485.60N already takes into consideration the angles of the swing arm, which means I should be able to apply it directly to the free model of the shock.

You guys are telling me I can NOT calculate vertical FM unless I consider the angle of FS as part of the FM calculation. But in fact FS is not even in the free body diagram and surely I can still calculate FM. Then why can't I take that vertical value and apply it to the next free body diagram of the shock as FM, the vertical force from the bottom?

 

[haruspex]

Apparently my method is wrong so I'm trying to understand why

This is what I have tried to explain in posts #42 and #45.
Yes, you can calculate F_M from F_R the way you have, but you are wrong to then take the component of this along the shock as the force it exerts on the shock. The tension in the arm adds to it, leading to a much greater force on the shock. You can calculate that tension, or avoid the need to by taking moments about the pivot.

 

[Builder89]

FS cos(52) (102)=FR cos(12) (495).
So FS=7.7 FR.

OK then. I understand what you are doing here. It makes sense to me. I don't understand why the other one does NOT work but I DO understand what is going on here with the opposing torques being equal and the moments multiplied by distance from fulcrum.

But this is simple with just the shock.

How do I do the same thing for the linkage triangle in post 46? I'm not sure where to start to get FM.

I'd use torques to get a relationship between FL and FR.

FL(sin80)(15) = FR(sin70)(30)
FL(14.77) = FR(28.19)
FL = 1.91(FR)

Then I'd use sum of forces to get FR:
0 = FM - FLV - FRV
FM = FLV + FRV
FM = FL(cos25) + FR(cos55)
FM = 1.91(FR)(cos25) + (FR)(cos55)
FM = 1.73(FR) + (.57)(FR)
FM = 2.30(FR)
FR = FM/2.30

But how to get FM since I can't take the vertical component from the swing arm calculation. Do I do a torque calculation somehow using both the 15 and 30 sides of the triangle together like FS was used in your quote at the top of this post? I'm lost here.

 

[haruspex]

How do I do the same thing for the linkage triangle in post 46? I'm not sure where to start to get FM.

In the drop scenario, you cannot compute F_M independently of the shock behaviour.
First, the forces vary during the landing, so you have to decide what force you want to know. I suggest the peak force - ok?
The graph of force against time (whether it's F_M, F_S or F_R) will depend on how the shock reacts. The more compression it allows, the longer the time, so the lower the forces.
Just how much do you know about the absorber? In post #22 I listed some possible features and parameters.

 

[Builder89]

Information about the shock isn't published in that level of detail. Basically you buy the shock and then you buy the spring for it that compresses 25% with the rider on the bike sitting still. The shock is built with this configuration method in mind. That's a starting point but rider preference and terrain could result in the rider choosing a different spring for the occasion. Let's take advantage of that configuration method.

At this point in time represented by the diagrams, I have a load of 100kg (980N) from the rider and bike just sitting there. The swing arm is at this 12° location without movement. At this point, the shock is compressed 25%. The total possible travel (distance from fully uncompressed to fully compressed) of the shock is 3in (76mm) so this 980N at the axle is compressing the shock 19mm right now. The measurements in the diagrams are mm.

We should be able to calculate the force pushing on the bottom of the shock at this time. Then we can calculate the spring rate that would allow the spring on the shock to be compressed 19mm by this force at this time.

Once we have the spring rate, we could calculate the other forces, such as the force it would take to compress the spring 100% and how much force would be applied to the frame, etc.

 

[JBA]

The spring and the shock are going to act in concert to absorb the mgh energy of the falling mass so in order to calculate the total shock/spring travel and the final force on the shock/spring connection points it is necessary to know how much energy the shock absorbs per unit distance of travel.

i.e. mgh = 1/2*k*x^2 + (shock E/dx) * x

If you can determine the energy absorbed by the shock at a desired x travel, then that x value can be plugged into the spring section in that formula to solve for the required k (spring rate) for the spring.

 

[Builder89]

JBA, as stated in post #53, we are analyzing the suspension without motion, sitting still, in equilibrium with the shock compressed 25% and need to calculate the spring rate at that point that allows FR to push said spring 25% (76mm).

At this point, as I said in post #51, I'm unsure how to get the FM going into the bottom of the triangle and I'll need that in order to calculate FR.

 

[JBA]

Your first post #3 bicycle figure best illustrates the force vectors associated with your spring mounting; and based upon that figure and succeeding angles and connection dimensions for the swing arm and shock/spring and FR load points, it appears to me that we have already done that by calculating FS = 7.71 FR, from that k*x = 7.71 FR or k = 7.71 FR / x so if you want the spring compression to be 3" then k =7.71 FR / 3 (lb/in).

Edit: or the appropriate mks value that is std for springs.

 

[Builder89]

We calculated FS when the shock was connected directly to the swing arm. I explain in post #51 the problem that left me with as the method we used didn't apply to calculating the force on the linkage triangle when it was between the swing arm and shock, which is my final goal. Originally I left the triangle out because I figured I could calculate FM from post #3 and apply it directly as FM vertical component at the bottom of the triangle. But apparently I was wrong about that.

You can see in post #51 how I intended to use the vertical component of FM from post #3. But instead we got FS, which doesn't apply to the bottom of the triangle.

 

[JBA]

OK, at this point it would help me understand what you need if you would just provide a bicycle figure like that in post #3 with the added linkage you want to use. All of these partial linkage diagrams you have given do not allow me to see the context of the complete structure.

 

[Builder89]

Here is a rough representation of the context it goes in. All points on the triangle are pivots. FL is a rod attached to the frame on the bottom via another pin (pivot). The shock is only connected at the right point on the triangle and the frame up and left. Remember, we are doing the calculation in equilibrium at this point with the shock compressed 76mm. As such, this is how it looks. All numbers are mm.

And, as stated before, the FA is just the portion of weight on the bike that is at the rear tire.

 

[haruspex]

Here is a rough representation of the context it goes in. All points on the triangle are pivots. FL is a rod attached to the frame on the bottom via another pin (pivot). The shock is only connected at the right point on the triangle and the frame up and left. Remember, we are doing the calculation in equilibrium at this point with the shock compressed 76mm. As such, this is how it looks. All numbers are mm.

And, as stated before, the FA is just the portion of weight on the bike that is at the rear tire.

I believe your force diagram for the linkage in post #46 is wrong. (If correct it would be overspecified, with two unknowns, the magnitudes of the two forces, but three equations.)
In your diagram just above, which of the connections in the linkage are free joints? Even if they are all free joints, the F_R force is not constrained to act along the rod that goes to the rear wheel. That is a rigid rod, so it can generate a restraining force in any direction.
If we go back to your post #46 diagram, we can make the angle you show as 110 degrees an unlnown. That gives us the right number of unknowns. From here, it is just a matter of writing out the horizontal and vertical force balance equations (ΣF=0) and the torque balance equation. The easiest is probably to use the joint with the unknown angle as axis for torque. That allows us to determine F_L immediately.

(It is unfortunate you have reused "F_R". This might lead to confusion.)