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Old Nov 14th 2013, 07:49 AM   #1
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Calculating forces on a platform

Hi there,
I've been struggling with this problem for a little while, I'm not sure quite how to solve it. Essentially it is this:

A platform is suspended at points A, B, & C and held in equilibrium. a downward Force F is applied to the platform at point P. Calculate the distribution of force F that is applied to points A, B, & C.

The distribution of force will be used in the equations below:

(Force at A) = F * (Ratio at A)
(Force at B) = F * (Ratio at C)
(Force at C) = F * (Ratio at C)

I'm assuming the sum of distributions is equal to 1, as any object added to the platform cant weigh more than it does ^^ So:

1 = (Ratio at A) + (Ratio at B) + (Ratio at C)

I've tried a few things to work out the correct ratios using this as my main sanity check. If anyone can explain how I can work this out, or point me in the right direction I would be very grateful. It would also be great to know how to find the distribution given any number of support points on the platform.

Thanks in advance!

Last edited by Marni; Nov 14th 2013 at 09:30 AM.
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Old Nov 14th 2013, 08:30 AM   #2
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You are correct that the sum of reaction forces at A plus B plus C must equal F. The other equations(s) that determine the actual forces have to do with moments. You have three points of interest - A, B, C and P - and you can use the fact that sum of moments at any two of them must equal zero in order to determine the the forces at A, B and C.

This assumes that the supports at A, B and C are "simple supports," meaning the platform is free to rotate at the support point (such as if hanging from a cable). If the support is fixed - such as a cantilever from a wall - then the problem becomes much more complicated; you would have to use beam and plate theory to take into account bending of the platform to solve such an indeterminate problem.

Last edited by ChipB; Nov 14th 2013 at 08:48 AM.
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Old Nov 14th 2013, 08:54 AM   #3
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Hi Chip, thanks for the reply.

We can assume the supports are "simple" and the platform is perfectly rigid.

Perhaps a simpler problem may be better, how would you calculate the moment forces for points A, B, and C? I have no idea on how to calculate moments on anything other than a 2 dimensional bridge, and I can't seem to find anything useful from Google searches.

Last edited by Marni; Nov 14th 2013 at 10:19 AM. Reason: typo
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Old Nov 14th 2013, 11:47 AM   #4
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Thinking about this some more and realizing that points A, B and C are not co-linear I have a slightly different suggestion. Let's call the tensions of the supporting cables T_A, T_B, and T_C. Draw a line from point A to point B, and determine the perpendicular distance from that line to points P and to C. Let's call those distances D_P and D_C respectively. The sum of moments about the line equals D_F times F minus D_P times T_C, which equals zero. Thus T_C = (D_P/D_C)F.

Now do the same using the line from points A to C, or form B to C.

These two equations plus the equation T_A+T_B+T_C = F gives you three equations in three unknowns, which you should be able to solve.

Last edited by ChipB; Nov 14th 2013 at 01:38 PM.
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Old Nov 15th 2013, 01:49 AM   #5
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Hi Chip,

Thanks for the advice! When testing that method, the sum of the ratios doesn't add to 1. I've had a think about it, and I was wondering should there be another component to the equation?

It makes sense that if points A & B are both equidistant from C ten it would follow this logic, but say if point B was infinity far away C, then C should get a larger force than otherwise. So I'm guessing length A to B comes into play.

Also would distances not be be needed from the line from C going through P and intersecting line AB?


Last edited by Marni; Nov 15th 2013 at 02:53 AM. Reason: logic
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Old Nov 15th 2013, 07:23 AM   #6
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I've tried using your method, but it seems I'm still getting the wrong answer. I'm not sure whether it's the logic or the execution. An example that I've tried to work through is as follows:

Given the points:

A = (0,0)
B = (9,0)
C = (6,6)
P = (4,2)

I get the lines:

AB -> y = 0
BC -> y = -0.5x + 9
AC -> y = x

I then calculate their perpendicular lines while intersecting the points:

A_BC -> y = 2x
P_BC -> y = 2x + 6

B_AC -> y = -x + 9
P_AC -> y = -x + 6

C_AB -> x = 6
P_AB -> x = 4

From these I then work out the intersecting points of the lines with their perpendicular lines:

I_A_BC = (3.6, 7.2)
I_P_BC = (1.2, 8.4)

I_B_AC = (4.5, 4.5)
I_P_AC = (3, 3)

I_C_AB = (6, 0)
I_P_AB = (4, 0)

I then use the intersecting points to calculate the distance from their relevant point (using Pythagoras theorem):

D_A_BC = sqrt(64.8)
D_P_BC = sqrt(48.8)

D_B_AC = sqrt(40.5)
D_P_AC = sqrt(2)

D_C_AB = 6
D_P_AB = 2

Then assuming the equation to calculate the ratios is simply: (the perpendicular distance form opposite line to the point force / the perpendicular distance from the opposite line to the corner point) Then I get:

Ratio_A = 0.867....
Ratio_B = 0.222...
Ratio_C = 0.333...

Which is incorrect as they don't total 1. /sadface

Can you see where I'm going wrong? Or is there a simpler way of going about it?
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Old Nov 16th 2013, 06:07 AM   #7
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Your equation P_BC should be y= 2x-6. Make that correction and ratio A turns out to be 0.444, and the ratios do indeed add to 1, as expected.
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Old Nov 16th 2013, 06:34 AM   #8
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Originally Posted by Marni View Post

It makes sense that if points A & B are both equidistant from C ten it would follow this logic, but say if point B was infinity far away C, then C should get a larger force than otherwise. So I'm guessing length A to B comes into play.


It depends on the arrangement. If B moves to infinity along the AB line then the reaction force at C stays the same, and the force at A increases while the force at B goes to zero. If B moves to infinity along the CB line, then the reaction force at A stays the same and the force at C would increase.

Originally Posted by Marni View Post
Also would distances not be be needed from the line from C going through P and intersecting line AB?
No.

Last edited by ChipB; Nov 16th 2013 at 06:46 AM.
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Old Nov 16th 2013, 07:56 AM   #9
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Excellent thanks so much!
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