Calculating Force on Cushioned Legs from Dropped Mass

  • #1
btbam91
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Say we have a platform with 4 legs attached. Attached to each leg is a small cushion.

If we drop our object of mass m a distance h, the small cushions compress completely in time t.

So we have variables: m,h,t

My analysis went has follows:

1. Energy conservation to find the max velocity the instant before our object hits the ground.

mgh = (1/2)mv^2

v = sqrt(2gh)

2. To find the impact force, set impulse equal to the change in momentum.

J = mv2 - mv1 = F*t

v2 = 0 because our object will stop when it is fully compressed, and v1 is the velocity calculated earlier.

F*t = -m*sqrt(2gh)

|F| = (m*sqrt(2gh))/t

3. In order to find the force in each leg...

|F| = (0.25)*(m*sqrt(2gh))/t


Is this correct?

Thanks for the help!
 
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  • #2
3. In order to find the force in each leg...

|F| = (0.25)*(m*sqrt(2gh))/t


Is this correct?

Thanks for the help!

Assuming that the object of mass m is dropped on the dead center of the platform, the force you are calculating by your formula is the NET force acting on each leg. This is always true when using Newtons 2nd law. You might want to separate out the weight and normal force acting on the leg.
 
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