Designing a Pneumatic Arm to Lift 5kg Load 10m

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MadaraUchiha
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Hey
I want to build a pneumatic arm that can throw stuff around (see attachment image).
Before actually ordering the stuff i want help with some calculations.
i want the arm to be able to push 5kg load (rectangular shaped) straight up into air.
the arm/plunger area that makes contact with load = 0.0050m^2
i want this arm to throw the load to atleast 10 metres off the ground.
How much force/pressure would that require ?
 
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A five kilogram load will be pulled by the Earth by 5*9.8=49N of force. So this much is minimum. To throw it 10m high, you need to calculate the acceleration, which requires information on how long the plunger will be in conact with the mass, or, upto what distance. That will be equal to the length of the stroke of the piston which you should decide.
 
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PhysicoRaj said:
A five kilogram load will be pulled by the Earth by 5*9.8=49N of force. So this much is minimum. To throw it 10m high, you need to calculate the acceleration, which requires information on how long the plunger will be in conact with the mass, or, upto what distance. That will be equal to the length of the stroke of the piston which you should decide.

Ill decide that later according to portability and availability of cylinders but let's say the stroke is 300mm.
Then how do we go on about calculating it ?
 
Let me take the stroke to be x.
Now we will calculate the velocity with which it has to be ejected so that it ends up 10m high.
v2=u2-2gh
v is the velocity at 10m so it is = 0
u is what we have to find out.
h=10m
solving for u we get:
u=√2gh
u=√2*9.8*10
u=14ms-1

Now to calculate the acceleration required to produce that initial velocity:
u2=2ax
u is 14
a is what we will find
x is your piston stroke.

Solve for a:
a=u2/2x
a=196/(2*x)
a=98/x

Now the force that causes this acceleration is:
F=ma
F=5*98/x
F=490/x N

Either choose a convinient x value or F value.

Edit: 300mm implies force=1633N.
 
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Got it !
thanks again.
 
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