huangyhg

force due to falling objects
my physics information are not that fresh so i need some help please.
i’d like to know how to calculate the force of a falling 30 kg object from 5m high.
is it 1/2*m*v^2 where v^2=g.h
thanks
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the equations you described are for kinetic energy and potential energy. perhaps you can elaborate upon your paricular design problem. is your falling object going to strike a beam or some other structural element? lets give you an example:
potential energy from a suspended object will convert to kinetic energy when the object falls and then convert to spring energy when it strikes the beam.
spring energy=0.5kx^2
x=deflection
k=48ei/l^3 (spring constant for a simple beam with concentrated load at the center)
p=kx (concentrated load)
you can refer to design of welded structures by omer blodgett for more info.
some guy named newton suggested f=ma
whilst you can't disagree with the previous post (watch, someone will ! ... actually someone already has, einstein), anyways, i think the op is asking about the impulse of the mas ?
frusteng, over to you, 'cause the force is independent of the height. are you dropping something onto something else ? the force felt by the receipent will depend on the height (the velocity achieved) and the time it takes to stop it.
good luck
motorcity. thats for an object that isn't falling. impact forces resulting from falling objects is a different story.
impact equations are not an easy thing to figure out. the defining factor is how long does it take the object to come to a complete rest once it hits the stationary object. there is a finite, however usually quite large deceleration of the object when it hits the ground. it will all depend on how much of a distance the object is going to decelerate. the longer the distance, the less the deceleration and thus the less the force that is a result.
in reality, it is going to be impossible to calculate the impulse force, without guessing at some point. there is surely some guide out there with something along the lines of multiplying times the weight (f=ma*3), but i don't know of it.
well, i rememer that f=ma is the force needed to keep (m) moves at a rate of acceleration (a). this force obviously is independant of the height.
my problem is that i am required to account for the load applied on a beam due to the fall of the object,( 30kg from 5m). what is that load in newton or kn or pound...
thanks
frusteng,
you have defined your case further. you know the following:
a) i (moment of inertia)
b) e (youngs modulus)
c) l (beam span)
d) h (height of object above beam)
e) g (gravity)
f) m (mass of falling object)
now calculate:
k (spring constant of your beam) = 48ei/l^3
now calculate:
x (deflection of beam) = (2mgh/k)^0.5
now calculate:
p (load due to falling object hitting beam) = kx
there you go.
eddyc,
sounds too complicated especially that i don't know the size of the beam. a trial and error is required i guess. any practical conservative solution?
this is a temporary overhead canopy for the enterance of a building where some constuction is being done. the structure is wood.
thanks
"sounds too complicated especially that i don't know the size of the beam. a trial and error is required i guess. any practical conservative solution?"
i don't think a simple conservative solution exists. i think you have three choices:
1. use your judgement to determine that you shouldn't worry about 30kg from 5m. i hate monkeymeters, but i think that's 66 lbf falling from about 15 ft. i don't know enough about your roof system to know if that's a lot. intuitively, i think i'd be worried about it landing between two beams and crashing through the deck a lot more than the effect of landing on a beam.
2. tell them they can't do it.
3. calculate the force imposed on the beam from this load as suggested earlier. break out that physics book. this calculation is not difficult. calculate the kinetic energy of the object when it contacts the beam. the kinetic energy is transferred to spring potential energy, allowing you to calculate how far the beam deflects. from this deflection, you should be able to get the beam internal forces.
dbd
hi frusteng
whilst i agree with the majority of the previous posts here is my ten pennies worth.
firstly a collision which involves a fixed surface falls outside the theory of "conservation of linear momentum" as
the fixed surface would be an external impulse.
so in the case of a moving and fixed object collision newtons law of restitution should be used.
this however would be difficult as stated previously as the time of contact between the two objects would be difficult
if not impossible to determine, as would whether there was
any bounce of the moving object from the fixed surface.
however if we assume that there is no elasticity in the collision (no bounce of object from fixed surface)and therefore the coefficient of restitution is zero then we can
say that :-
impulse (j)= final momentum - initial momentum
the final momentum will be zero as the coefficient of restitution is zero however the initial momentum is that of the falling object which can be calculated as follows:-
conservation of energy:- mgh= 0.5*m*v^2
30*9.81*5=0.5*30*v^2
v= 9.9 m/s
therefore impulse= final momentum - initial momentum
= 5*9.9-0 = 49.52ns (impulse force)
regards desertfox
that's probably fine as far as it goes, but that is not a conservative solution.
cheers
greg locock

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