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Drop forceby rampage11
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#1
Mar2013, 04:16 PM

P: 6

im trying to calculate the drop force of falling object. we do drop test at my work and believe the figures are wrong. we drop a 24.75lb weight from different heigths to check parts . they have it now set that dropping the given weight from 1 ft only produces 24.75 ft/lb force and they add 24.75 for each foot after that. I came up with around 74.86ft/lb at 1 foot. 24.75 seems low at 1 foot. any help is appreciated. brandon



#2
Mar2013, 04:24 PM

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#3
Mar2013, 04:38 PM

P: 6

the parts has to hold 130lb of sheer force



#4
Mar2013, 06:52 PM

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Drop force
The numbers in the OP don't make any seise to me. The force you apply during the impact depends very much on the deceleration of the mass, as well as the height you drop it from.
For an impact between two stiff and "hard" objects, the maximum impact force could be hundreds or thousands of times the weight of the dropped object. Aside from that, the unit ft/lb isn't a "force"  maybe you meant ftlbf, which is the energy of the dropped object, as Doc Al said. 


#5
Mar2013, 07:12 PM

P: 6

ftlbf is what I meant..the part moves about 1/2 inch at impact



#6
Mar2013, 08:25 PM

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Potential energy then is weight times height.



#7
Mar2013, 08:58 PM

P: 6

130 psi of shear force from drop test is what our prints say



#8
Mar2013, 11:18 PM

P: 105

It is impossible to predict the force (average or peak) during the impact without knowing what material you are dropping the weight onto. I think you said that your specimen deformed a maximum of 1/2 inch during the test? That could be useful information, if we knew the area and thickness of the specimen, and if we knew how elastic the collision is.



#9
Mar2113, 05:33 AM

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The test sounds a little like a charpy impact test, which measures fracture energy:
http://en.wikipedia.org/wiki/Charpy_impact_test But I'm sorry, there is no such term as "drop force" used in engineering. 


#10
Mar2313, 05:17 AM

P: 17

Hey,
Since the object is having sudden impact, the impact force will be more twice the load. The general equation of the impact load is : W+[itex]\sqrt{(W^{2} + 2AWhE/L)}[/itex] where, W  weight of the object A  surface area of impact L Length of the body perpendicular to surface of impact h height from which body is dropped E Modulus of elasticity of the object Just put in the material and geometrical properties.. So the value comes out to be greater than twice the weight of the body. So 74.86 seems kinda ok. But I don't know the dimensions or the material dropped. So I can't tell.. If you want, just give me the details.. I'll try to work out the values 


#11
Mar2313, 10:33 AM

P: 6

ill have to remeasure it..but I believe its 4'' round stock steel..12'' long dropped from 1 ft increments..weighs 24.73lb



#12
Mar2313, 12:07 PM

P: 105

Is the specimen being dropped onto something "rigid?" Or are you dropping something "rigid" onto the specimen? Your specimen is 4" diameter steel?



#13
Mar2513, 07:23 PM

P: 6

rigid..dropping specimen onto a 5/8 steel rod in 1ft increments



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