# Drop force

1. Mar 20, 2013

### rampage11

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. Mar 20, 2013

### Staff: Mentor

Are you sure it's not 24.75 ft-lbs of kinetic energy per foot drop, not force? That would make sense.

3. Mar 20, 2013

### rampage11

the parts has to hold 130lb of sheer force

4. Mar 20, 2013

### AlephZero

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 ft-lbf, which is the energy of the dropped object, as Doc Al said.

5. Mar 20, 2013

### rampage11

ft-lbf is what I meant..the part moves about 1/2 inch at impact

6. Mar 20, 2013

### Staff: Mentor

Potential energy then is weight times height.

7. Mar 20, 2013

### rampage11

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

8. Mar 20, 2013

### afreiden

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. Mar 21, 2013

### Staff: Mentor

10. Mar 23, 2013

### sci-phy

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+$\sqrt{(W^{2} + 2AWhE/L)}$

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. Mar 23, 2013

### rampage11

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

12. Mar 23, 2013

### afreiden

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

13. Mar 25, 2013

### rampage11

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