A man throws a 2.7kg object at a gate 7m away; how much force hits the gate?

Question

The man releases the object at a height of 1.2 m, it rises another .3m to a peak height of 1.5m, then hits the gate at a height of .3m from the ground. The horizontal distance between the point of release of the object and the gate is 7m. With what force does the 2.7kg object hit the gate? Thank you for any help calculating this. The only ballpark answer I could come up with (for being an extremely Physics-handicapped person) is 104.5 pound/ft. This is from an event that actually happened, with someone throwing an object and damaging our gate; said person does not believe a 2.7 kg object could do this being thrown from the given distance.

perhaps w=mgh?

Object is a phone book: 11cm high, 27cm long, 22cm wide.

gatekeeper doesn't like phonebooks now...

Actually it's two phone books, one larger than the other - I've just approximated the total size. It was also in a plastic bag - swung like a slingshot; Perhaps it hit right on the bound edge, which is rather hard. Certainly would feel brick-like enough if it hit someone's head. In any event, it seems to have hit just right to whack the chain link gate bolts out of alignment.

Answer 1

The force cannot be calculated from the information you have given. The above formulas (weight=force*displacement, etc.) are just wrong. The force will be a function of the object's mass and velocity, and the time it takes for the gate to decelerate the object to zero velocity. That last item is what we don't know. You can either guess at the deceleration time (and we would have to simplify, and lose accuracy, by assuming a constant deceleration over that time), or you would do a very fancy analysis taking into account the mass properties, stiffness and shape of the object and the gate.
EDIT: The equations for the simple but not-so-accurate assumption would be
acceleration = velocity / deceleration time
force = mass * acceleration
(Note that to a physics type, deceleration is the same as acceleration.)
The velocity and kinetic energy could be calculated from the information you have given, but that alone isn't sufficient as I mentioned. But what the heck, it's easy so let's do it. Find the total flight time, and from that the horizontal velocity. Also find impact velocity. For flight time, use the famous formula s=0.5*a*t^2 thus t = sqrt(2s/a). Break the object's flight into two phases, phase 1 the 0.3 m rise to peak height and phase 2 the 1.2 m fall to the collision point. (Don't believe the answer that lumped both distances together as a single fall; it doesn't work that way!)
t1 = sqrt(0.6/g) = 0.247 s
t2 = sqrt(2.4/g) = 0.495 s
Time of flight t = t1 + t2 = 0.742 s
Horizontal velocity = hor. distance/t = 7/0.742 = 9.43 m/s
Final vertical velocity = g*t2 = 7.28 m/s
Impact velocity v = sqrt(9.43^2 + 7.28^2) = 11.91 m/s
Kinetic energy = 0.5*m*v^2 = 191.6 Newton-meters
Impact angle from horizontal = arctan(7.28/9.43) = 37.7 deg
BTW, it takes a special kind of brass to "not believe" one caused damage when flinging 2.7 kg of anything at someone else's property. But I hope this isn't about a 2.7 kg pillow!
EDIT again: Phone book? Not exactly like a brick, is it?

Answer 2

The equation force = Mass * Acceleration shows that it depends on how quickly the object decelerates. For instance, getting hit by a beanbag hurts less than getting hit by a baseball, even if they weigh the same and are going the same speed. Similary, the force is less if the gate can flex.

The distance from the gate is irrelevant if the object is small enough to not be affected by wind resistance, although, as you state the problem, the distance to the gate and the vertical travel will let us calculate the speed of the object as it hits the gate. Acceleration due to gravity is 9.8m/s/s, and the distance fallen is 1.5m, which it covers in 0.55s. In this time over 7m, it must be moving horizontally at 12.7m/s which is 45km/h, and the force to the gate is the same as if the gate was horizontal and you dropped the object on it from 8.2m.

Answer 3

I don't want to work through the entire math of the problem but if you read through it says that the object raises to 1.5m and then hits the gate at .3m so you have gravity pulling it down a distance of 1.2m so you can figure out the downward velocity from that. Then you can also find out how fact the horizonal velocity is by pluging it into the equation that should be in your physics book or just use differental eq to find it out. So now you can take
V(total)=[V(down)^2 + V(horiz)^2]^(1/2)
where V total is the magnitude of velocity and use the to figure out the kenetic energy.

Answer 4

Now what if the gate is open? It might hit the gate keeper and kill the poor bastard.

Answer 5

weight=force*displacement
force=weight/displacement
that is force= 2.7/7=0.3857 newtons