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Started By
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Physics question
Posted on 4/11/18 at 7:59 am
Posted on 4/11/18 at 7:59 am
Take three coins and put into a triangle (with them touching each other). Strike perfectly against two coins with a fourth coin. Does the opposite coin (that's on other side of where you struck) move or do just the two coins you hit only move and the opposite coin doesn't move? The coin you strike with does not continue forward, striking the third coin.
9
Posted on 4/11/18 at 8:01 am to theantiquetiger
Couldn't this be solved using a simple statics equation?
Posted on 4/11/18 at 8:01 am to theantiquetiger
You didn't do your homework did you?
Posted on 4/11/18 at 8:01 am to theantiquetiger
frictionless table?
Posted on 4/11/18 at 8:02 am to theantiquetiger
nvm I read wrong.
This post was edited on 4/11/18 at 8:06 am
Posted on 4/11/18 at 8:02 am to Mingo Was His NameO
quote:
Physics question by Mingo Was His NameO
You didn't do your homework did you?
Sitting here at work bored, have four pennies sitting on desk. I was testing my question.
Posted on 4/11/18 at 8:04 am to bamafan1001
quote:
what? You need to reword that. Are you striking one or two coins?
You are striking two coins on the bottom of the triangle at the exact same time. The striking coin does not continue forward striking the third coin. Does the third coin (top of the triangle) move?
This post was edited on 4/11/18 at 8:06 am
Posted on 4/11/18 at 8:04 am to theantiquetiger
quote:
Take three coins and put into a triangle (with them touching each other). Strike perfectly against two coins with a fourth coin.
And shove it up your butt! hahahah
Posted on 4/11/18 at 8:11 am to theantiquetiger
What's the coefficient of restitution of the coins?
Posted on 4/11/18 at 8:23 am to theantiquetiger
quote:
Does the opposite coin (that's on other side of where you struck) move
Yes
Posted on 4/11/18 at 8:23 am to TigerstuckinMS
quote:
What's the coefficient of restitution of the coins?
Damn, everyone and everything wants restitution these days.
Posted on 4/11/18 at 8:25 am to theantiquetiger
You expect a treatise on Inelastic vs elastic collisions on the OT. You’ve clearly come to the right place.
Posted on 4/11/18 at 8:25 am to 50_Tiger
quote:
Couldn't this be solved using a simple statics equation?
So you want to solve a dynamic movement problem with statics? Sounds like a plan....
Posted on 4/11/18 at 8:33 am to KG6
quote:
So you want to solve a dynamic movement problem with statics? Sounds like a plan....
Yeah the striking makes it a dynamic equation.
The original thought in my head was to make a force diagram and determine if the striking force is enough to overcome the frictional force. Which is not static but dynamic.
That's my mistake haha.
Posted on 4/11/18 at 8:51 am to 50_Tiger
The way the question was posed can be solved as a vector problem. You aren’t looking to see how fast or how far the coin moves. You have three static objects with an external force applied - you can find a resultant vector with a magnitude and direction that would determine if the third coin would move.
Coins of equal mass on a rather smooth table with the speed of a finger flick gives you pretty good assumptions and making it into a dynamics problem is beyond the scope of what was asked, imo.
Coins of equal mass on a rather smooth table with the speed of a finger flick gives you pretty good assumptions and making it into a dynamics problem is beyond the scope of what was asked, imo.
Posted on 4/11/18 at 9:01 am to Engineer
I was bored and tried to figure this out. I normally deal with electrical problems so it's always a nice refresher in the mechanical realm.
Three pennies connected: 2 Forces (1 equal and opposite, other @ 45 deg)
mass of penny = 2.5g
g = 9.81 m/s^2
F1 is equal and opposite
F2 = Fx2 + Fy2 = cos(45)2.5*9.81 + sin(45)2.5*9.81 = 34.678 N
Coefficient of Friction on a wooden table ranges from (0.25-0.5)
Ff of penny on table = umg = (0.25-0.5)(2.5g)(9.81 m/s^2) = (6.131-12.26) N
Since F2 > Ff the penny should overcome the frictional force and move in the path of F2.
How did I do?
Three pennies connected: 2 Forces (1 equal and opposite, other @ 45 deg)
mass of penny = 2.5g
g = 9.81 m/s^2
F1 is equal and opposite
F2 = Fx2 + Fy2 = cos(45)2.5*9.81 + sin(45)2.5*9.81 = 34.678 N
Coefficient of Friction on a wooden table ranges from (0.25-0.5)
Ff of penny on table = umg = (0.25-0.5)(2.5g)(9.81 m/s^2) = (6.131-12.26) N
Since F2 > Ff the penny should overcome the frictional force and move in the path of F2.
How did I do?
Posted on 4/11/18 at 9:35 am to 50_Tiger
Given the coins are round and overlap, they would need to have very little forward component of the vector. Since they’ll fly out at 45 degrees, they’ll clip the third coin.
Posted on 4/11/18 at 9:36 am to theantiquetiger
The two coins you hit do not move, just the third coin. The coin you strike with does not move once it makes contact with the two coins. The reason is because the force from the 1st coin that would push the two coins to move away from each other (y direction) goes directly into the 3rd coin equally and opposite direction, transferring all energy to third coin.
This post was edited on 4/11/18 at 9:44 am
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