City-Data Forum Newton's Second Law of Motion (cable, crash, analog, connection)
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01-19-2015, 06:37 PM
 Location: Proxima Centauri 3,786 posts, read 1,609,142 times Reputation: 4194

Force = acceleration * mass

mass is a constant at 4000 lbs
acceleration is defined as V2 - V1 / time

V1 is usually 0
Lets say V2 is 500 mph

Event A - the 4000 pound mass takes 5 seconds to hit the wall at 500 mph
Event B - the same 4000 pound object takes 20 seconds to to hit the wall at 500 MPH

The problem: Since this 4000 pound object hits the wall at 500 mph in both events, why is the force less in event B?

01-19-2015, 07:52 PM
 Location: Westwood, MA 3,502 posts, read 4,379,683 times Reputation: 4508
Why would the force be the same? Think of an opposite problem, pushing something to a start. If you push hard you'll accelerate quickly and reach the desired velocity quickly. If you push gently you will accelerate slowly but if you keep pushing long enough eventually you'll get up to the same speed (ignore friction if it makes it easier to understand). Same mass and same final velocity, but wildly different forces.

Mathematically the velocity and mass give you a momentum. A momentum change is an impulse. The impulse in the two situations is identical, not the force. Force is the rate of change of impulse, so the shorter event necessarily has the larger force:

F = dp/dt = pf - pi / tf - ti

01-19-2015, 08:16 PM
 7,206 posts, read 5,286,133 times Reputation: 7862
The force of the mass on the wall when it hits? Has nothing to do with how long it took the mass to get to 500 mph and everything to do with how quickly it went from 500 to 0. The force would be the same for the impact.

01-20-2015, 05:54 AM
 Location: Westwood, MA 3,502 posts, read 4,379,683 times Reputation: 4508
Quote:
 Originally Posted by JasonF The force of the mass on the wall when it hits? Has nothing to do with how long it took the mass to get to 500 mph and everything to do with how quickly it went from 500 to 0. The force would be the same for the impact.
Of course it has nothing to do with how long it took to get to speed. You sure think I'm stupid. I was using an analogous problem (acceleration) to describe a less intuitive phenomenon (deceleration).

It's not clear to me what the conceptual problem is. I thought that 5 and 20s were the deceleration times for two different kinds of walls. If that's not the case the problem isn't very clear.

Also, I should note that the definition of acceleration in the OP is only correct for constant acceleration. The real definition involves instantaneous changes not cumulative ones. The formula you give is for average acceleration. The actual acceleration is given by:

a = dv/dt

It's doubtful any wall would give constant deceleration, so the force will likely vary as the object comes to a stop.

Last edited by jayrandom; 01-20-2015 at 06:19 AM..

01-20-2015, 10:37 PM
 7,206 posts, read 5,286,133 times Reputation: 7862
Quote:
 Originally Posted by jayrandom Of course it has nothing to do with how long it took to get to speed. You sure think I'm stupid.
Actually I was just answering the OP's question. They seemed to be conflating what are actually independent problems. Sorry that you think everything is all about you.

Quote:
 I was using an analogous problem (acceleration) to describe a less intuitive phenomenon (deceleration).
They're exactly the same concept. Your post did nothing but state what the OP clearly already knew, confuse the issue and drag in other concepts that are completely unnecessary to explain the problem. And then you never actually addressed what the OP seemed to be confused about.

Quote:
 It's not clear to me what the conceptual problem is. I thought that 5 and 20s were the deceleration times for two different kinds of walls. If that's not the case the problem isn't very clear.
It's not terribly confusing, either. v1 is zero. v2 is 500 mph. Then the forces are calculated based on the time it takes to get from v1 to v2. The object presumably doesn't speed up when it smashes into a wall.

01-21-2015, 05:21 AM
 Location: Westwood, MA 3,502 posts, read 4,379,683 times Reputation: 4508
Quote:
 Originally Posted by JasonF Actually I was just answering the OP's question. They seemed to be conflating what are actually independent problems. Sorry that you think everything is all about you. They're exactly the same concept. Your post did nothing but state what the OP clearly already knew, confuse the issue and drag in other concepts that are completely unnecessary to explain the problem. And then you never actually addressed what the OP seemed to be confused about. It's not terribly confusing, either. v1 is zero. v2 is 500 mph. Then the forces are calculated based on the time it takes to get from v1 to v2. The object presumably doesn't speed up when it smashes into a wall.
Acceleration and deceleration are similar but there are a lot of people who don't make the connection. I misread the problem as asking the force of impact at the wall, which depends on the stopping time. And which can vary depending on the kind of wall. My mistake. .

01-22-2015, 11:37 PM
 5,783 posts, read 3,069,764 times Reputation: 15193
Well, the problem as stated makes no sense. It says it takes either 5 seconds to hit a wall or 20 seconds to hit a wall. Are we talking distance to wall or impulse on impact or a really poorly written acceleration problem (slowing down is acceleration, just negative)?

01-26-2015, 11:44 PM
 Location: Heart of Dixie 12,448 posts, read 10,186,344 times Reputation: 28075
The forces are different because they are the forces required to accelerate a 4000lb object from 0 MPH to 500 MPH, over two different time intervals.

01-29-2015, 08:08 AM
 5,092 posts, read 4,378,235 times Reputation: 4337
This has been an interesting thread. It exposes the difference between force and kinetic energy.

While Newton's Second Law is Force is the product of mass and acceleration (F = ma), it's not applicable to the question of an impact event. What is applicable is the equation E = ½mv², where E is kinetic energy, m is the mass of the object, and v is the velocity of the object. When engineers study automobile crashes, they may consider the forces involved, but when they talk about collisions, they measure primarily kinetic energy and then subsequently the forces that result from changes of that kinetic energy.

Force has the same relationship to kinetic energy that acceleration has to velocity. Force increases or decreases kinetic energy in the same way that acceleration/deceleration changes velocity. For a real mind twist, consider that Work is force over distance and that Work is what actually changes Kinetic Energy. Or something like that.

I'm probably doing a bad job of trying to explain the concept, please look at this for a better explanation Kinetic energy - Wikipedia, the free encyclopedia

01-30-2015, 01:34 PM
 12 posts, read 10,299 times Reputation: 22
Quote:
 Originally Posted by Dirt Grinder The forces are different because they are the forces required to accelerate a 4000lb object from 0 MPH to 500 MPH, over two different time intervals.
What he said
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