Calculating Stopping Force for an Object in Motion

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SUMMARY

The discussion centers on calculating the stopping force exerted on a 20,000 lbs object moving at a constant velocity of 5 ft/s that comes to an instantaneous stop. Participants clarify that an instantaneous stop would require infinite force, but they provide a practical approach by estimating deceleration over a small distance, such as a 1/4" divot in the ground. Using the formula F=ma, they derive that the acceleration is 600 ft/sec², leading to a calculated force of approximately 13,818.16 N when converted to metric units. The conversation emphasizes the importance of understanding deceleration distance and surface conditions in force calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Basic knowledge of units of force (pounds, Newtons)
  • Familiarity with concepts of acceleration and deceleration
  • Awareness of the impact of surface material on stopping force
NEXT STEPS
  • Learn about calculating deceleration distances in impact scenarios
  • Explore the effects of different surface materials on stopping force
  • Study the implications of instantaneous stops in physics
  • Investigate the conversion between imperial and metric units for force calculations
USEFUL FOR

Engineers, physics students, and anyone involved in mechanics or impact analysis will benefit from this discussion, particularly those interested in force calculations and material interactions during collisions.

KelseyG
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Can anyone help me? I'm pulling my hair out on a simple problem that I should know by heart.

I have an object in downward motion that weighs 20,000 lbs moving at a constant velocity of 5 ft/s. It contacts a solid object (say the ground) and is stopped instantly. What force was exerted to stop the object?

Thanks,
Kelsey
 
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Force is the rate of change of momentum with respect to time. If the object stops instantly, you are going to have some problems.
 
KelseyG said:
and is stopped instantly
The ball stopped in zero time, meaning the force is infinite.
 
KelseyG said:
I have an object in downward motion that weighs 20,000 lbs moving at a constant velocity of 5 ft/s. It contacts a solid object (say the ground) and is stopped instantly.
As the previous posts point out, the object didn't really stop instantly - that would require an infinite force. But we can estimate the actual forces involved. For example:

Let's say the object makes a divot 1/4" deep in the surface of the earth. That means the speed object went from 5 ft/sec to 0 ft/sec as it moved 1/4" from the point where the surface of the Earth was to where it ended up. Assuming constant deceleration (an approximation that simplifies the calculation no end) the object moved through that distance at an average speed of 2.5 ft/sec or 30 in/sec. Thus, it covered that distance in 1/120 seconds (1/4" at 30 in/sec). Its speed went from 5 ft/sec to 0 ft/sec in 1/120 seconds, so the acceleration was 600 ft/sec^2.

Now we can use F=ma to calculate the force needed to produce that acceleration. Just remember that if the object weighs 20000 lbs its mass is 20000/32 slugs. However, I'm done doing arithmetic in my head for this problem... There's a reason why scientists and engineers use the metric system these days :)
 
Well, sadly to say, I'm an engineer for forgot this stuff, lol. Thanks for the input. It makes me feel better about it knowing that it's not that simple and that I don't have all the variables I need in order to solve my situation.
 
KelseyG said:
Well, sadly to say, I'm an engineer for forgot this stuff, lol. Thanks for the input. It makes me feel better about it knowing that it's not that simple and that I don't have all the variables I need in order to solve my situation.

The basic principle is that the more abruptly the object is stopped (the shallower the divot in the earth, the less distance it travels as it decelerates) the greater the force. That's why a glass will break if you drop it on a concrete floor but not if you drop it onto a mattress.
 
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There are several things here that need to be addressed... 1) how far did it drop ? 2) There is no way that anyone can comment on the crater impact
component without knowing what state the surface is in... clay, cement, dirt that hasn't seen water for years... At best F = ma
9090.9 Kg x 1.52 m/s squared =13,818.16 N = Kg x m/s squared...
 

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