Determining the 'Instantaneous Force' during impact

  • Thread starter Thread starter structengine
  • Start date Start date
  • Tags Tags
    Force Impact
AI Thread Summary
Calculations of impact force on a conveyor belt involve understanding the average force (Favg) as the change in kinetic energy divided by the distance after impact. When a weight lands, it experiences slipping against the belt, with friction acting to accelerate it to the belt's speed, causing the force to start high and decrease to zero. The average force is essential for calculating energy gain, and if the force decreases uniformly, the maximum force can be twice the average force. The discussion also highlights the importance of knowing the force curve, especially when using materials like crushable honeycomb to absorb impact. Further insights from the material manufacturer could clarify the force dynamics during deceleration.
structengine
Messages
2
Reaction score
0
I have been doing some calculations on a falling object and its impact force. We use heavy weights on our belt conveyors to maintain tension on the belts. One of our calculations relates to the amount of force placed on the structure that one of these weights lands on. I have the calculations and the answers, but as a matter of curiosity, I started looking at the formulas a little closer. The force of impact is Favg. Since this force is the change in Kinetic energy divided by the distance after impact, it seems to me that this is the average force over that distance. Is there a way to graph this force during deceleration? How can the instantaneous force (for lack of a better term) be found? Is there a formula that I am overlooking? Not having to work very often with the dynamics of the structures I engineer, my physics in this area is very rusty, and being a pretty new engineer, I like to dig a little deeper into the things I am doing. Can anybody help? Thanks
 
Physics news on Phys.org
Your logic is correct since work done = gain in KE = Force x distance
When a weight lands on the conveyor belt there is slipping between the weight and the belt.
This friction force is what acts on the weight and speeds it up to the speed of the conveyor belt. When the weight is up to speed there is no resultant force on the weight. So the foirce on the weight starts at some 'high' value and decreases to zero.
You must use the average force to calculate the gain in energy.
If the force decreases uniformely then the maximum force will be 2 x the average force.
If there was no friction (ice covered conveyor !) I think you could say that the weight would not be accelerated up to the speed of the conveyor belt.
This is not a particularly rigorous analysis but I hope it goes some way to explain why an average force is needed in any calculation.
 
That makes sense for using the average force. Since the force does change between the initial energy and the final energy with respect to the distance traveled it would have to be the average over that distance. I don't know if the force changes uniformely over that distance though. I am using a crushable honeycomb product to 'catch' the takeup weight, and so I can probably find out from them what the force curve looks like. Thank you
 

Thread 'Average velocity as a weighted mean'

Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
View full post »

Thread '1:1 versus 2:1 Mechanical Advantage debate'

The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
View full post »
Thread 'Beam on an inclined plane'

Thread 'Beam on an inclined plane'

Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
View full post »

Similar threads

I Momentum, impact force and Earth
Replies
9
Views
2K
Comparing force of static object to that of an impact force
Replies
4
Views
3K
How is initial acceleration accounted for in the impact of two bodies
Replies
4
Views
2K
Finding the 'g-force' of a decelerating falling mass.
Replies
1
Views
2K
Having Trouble Finding G-Force of an Impact -- Details in post
Replies
6
Views
3K
Calculating Impact Force of a Schmidt Hammer
Replies
1
Views
2K
B Understanding Friction: The Impact of Inter-Molecular Bonding on Movement
Replies
2
Views
1K
Impact of trolley - hand calculation method
Replies
3
Views
3K
Impact forces feet can receive without get injured
Replies
1
Views
1K
Estimating the damage to a material based on force or kinetic energy
Replies
12
Views
4K

Hot Threads

Recent Insights

Back
Top