How to Calculate Impact Force Without Knowing Stopping Time or Distance?

AI Thread Summary
Calculating impact force requires specific information about the stopping time or distance, which is not provided in the initial query about a 180-pound object dropped from 11 feet. The discussion also touches on the concept of Momentary Maximum Muscle Tensions (MMMs) during strength training, emphasizing the increased forces experienced during rapid transitions in weight lifting. A user seeks to determine the force exerted when a 90 kg weight is dropped over a distance of 1 meter in 0.5 seconds, highlighting the complexities of measuring forces in dynamic movements. Accurate calculations depend on understanding both the falling and stopping phases of the object's motion. Overall, more detailed information is essential to accurately calculate the impact force in both scenarios.
alexa23
Messages
1
Reaction score
0
calculating impact force??

I need help calculating impact force. Here's all the information I have:

the object weighs 180 pounds
being dropped from a height of 11 feet

I'm not completely sure if the force can be determined from that. I'm having a hard time figuring it out for some reason.
 
Physics news on Phys.org


Hi Alexa
I recommend you to read forum rules before posting because your thread seems like a homework and homeworks are discussed in other forum.
By the way welcome to PF
 


alexa23 said:
I need help calculating impact force. Here's all the information I have:

the object weighs 180 pounds
being dropped from a height of 11 feet

I'm not completely sure if the force can be determined from that. I'm having a hard time figuring it out for some reason.

Hi all,

I would like to ask the same thing, and first to prove its not homework, first I am 48, and second, we were wondering how mush force the muscles would be taking on the transition from negative to positive, like in strength training, and repping a barbell up and down. We call this transition MMMs {Momentary Maximum Muscle Tensions}

These tensions or forces are as you imagine the highest there is in the whole rep. {repetition} As when you are on the eccentric of the rep, you are still controlling it down, but when a given load is lowed quite fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load.

Would it be possible to work out the force of ? Let’s say a 90kg weight, dropping at .5 of a second for 1m, please ?

Wayne
 


More information is needed. Specifically, either how long or over what distance is the object brought to a stop? Note, this is different than the time or distance that the object is falling.
 


Redbelly98 said:
More information is needed. Specifically, either how long or over what distance is the object brought to a stop? Note, this is different than the time or distance that the object is falling.

Hmm, as you can see this is a bit tricky to say, as when you are repping up and down with a barbell it quite hard to tell.

Therefore, what we say the last fifth of the eccentric, and that would also be one fifth of .5 of a second.

And thanks for you time.

Wayne
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top