How do I Find Average Force in a Force vs Time graph?

AI Thread Summary
To find the average force in a Force vs. Time graph, focus on the area under the curve, which represents impulse or change in momentum. The average force can be calculated by dividing the total impulse by the total time. The graph typically consists of sections where the force increases, remains constant, and then decreases. The "time-averaged force" is the constant force that would produce the same change in momentum over the same time period. Thus, the average force is derived from the impulse divided by the duration of the force application.
haroldham
Messages
6
Reaction score
0
This is really simple, but I can not remember how to find the average force in a Force vs TIme graph.
 
Physics news on Phys.org
Think about the height of a "[certain] rectangle." Think area.
 
robphy said:
Think about the height of a "[certain] rectangle." Think area.
Ok, you completely lost me, but I think I figured it out. This certain graph increases, remains constant, then decreases. Do I take the final force of each section, add it to the initial force of each section, divide that value by the time of that section, then average those together?
 
Presumably, you mean "time-averaged force".
The area under your graph is equal to the impulse (ie, the change in momentum).
The "time-averaged force" you seek is the constant force you would apply in the same amount of time in order to obtain the same change in momentum. Translate the last sentence into facts about the graph.
 
Oh, ok I get it. It is impulse divided by time.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top