Explain Force of Thrust Equation: 6N/3.48s

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SUMMARY

The discussion centers on the Force of Thrust equation, specifically the calculation of thrust using the formula Ft/ΔT, where Ft equals 6N and ΔT equals 3.48s, resulting in a thrust of 1.7N. Participants clarify that the equation for Force of Thrust can also be expressed as F(thrust) = Δp/Δt, indicating that thrust is the change in momentum over time. The conversation emphasizes the importance of understanding momentum in relation to thrust calculations, as the units of Newtons cancel appropriately when divided by time.

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Please Explain this?!

hey, ok so i was going tthrough my textbook and following some examples hwne i cam up to this:

Force of Thrust = Ft/ delta T

= 6N / 3.48s

=1.7 N

but it doesn't say what the equation means. can someone please help me and explain what is happeing.
 
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A little more information regarding the problem would be helpful. However, how can one divide Newtons by time and wind up with Newtons?
 


Have you covered momentum yet? I don't remember how to calculate force of thrust without using momentum, so I hope this is helpful to you.

The equation for Force of Thrust I know, is
F(thrust) = ((m * V)2 - (m * V)1) / (t2 - t1)

This can also be written as
F(thrust) = Δp / Δt
Force of thrust = change in momentum / change in time

Which would give you
N = N*s/s
meaning your seconds would cancel to give the correct units.
 


the doth prevous questins are pretty good. But i don't understand why did u posted a quastion that is already answered.. If you don't understand it just read the 3-4 prevous pages of your book! That is not so difficult (sorry about my english and wrong words sometimes,,, ist because I am writing from my mobile)
 

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