English units conversions

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Homework Statement



Hello guys,

I just have a very quick question about English units. On my fluids homework, I got the answer 9714.806 slugs*ft/(s^2) which seems to be right according to the solutions manual. The homework problem is asking for the force , and the solution manual is dividing 9714.806 slugs*ft/(s^2) by 1.059 (lb*s^2)/(slug*ft) to get an answer of 9714 lb. Why is this right? I thought Slugs*ft/(s^2) = lbf.

Thank you for the help.
 

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SteamKing
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Homework Statement



Hello guys,

I just have a very quick question about English units. On my fluids homework, I got the answer 9714.806 slugs*ft/(s^2) which seems to be right according to the solutions manual. The homework problem is asking for the force , and the solution manual is dividing 9714.806 slugs*ft/(s^2) by 1.059 (lb*s^2)/(slug*ft) to get an answer of 9714 lb. Why is this right? I thought Slugs*ft/(s^2) = lbf.

Thank you for the help.
We really can't answer all your questions unless you post your HW.

1 lbf = 1 slug-s2 / ft

In more familiar F = ma terms, slugs = lbf / g, and g = 32.174 ft/s2

I don't recognize the factor 1.059 lb-s2 / (slug-ft), unless it's some weird density unit.

https://en.wikipedia.org/wiki/Slug_(mass)
 
  • #3
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We really can't answer all your questions unless you post your HW.

1 lbf = 1 slug-s2 / ft

In more familiar F = ma terms, slugs = lbf / g, and g = 32.174 ft/s2

I don't recognize the factor 1.059 lb-s2 / (slug-ft), unless it's some weird density unit.

https://en.wikipedia.org/wiki/Slug_(mass)

Hello Steamking,

Thanks for the reply,

The homework problem is the following.

Upon landing, airplanes exhaust the air from the engine through a reverse thrust mechanism in order to slow the plane. In this configuration, air flows into the jet engine shown at a rate of 9 slugs/s and a speed of 300 ft/s. Upon landing, the engine exhaust exits through the reverse thrust mechanism with a speed of in the direction indicated. Determine the reverse thrust applied by the engine to the airplane. Assume that the inlet and exit pressures are atmospheric and that the mass flow rate of fuel is negligible compared to the air flow rate through the engine

I was able to derived the equation for the reverse thrust :

300*9 + 900*cos(30)*4.5 + 900*cos(30)*4.5 = 9714 slugs*ft/(s^2)

Which seems to be right according to the solutions manual . However, I don't understand why they are dividing 9714 slugs*ft/(s^2) by 1.059 (lb*s^2)/(slug*ft) .

Any thoughts?

Thank you.
 

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