The gravitational attraction between two masses varies inversely as the square of the distance between them. If the attraction force between them is 75 lbs. when the bodies are eight feet apart, find the attraction when the masses are twelve feet apart.
this is just a ratio and proportion problem...do you know the equation for inverse proportion?
no my notes for this is just way too confusing
i'll make it simple... \(\Large F_{g} = \frac{k}{d^2}\) where \(F_{g}\) = gravitational force k = proportionality constant d = distance between the objects you understand that so far?
yea
good...so in the first part of the problem..."If the attraction force between them is 75 lbs. when the bodies are eight feet apart," what is given to you?
yes
yes? i was asking what is the given in the first part of the problem :P hahaha..so what's given? gravitational force? proportionality constant? distance? Hint: there are 2 given values
uhh gravitational force
right! what else?
between 2 masses inversely as a square
no no..that's the condition for the proportion..just look here ""If the attraction force between them is 75 lbs. when the bodies are eight feet apart,"
oh yea...we have to find the attraction when they are 12 feet apart
physics :)
focus on "If the attraction force between them is 75 lbs. when the bodies are eight feet apart," first...don't look at the other parts of the problem
@AravindG it's ok...it's proportion haha
ok
so what are the 2 given values in there?
71 and 8
75 and 8* and what is 75? what is 8? by that i mean if it is attraction force, distance or proportionality constant
distance?
which is the distance?
75 or 8?
75
check the problem again.. "If the ATTRACTION FORCE between them is 75 lbs. when the bodies are eight FEET APARTt,"
oh sorry haha 8
right! and 75 is?
the proportionality constant
check what i capitalized there...
the Force
good! so we substitute force and distance to the equation i said earlier... \(\LARGE F_{g} = \frac{k}{d^2} \rightarrow 75 = \frac{k}{(8)^2}\) you're gonna look for k..how do you think you can do that?
Hint: "cross" multiply
uh multiply by its reciprocal??? lol ireallydk
hmm close...whose reciprocal?
k/(8)^2
and what do you mean multiply by reciprocal? multiply by 8^2/k?
yes
hmm well that would cancel out the k right? but you want k...so you'll just multiply both sides by 8^2 got it?
right
so your k = 72(8^2) you might wanna use the calculator
4608
so that becomes your k...moving on to the second part of the problem..." find the attraction when the masses are twelve feet apart." what is the given value this time?
12
but wasnt that 72 supposed to be 75
what 72?
72(8^2)
oh..sorry typo!!! :))) do it (75)(8^2) hehe sorry
so..on to the second part of the problem?
k=4800
okay..we'll need that k for later...so back to what we were talking about...what do you think is 12? distance? attraction? proportionality constant?
12 is definitely the distance
right! and what are we looking for?
the attraction force
GOOD! :D and we have proportionality constant already right? so we substitute! \(\LARGE F_{g} = \frac{k}{d^2} \rightarrow F_{g} = \frac{4800}{(12)^2}\)
ohh koolio understandable
got it now? :)
so now we have to divide?
yuppp :)
so that going to be 33.33
you got it! :)
THANK YOU SOOOOOO MUCHHH you a LIFESAVER
hehe it was nothing :"> haha
have a blessed day
you too ^_^
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