Can someone explain this equation?

  • Thread starter lionely
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  • #1
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F= (mg/sin θ) + K

I don't understand how this equation came about. I have a physics practical to finish and I need to use this to finish it.

I plotted a graph of F against 1/sin θ
I got a straight line passing through the origin.

So i tried to see if i could get the formula from

y=mx + c

Could someone tell me where mg comes from in this?
 

Answers and Replies

  • #2
Simon Bridge
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Take your equation written at the top - erase the ##F## and put a ##y## in it's place. Erase the ##1/\sin(\theta)## part and put an ##x## there instead. Now compare what you have with the equation ##y=mx+c##.

m=?
c=?

which part of mx+c is the slope of the graph and which the y-intercept?
 
  • #3
576
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m is the slope c is the intercept.

but how would mg be the gradient?
 
  • #4
576
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from my graph c = 0

and for m I got 1.92, but how is that related to mg?
 
  • #5
SammyS
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from my graph c = 0

and for m I got 1.92, but how is that related to mg?
mg = 1.92
 
  • #6
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Oh, I'm sorry I still quite fully understand. so Force/sin X is equal to mass x gravity?
 
  • #7
haruspex
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Oh, I'm sorry I still quite fully understand. so Force/sin X is equal to mass x gravity?
You determined that K (=c) = 0. Put that in your original equation.
 
  • #8
SammyS
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Oh, I'm sorry I still quite fully understand. so Force/sin X is equal to mass x gravity?
How about [itex]\ \ F\times\sin(\theta)=mg\ ?[/itex]
 
  • #9
Simon Bridge
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m is the slope c is the intercept.

but how would mg be the gradient?
How can it be anything else?

You know from the theoretical relation that:
F=mg/sin(θ)+K

(Presumably you started with Fsin(θ)=mg+Ksin(θ) or something - then rearranged it so you can get a graph that is a straight line.)

Since x=1/sin(θ) and y=F, this is the same as:

y = mgx + K

This has to be the same as y=μx+c
(using μ for the gradient because m is already taken for "mass".)

So what do μ and c have to be to make these two lines the same?

hint - if the lines are the same then: mgx + K = μx+c

---------------------

Note: starting from Fsin(θ)=mg+Ksin(θ) ... you could have plotted y=Fsin(θ) vs x=Ksin(θ): then the slope would have been 1 and the y-intercept would have been mg. But it is best practice to find the thing you want in the gradient of something.
 
  • #10
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mu would have to be 1.92 and c 0.
 
  • #11
Simon Bridge
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mu would have to be 1.92 and c 0.
No - using only mgx+K = μx+c.
 
  • #12
576
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mgx+ k = μx+c

1.92/sin + 0 = 1.92x + 0

right?
 
  • #13
Simon Bridge
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You did one step too many.

If Ωx+ζ=μx+c
then μ=Ω
and c=ζ

If <carrot>x+<banana>=μx+c
then μ=<carrot>
and c=<banana>

If (a+b)x+qwerty=μx+c
then μ=(a+b)
and c=qwerty

If (a/b)x+A=μx+c
then μ=(a/b)
and c=A

If mgx+K=μx+c
then μ=
and c=
 
  • #14
576
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μ= mg
c=K
 
  • #15
Simon Bridge
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Great - in relation to your question... "how is mg the slope":

You have a theory that predicts that if you plot y=F vs x=1/sin(θ) you will get a line whose slope is mg - namely: y=(mg)x+K. You did the experiment and came up with a line ... the slope of which you found to be 1.92units (is that correct? what are the units?): namely: y=(1.92)x+0.

Therefore:
1. The theory has successfully predicted the line.
2. If the theory is correct then (mg)x+K=(1.92)x+0 ... which means that mg=?, K=?

This is a common technique: you rearrange the theory so that some experiment - if the theory is right - will produces a line. Then you measure the line and compare with the theoretical prediction.
 
  • #16
576
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1.92 Newtons
 
  • #17
576
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Oh I think I understand now.
 
  • #18
Simon Bridge
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Great. 1.92N suggests to me a mass something like 200g.
Has your question been answered?
 
  • #19
576
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Yes! Cause when I worked out the mass I got .20kg. The mass would be equal to

m= gradient/9.81 right?
 
  • #20
576
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Well thank you for the help, I finished everything.
 
  • #21
Simon Bridge
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That would be all correct - well done.
 

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