# Can someone explain this equation?

1. Nov 3, 2012

### lionely

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?

2. Nov 3, 2012

### Simon Bridge

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. Nov 3, 2012

### lionely

m is the slope c is the intercept.

but how would mg be the gradient?

4. Nov 3, 2012

### lionely

from my graph c = 0

and for m I got 1.92, but how is that related to mg?

5. Nov 3, 2012

### SammyS

Staff Emeritus
mg = 1.92

6. Nov 3, 2012

### lionely

Oh, I'm sorry I still quite fully understand. so Force/sin X is equal to mass x gravity?

7. Nov 4, 2012

### haruspex

You determined that K (=c) = 0. Put that in your original equation.

8. Nov 4, 2012

### SammyS

Staff Emeritus
How about $\ \ F\times\sin(\theta)=mg\ ?$

9. Nov 4, 2012

### Simon Bridge

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. Nov 4, 2012

### lionely

mu would have to be 1.92 and c 0.

11. Nov 4, 2012

### Simon Bridge

No - using only mgx+K = μx+c.

12. Nov 5, 2012

### lionely

mgx+ k = μx+c

1.92/sin + 0 = 1.92x + 0

right?

13. Nov 5, 2012

### Simon Bridge

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. Nov 5, 2012

### lionely

μ= mg
c=K

15. Nov 5, 2012

### Simon Bridge

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. Nov 5, 2012

### lionely

1.92 Newtons

17. Nov 5, 2012

### lionely

Oh I think I understand now.

18. Nov 6, 2012

### Simon Bridge

Great. 1.92N suggests to me a mass something like 200g.

19. Nov 6, 2012

### lionely

Yes! Cause when I worked out the mass I got .20kg. The mass would be equal to