Ok, I have no idea where you got the idea the angles have to be equal.
Theta2 and theta3 are complementary, therefore theta1 and theta3 are equal (theta2 and theta3 are complementary).
Now, your picture is wrong if it is supposed to be accurate lengths in vectors. Fg is the real vector, so the normal should be shorter (it is a component). The normal and the force down the plane have to add to Fg.
So see this http://dev.physicslab.org/Document.aspx?doctype=5&filename=Compilations_CPworkbook_InclineForceVectors.xml
If you do that, then the two vectors should be FgsinTheta+FgcosTheta, and as vectors, they add to Fg.
h t t p : / / i m g 4 . i m a g e s h a c k . u s / i m g 4 / 9 2 9 0 / p i c t u r e d q e . j p g
6 answers
I understand that the vectors should add up I just did a rough sketch sorry
so how come when we know the angle the ramp makes with the surface, theta 3, it sits on
say 37 degrees we can find
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
If these equations are ture (the angle I used was theta 3 the one the ramp makes with the surface it sits on) then how come I placed theta 3 in intead of 2 or 1
I've always been told to put it in that way just use the angle the theta the ramp makes with the surface it sits on however this can't be true because then the rules of trigonmetric identies, opposite, hypotenuse, adjacent, when you use cosine and sine to find the components are clearly different triangles and use different angles other than the one that the ramp makes witht the surface it sits on...
if theta 3 does not equal theta 2 or theta 1 but theta 2 equals theta 1
If these equations are ture
"Theta2 and theta3 are complementary, therefore theta1 and theta3 are equal (theta2 and theta3 are complementary).
"
ok I remeber what complementary means they add up to 90 degrees but how do we know they are complementary
one we establish that how do we know that theta 1 and theta 3 are equal to each other can't one be like 60 and the toher 30 and still be complementary
so how come when we know the angle the ramp makes with the surface, theta 3, it sits on
say 37 degrees we can find
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
If these equations are ture (the angle I used was theta 3 the one the ramp makes with the surface it sits on) then how come I placed theta 3 in intead of 2 or 1
I've always been told to put it in that way just use the angle the theta the ramp makes with the surface it sits on however this can't be true because then the rules of trigonmetric identies, opposite, hypotenuse, adjacent, when you use cosine and sine to find the components are clearly different triangles and use different angles other than the one that the ramp makes witht the surface it sits on...
if theta 3 does not equal theta 2 or theta 1 but theta 2 equals theta 1
If these equations are ture
"Theta2 and theta3 are complementary, therefore theta1 and theta3 are equal (theta2 and theta3 are complementary).
"
ok I remeber what complementary means they add up to 90 degrees but how do we know they are complementary
one we establish that how do we know that theta 1 and theta 3 are equal to each other can't one be like 60 and the toher 30 and still be complementary
I guess I just don't get it
I understand resolving into components
I agree with this
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
i see were that comes from
my only concern is if you use the angle that the ramp makes with the surface it sits on as the angle you use then the triangle parts
adjacent hypotenuse opposite
don't match up so that would lead you to believe that those equation refference a different triangle and use a different angle
but yet I have always been told that you just use the angle the ramp makes with the surface it sits on which makes no sense when you use that angle because the
force of gravity
is clearly the hypotenuse in both equations which would make no sense if you use the angle the ramp makes with the surface it sits on...
I understand resolving into components
I agree with this
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
i see were that comes from
my only concern is if you use the angle that the ramp makes with the surface it sits on as the angle you use then the triangle parts
adjacent hypotenuse opposite
don't match up so that would lead you to believe that those equation refference a different triangle and use a different angle
but yet I have always been told that you just use the angle the ramp makes with the surface it sits on which makes no sense when you use that angle because the
force of gravity
is clearly the hypotenuse in both equations which would make no sense if you use the angle the ramp makes with the surface it sits on...
because if you used that angle then the force of gravity wouldn't be the hypotenuse which it clearly is in the equations
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
I understand components and why you have to resolve into components and that they have to add up I just did a rough quick sketch
I understand the trig identies the triangles that you use just don't make sense and why you can just use the angle the ramp makes with the surface
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
I understand components and why you have to resolve into components and that they have to add up I just did a rough quick sketch
I understand the trig identies the triangles that you use just don't make sense and why you can just use the angle the ramp makes with the surface
I saw this picture before I think you posted it or someone on here
h t t p : / / w w w . l o w e l l p h y s i c s . o r g / b e t a / T e x t b o o k % 2 0 R e s o u r c e s / C h a p t e r % 2 0 1 1 . % 2 0 R o l l i n g , %2 0 T o r q u e , % 2 0 a n d % 2 0 A n g u l a r % 2 0 M o m e n t u m / 1 1 - 4 / c 1 1 x 1 1 _ 4 . x f o r m _ f i l e s / n w 05 7 7 - n . g i f
which showed the traingle i thought u yoused and the angles I thought you used which lead me to belive that angle 1 angle 2 and angle 3 were always equal to each other and that's why i was always told to just the angle the ramp makes with the surface it sits on
but the angles are not equal to each other so what allows us to use that angle when we use these equations
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
h t t p : / / w w w . l o w e l l p h y s i c s . o r g / b e t a / T e x t b o o k % 2 0 R e s o u r c e s / C h a p t e r % 2 0 1 1 . % 2 0 R o l l i n g , %2 0 T o r q u e , % 2 0 a n d % 2 0 A n g u l a r % 2 0 M o m e n t u m / 1 1 - 4 / c 1 1 x 1 1 _ 4 . x f o r m _ f i l e s / n w 05 7 7 - n . g i f
which showed the traingle i thought u yoused and the angles I thought you used which lead me to belive that angle 1 angle 2 and angle 3 were always equal to each other and that's why i was always told to just the angle the ramp makes with the surface it sits on
but the angles are not equal to each other so what allows us to use that angle when we use these equations
fg x = fg sin theta 2 = fg sin 37
fg y = fg cos theta 1 = fg cos 37
Draw the ramp at not close to 45 degrees, say 10 degrees. You will see it clearer I hope. The relationship I led you through is based on geometry of the ramp, horizontal and triangles constructed at the sliding mass