ok well obviously mg is the hypotenuse and the force in the x direction is the opposite
obviously but what triangle is this????
how is mg the hypotenuse and the force in the x direction the opposite???
what triangle gives us this???
with reference to what angle were only given one the angle of the ramp it makes with horizontal
I'm tyring to undersatnd uh motion of a particle at an incline on a ramp with a certain angle with the horizontal.
the force pointing downwards is mg
the force that is pointing upwards exerted by the ramp onto the the particle is the normal force which is not pointing perfectly up.
In the opposite direction of the normal force is the mg cos theta
ok i understand this because the angle formed by these two forces, mg cos theta, and mg, is the same angle that the ramp makes with the horizontal
ok i get it
def of cos theta is a^-1 h
that's why the name of this force is mg cos theta sense because the adjacent peice is the force mg and the hypotenuse is equal to mg cos theta
ok i get it
but now the force in the x direction if the force Fn is considered to be the force in the positve y direction is mg sin theta... ok why
well theta is the angle that the ramp makes with the ground
defintion of sin theta is h^-1 O
ok the hypotenuse is the force in the x direction and the opposite is the mg
so I really don't see were the term comes from mg sin theta
why is it mg sin theta?
defintion of sin is theta is h^-1 O
what force is the hypotenuse what force is the opposite and what angle are they making reference to
again this is just a normal particl or in this case a block setting up ontop of a ramp and the force in the positive x direction, if the normal force is considered the positive y direction, is mg sin theta why????
2 answers
THe only forces are the gravity force, mg downward, and the force the plank is exerting, which is indeed normal to it's surface. But any vector can be broken into components. Break mg into two forces, one down the ramp, and one normal to the ramp. Why do this? It is convenient to have it as such. If you do this, then the normal component is mgcosTheta, which of course is the component of gravity in the direction normal to the ramp. The other component of mg is down the plank, mgsinTheta.
That component of force down the plank is what makes things go downhill.
http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%2011.%20Rolling,%20Torque,%20and%20Angular%20Momentum/11-4/c11x11_4.xform_files/nw0577-n.gif
That component of force down the plank is what makes things go downhill.
http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%2011.%20Rolling,%20Torque,%20and%20Angular%20Momentum/11-4/c11x11_4.xform_files/nw0577-n.gif