February 13, 2016

Homework Help: Physics

Posted by Physics on Saturday, July 11, 2009 at 5:24pm.

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????

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