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????
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
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
To understand why the force in the x direction is mg sin theta, let's break it down step by step.
1. Consider the particle on the incline ramp. The force pointing downwards is the gravitational force acting on the particle, which is mg (the weight of the particle).
2. Now, let's resolve this gravitational force into two components. One component is along the incline ramp, and the other component is perpendicular to the ramp (normal to the surface).
3. The force that is pointing upwards exerted by the ramp onto the particle is the normal force. This force is perpendicular to the ramp and opposes the gravitational force. It can be split into two components: one component in the positive y-direction (opposite to the gravitational force), and another component in the positive x-direction (parallel to the incline ramp).
4. The component of the normal force in the x-direction is mg sin theta. Here's why:
- We know that the angle that the ramp makes with the horizontal is theta.
- The gravitational force (mg) is the hypotenuse of the right triangle formed with the x-axis and the component of the normal force in the x-direction.
- The mg sin theta component is opposite to the angle theta in this right triangle. So, according to the definition of sine (sin theta = opposite/hypotenuse), the opposite side is mg sin theta, which represents the force in the positive x-direction.
In summary, the force in the positive x-direction (parallel to the incline ramp) is mg sin theta because it represents the component of the normal force that opposes the gravitational force along the ramp.