Natural force down incline plane

[PhizKid]

Homework Statement

Find the force and acceleration of the block down the inclined plane

Homework Equations

F = ma

The Attempt at a Solution

We have two unknowns in F = ma

I know there is the gravitational force going straight down, and there is the normal force that is perpendicular to the incline.

What I don't know if is if the force of gravity acting on the block and the normal force are active/reactive for pairs (do they cancel each other out like an object would at rest?) because I'm told every object has at the least two forces acting on it (the force pairs according to Newton's 3rd law).

So I know the force pulling down on the object is gravity, but what other forces are there that I can calculate the force of the object sliding down the incline?

 

[Spinnor]

"What I don't know if is if the force of gravity acting on the block and the normal force are active/reactive for pairs (do they cancel each other out like an object would at rest?)"

The forces would cancel only if the plane were horizontal.

For examples see,

http://www.physicsclassroom.com/class/vectors/u3l3e.cfm

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html

Both links from,

https://www.google.com/search?num=1...e=UTF-8&sa=N&tab=iw&ei=UbR1ULWOH4fE0AGbn4DICQ

 

[PhizKid]

So what are the active/reactive force pairs that cancel each other out on an inclined plane?

 

[SammyS]

So what are the active/reactive force pairs that cancel each other out on an inclined plane?

The active/reactive force pairs you talk about, as they are described in Newton's 3rd law, don't cancel each other! Why? Because the two forces making up that pair act on two distinct objects.

Their are two pair of active/reactive forces in this problem.

The Earth attracts the block via gravity / the block attracts the Earth via gravity.

The block exerts a force perpendicular to the inclined plane / the block exerts a force perpendicular to the inclined plane .​

The two forces you show in your figure cannot cancel because they don't act along the same line.

 

[PhizKid]

So how can we use these 4 forces to solve for the original problem? I don't understand how they interact to form the force going down the inclined plane.

 

[SammyS]

So how can we use these 4 forces to solve for the original problem? I don't understand how they interact to form the force going down the inclined plane.

Only two of those forces act on the block.

One of the others acts on the earth, the fourth acts on the incline plane. These two do not affect the block.

To solve the problem, first set up a coordinate system.

I suggest one axis should be parallel to the inclined plane, the other axis perpendicular to the inclined plane.

 

[PhizKid]

So then it's the normal force of the surface pushing against the block, and the force of gravity pulling down on the block?

Why do you suggest setting up a coordinate system in this way? Why not just straight up and down (with the y-axis parallel to the side of the inclined plane)?

 

[SammyS]

So then it's the normal force of the surface pushing against the block, and the force of gravity pulling down on the block?

Why do you suggest setting up a coordinate system in this way? Why not just straight up and down (with the y-axis parallel to the side of the inclined plane)?

Work the problem both ways to see.

With this set-up, what is a_y, y component of acceleration ?

 

[PhizKid]

Well, the acceleration is in the direction of the negative x-axis, so I guess the positive y-axis would be the acceleration's a_y direction, but acceleration doesn't have a y-direction in this case since it's only moving along the x-axis with this set up. So the y-component of the acceleration here is just 0.

 

[SammyS]

Well, the acceleration is in the direction of the negative x-axis, so I guess the positive y-axis would be the acceleration's a_y direction, but acceleration doesn't have a y-direction in this case since it's only moving along the x-axis with this set up. So the y-component of the acceleration here is just 0.

Yes, a_y is zero, which is a good reason to set-up a coordinate system in this way. If you would need to know the normal force, N, for any reason, this coordinate system would be a big help.

So, what is the y-component of the net force?

What is the x-component of the net force?

 

[PhizKid]

I cannot seem to find the angle...the 30 degrees is situated on the coordinate plane awkwardly and I do not know how to move it appropriately/legally to get it into standard form.

 

[SammyS]

I cannot seem to find the angle...the 30 degrees is situated on the coordinate plane awkwardly and I do not know how to move it appropriately/legally to get it into standard form.

I assume a horizontal base for the wedge which forms the inclined plane.

The force of gravity is vertical, so you have a right triangle with one acute angle of 30°.

What's the other angle?

 

[PhizKid]

I don't understand "horizontal base for the wedge which forms the inclined plane."

How is the force of gravity vertical? It is a vector in the 3rd quadrant pointing somewhere southwest of the origin.