Solving Inclined Plane Problems: Explaining the Triangle Relationship

[Peter G.]

Hi,

I have a terrible time when solving Inclined Plane problems when it comes to dividing the weight component into two. I can't see why, to find the perpendicular component I do the cos Theta of the weight and the parallel component I do the sin Theta of the weight.

Can anyone, (maybe on a diagram) show how those relations drawing triangles?

Thanks,
Peter G.

 

[rock.freak667]

Excuse my bad handwriting with a mouse: Here is the diagram

http://img43.imageshack.us/img43/1571/diagra.png

You should see that the triangle with the (inclination angle-> leftmost triangle) has angles 90°,θ and 90°-θ.

The normal is drawn such that the angle made by the inclination and the normal is 90°(that's what a normal is right?). So the angles here must add up to 90°. Therefore the angles must be 90°-θ and θ. That's what in the rightmost triangle, that angle there is θ.

In this same triangle, the weight,w, is the hypotenuse. The side opposite to θ is the opposite side and the side adjacent is where the normal force would be.

so cosθ=adj/hyp OR cosθ= normal/w hence normal force = wcosθ or simply N=mgcosθ.

The side opposite would be the force due to gravity, sinθ=opp/w or force due to gravity = Wsinθ = mgsinθ.

Clearer now or is it still a bit confusing?

 

[Peter G.]

Hi!

Thanks, I think I got it now.

 

[anotherperson]

if this block was attached by a string how would you calculate the tension, cause I've seen many questions like this before but never knew how to do them. i don't have a specific example though.

 

[Fewmet]

if this block was attached by a string how would you calculate the tension, cause I've seen many questions like this before but never knew how to do them. i don't have a specific example though.

If the block is at rest the net force is 0 Newtons. The string's tension is the force up the incline, and the component of the weight down the incline is equal and opposite. So

T + mg sin (theta) = 0​

Does that do it for you?

 

[anotherperson]

yepp, thanks heaps!

 

[anotherperson]

& what about acceleration is the string is cut?

 

[Fewmet]

& what about acceleration is the string is cut?

Anytime you want acceleration, try
\SigmaF=ma.
If you cut the string, what forces act parallel to the slope? Add them up and divide by mass to get acceleration.

 

[rock.freak667]

& what about acceleration is the string is cut?

If your string is cut then T=0 and then do what Fewmet suggests.

 

[anotherperson]

Excuse my bad handwriting with a mouse: Here is the diagram

http://img43.imageshack.us/img43/1571/diagra.png

if i had to do a force diagram with this picture. the forces would be
down=Fgrav
up=Fnormal
and is left or right Ffric & i can't seem to work out what the other force would be?

 

[rock.freak667]

if i had to do a force diagram with this picture. the forces would be
down=Fgrav
up=Fnormal
and is left or right Ffric & i can't seem to work out what the other force would be?

If the force of gravity is making the block slide down the incline and friction opposes motion, then what direction would the frictional force act?

(in the diagram, there is the weight w and the components of the weight w if you didn't understand the diagram)

 

[anotherperson]

so the frictional force act to the right(up the incline) what is the oposing force, acting right? the force of gravity?

if so what is the force pulling the block down as in the opposing force to the normal?

 

[Fewmet]

so the frictional force act to the right(up the incline) what is the oposing force, acting right? the force of gravity?

There are two (at least) ways to think about that. One is that the force down the slope is the vector sum of gravity (i.e., the weight) and the normal force.

Alternatively, you can focus only on the motion parallel to the the incline. In that case friction opposes the component of gravity in that direction. That component is always mg*sin(theta), where theta is the angle of the incline to the horizontal.

if so what is the force pulling the block down as in the opposing force to the normal?

That would be the component of gravity that is perpendicular to the the incline.