Solving Newton's 2nd Law: Wx = W Sin θ

[sunflowerzz]

Hi,

Can someone please explain to me why Wx (in the attached diagram) will be W sin theta instead of normally as W cos theta?

I don't really understand the explanation that was given - "it is more useful to express the components of W in terms of the angle theta which the inclined surface makes with the horizontal"

The inclined surface meaning the hypotenuse of the triangle? And the horizontal meaning what exactly?

By the way, the question says "starting from rest, a 5 kg block takes 4s to slide down a frictionless incline. Find the normal force, the acceleration of the block, and the vertical height h the block starts from, if the plane is at an angle of 30 degrees"

I just don't understand why Wx is sin theta?

Thanks

 

[Johnahh]

W is always the hypotenuse due to it being the greatest value.
imagine completing the vector triangle between W and Wx. the value at W would be theta from the original triangle

 

[sunflowerzz]

W is always the hypotenuse due to it being the greatest value.
imagine completing the vector triangle between W and Wx. the value at W would be theta from the original triangle

But if you used the angle alpha from the diagram, wouldn't W also be the hypotenuse? Then Wx would be cos alpha - I just always assumed in these types of questions that the x-component is cos and not sin

 

[Johnahh]

yes W is always the hypotenuse, so if u want to use the angle alpha then yeah Wx would be W cos alpha. this is exactly the same as W sin theta.
always resolve the triangle like this. when its in the form shown in the book its harder to visualise as it makes it seem that Wx is the hypotenuse when really W is.
EDIT: edit: the reason W sin theta is easier as u are always given theta in these type of questions and rarely given alpha. this means another step in calculation just to find the value of alpha.

 

[Chestermiller]

But if you used the angle alpha from the diagram, wouldn't W also be the hypotenuse? Then Wx would be cos alpha - I just always assumed in these types of questions that the x-component is cos and not sin

Not necessarily. If you merely switched the the x and y axes, without changing anything else, what do think you would end up with? You need to use the diagram to determine the correct trig function to use. Here's a trick that might help. If the angle θ were zero, from the diagram, the x component of the weight would be zero. This should reveal to you whether it is sine or cosine that you should be using.

 

[technician]

One thing you can be sure of W is VERTICAL.
Usually you want the component of W ALONG THE PLANE and the component AT 90 to the plane.
Look at the component along the plane, when θ = 0 then the component along the plane = 0
When θ =90 then the component along the plane = W
Do you know that when θ=30 the component along the plane is W/2 ?
All of this is consistent with componenet along plane = W Sinθ
And component at 90 to plane is WCosθ

 

[sunflowerzz]

One thing you can be sure of W is VERTICAL.
Usually you want the component of W ALONG THE PLANE and the component AT 90 to the plane.
Look at the component along the plane, when θ = 0 then the component along the plane = 0
When θ =90 then the component along the plane = W
Do you know that when θ=30 the component along the plane is W/2 ?
All of this is consistent with componenet along plane = W Sinθ
And component at 90 to plane is WCosθ

I get W is vertical but I'm not understanding when you mean if θ = 0, the component along the plane = 0? When you say along the plane, you mean the incline?

There's a note on the side of the diagram that says: The angle of the incline equals the angle between the force of gravity and the line perpendicular to the incline - does this make sense? And is this the case for every incline question? Because sometimes the x-component is cos θ

 

[technician]

You are correct, when θ = 0 the plane (slope or incline) is horizontal.
The statement about the line of the force of gravity and the line PERPENDICULAR to the plane is correct...you might find it hard to visualise !
If you have difficulty I will try to do a sketch...but can't do that til tomorrow !

 

[sunflowerzz]

You are correct, when θ = 0 the plane (slope or incline) is horizontal.
The statement about the line of the force of gravity and the line PERPENDICULAR to the plane is correct...you might find it hard to visualise !
If you have difficulty I will try to do a sketch...but can't do that til tomorrow !

A sketch would be great! Thanks!

 

[sunflowerzz]

Not necessarily. If you merely switched the the x and y axes, without changing anything else, what do think you would end up with? You need to use the diagram to determine the correct trig function to use. Here's a trick that might help. If the angle θ were zero, from the diagram, the x component of the weight would be zero. This should reveal to you whether it is sine or cosine that you should be using.

When you say if θ is zero, the x component of the weight would be zero - how does it reveal whether it is sine or cosine?

 

[DrGreg]

When you say if θ is zero, the x component of the weight would be zero - how does it reveal whether it is sine or cosine?

What's the value of sin 0?
What's the value of cos 0?

 

[A.T.]

Maybe rotating your book clockwise by 90° would solve the confusion about vertical vs. horizontal?

 

[magaszag]

Think the sketch is little bit off...
http://imageshack.us/photo/my-images/812/och0.jpg/

Uploaded with ImageShack.us

 

[Dale]

Can someone please explain to me why Wx (in the attached diagram) will be W sin theta instead of normally as W cos theta?
...
I just don't understand why Wx is sin theta?

The way I always think of it is to think about what Wx would be if θ were very small, almost 0. If Wx would be almost 0 for small θ then Wx is W sin(θ) since sin is close to 0 for small θ, this is the case here. On the other hand if Wx would be almost W for small θ then Wx is W cos(θ) since cos is close to 1 for small θ, which is not the case here.