Boulder on a hill (Vectors and components)

[asqwt]

Homework Statement

A boulder of weight "w" rests on a hillside that rises at a constant angle "a" above the horizontal. The boulder's weight is a force on the boulder that has a direction vertically downward.

In terms of "w" and "a", what is the component of the weight of the boulder in the direction parallel to the surface of the hill?

What is the component of the weight in the direction perpendicular to the surface of the hill?


http://www.cramster.com/answers-jan-11/physics/physics-vectors-boulder-weight-rests-hillside-rises_1096914.aspx

Homework Equations

The Attempt at a Solution

i have no idea...

 

[tiny-tim]

welcome to pf!

hi asqwt! welcome to pf!

In terms of "w" and "a", what is the component of the weight of the boulder in the direction parallel to the surface of the hill?

What is the component of the weight in the direction perpendicular to the surface of the hill?
…
i have no idea...

oh come on, you must know something about components …

what do you know about the component of a vector (or a force) in a particular direction?

 

[asqwt]

i know a vector is made up of an x component, and a y component. but that is when the vector is in a diagonal direction.

 

[tiny-tim]

hi asqwt!

i know a vector is made up of an x component, and a y component. but that is when the vector is in a diagonal direction.

this is in a diagonal direction …

just turn the papaer round a little so that the slope is horizontal

(no seriously … it does work!)

ie use cos and sin just the way you usually would

 

[asqwt]

you want me to rotate the diagram till the vector of the rock... (which is pointing down) is horizontal? so rotate right 90 degrees?

am i solving for the length of the slope? sin a = w/length of slope.

 

[tiny-tim]

you want me to rotate the diagram till the vector of the rock... (which is pointing down) is horizontal? so rotate right 90 degrees?

no, until the slope is horizontal …

that'll then be a situation you're familiar with

am i solving for the length of the slope? sin a = w/length of slope.

let's see …

In terms of "w" and "a", what is the component of the weight of the boulder in the direction parallel to the surface of the hill?

What is the component of the weight in the direction perpendicular to the surface of the hill?

the length of the slope has nothing to do with it

(what made you think it had?)

the component is always the original magnitude time cos of the angle …

so for one direction it'll be cos(a), and for the other it'll be cos(90°-a) = sin(a) …

which is which? ​

 

[asqwt]

the component is always the original magnitude time cos of the angle …

so for one direction it'll be cos(a), and for the other it'll be cos(90°-a) = sin(a) …

how did you get those? i didnt read that in my book :(

 

[tiny-tim]

if you've done dot-products, it's the standard dot-product with a unit vector …

you know a.b = |a| |b| cosθ (where θ is the angle between a and b)

well, if b is the unit vector in, say, the x direction (we usually write that as i), then:

a.i = |a| |i| cosθ = |a| cosθ

(since by definition of a unit vector, |i| = 1)

anyway, it does work in the situations you're familiar with, doesn't it? ​

 

[asqwt]

woh i never did dot products. sorry I am making this hard for you when it is supposedly simple = \ i feel like a retard.

but can i get a few things straightened out?

so is the VECTOR of this problem is the weight of the rock going downward?

you want me to rotate my drawing so that the diagrams "hill" is parallell with the gruond?

then i lost you. :( can walk me through the process step by step by any chance? i have no idea why the answer is w sin(a)

 

[tiny-tim]

so is the VECTOR of this problem is the weight of the rock going downward?

yes

but before we go any further, can i check what you do know …

if i tell you a vector 16 is pointing at 20° above the horizontal, and ask you for the horizontal and vertical components, what are they?

 

[asqwt]

assuming you mean the magnitude is 16. vertical is 16 (sin 20) and horizontal is 16 (cos 20)

 

[tiny-tim]

yup!

ok, it's the same for any (perpendicular) axes …

choose one to be the x axis, and call the angle for the vector to that axis θ …

then the x component is the vector times cosθ, and the y component is the vector times cos(90°-θ), = sinθ

in this case, the vector is the weight, w, downwards …

so the component along the slope (going down) is … ? ​

 

[asqwt]

im scared i might be wrong... but..

so if i rotate it to make the slope my x-axis

is the x component w cos (90-theta)? since I am not using the angle given?

 

[tiny-tim]

(just got up :zzz: …)

im scared i might be wrong... but..

so if i rotate it to make the slope my x-axis

is the x component w cos (90-theta)? since I am not using the angle given?

that's correct!

and soon you'll get used to just looking for the angle, and not having to rotate the page

(btw, i personally think "not-cos" rather than "sin" … i always look for the "correct" angle, and it either is or isn't, so it's either cos or not-cos :biggrin;)