Introduction to Inclined Planes

 let's say if you have a box that

rests on the incline

the force that extends perpendicular to

the surface

this is the normal force the wav

e force

is in the negative y direction as always

so that's mg

and let's say we have an angle now what

you want to do is you want to draw

this line but in the other direction

and you want to draw a triangle making

this the hypotenuse of the triangle

so it's important to understand that

this angle is the same as this angle

for instance let's say

if this angle let me see if i can fit in

here let's say it's 30

and this we can see it's 90 which means

this has to be 60

and this is perpendicular

like this line is perpendicular to that

line so which means this

is 90 as well so therefore this part has

to be 30

which means theta is right there in that

triangle

so let's focus on that triangle

now let's say this is

let's call this x and let's call this y

in terms of mg what is x and what is y

what would you say

perhaps you heard of the term sohcahtoa

in trigonometry

so the first part means sine is equal to

the opposite side divided by the hypot

excuse me the hypotenuse and this is

cosine

is the ratio of the adjacent side of the

right triangle divided by the hypotenuse

and tangent is the ratio of the opposite

side to the adjacent side

so cosine theta is going to be equal to

the adjacent side divided by the

hypotenuse

so that's x over mg now if you cross

multiply and solve for x

you'll see that x is mg cosine theta

so therefore this portion

is equivalent to mg cosine theta

and because the block is not

accelerating upward or downward

in this direction we'll call it the y

direction let's call this the x

direction

in the y direction the net force has to

be zero which means that these two

must be equal to each other so for a

typical inclined problem

the normal force is equal to mg cosine

theta

so make sure you know this equation

now let's focus on sine theta sine theta

is the ratio of the opposite side which

is y

and hypotenuse so sine is y

over the hypotenuse which is mg

so if you cross multiply you'll see that

y is

mg sine theta

so that correlates to this portion of

the triangle

now it turns out that there is a force

that accelerates the block

down the incline so that force i like to

call it fg

and notice that it's parallel to this

part of the triangle

so it turns out that fg the component of

the weight force that accelerates the

block down the incline

is mg sine theta so that's another

equation that you want to use

when dealing with inclined problems

or incline plane problems

now let's say if you have a block

that rests on a frictionless incline

and you want to find the acceleration

how can we derive a formula to do that

so once again what we're going to do is

we're going to define this as the x

direction

and this says the y direction the only

force that's accelerating it in the x

direction is f g

so the net force in the x direction

is f g because that's the only force

there now according to newton's second

law

f is equal to ma the net force is always

the product of the mass and acceleration

and we know fg is mg sine theta

so to find the acceleration down the

incline we don't need to know the mass

of the block

it's independent of the mass of the

block so the acceleration down the

incline

is simply g sine theta it's dependent on

the angle

and so this is the equation you want to

use

now sometimes you may have to deal with

friction

so let's say the block is sliding down

it's moving in a positive x direction

where is friction located now we know f

g is going to be

down as well but friction always opposes

motion

so if the block is sliding down kinetic

friction will oppose it

and so it's in the opposite direction so

now if you want to find the acceleration

we need to start with this expression

the net force in the x direction

now this is in the positive x direction

and this is in the negative x direction

so this is going to be positive f g plus

negative f k or simply minus f k

now always you play assists of m a

now we know fg is mg sine theta

and fk the kinetic friction of force

is mu k times the normal force

so make sure you know this equation and

a static frictional force

is less than or equal to mu s

times the normal force

now we know the normal force as we

mentioned before

is mg cosine theta so therefore m

a is equal to mg sine theta

minus mu k times mg

cosine theta so once again we could

cancel

the mass in this problem so when dealing

with friction on an incline plane

the acceleration is going to be g sine

theta

minus mu k g cosine theta

and so this is the formula that you want

to use

now what about if we have a block that

is sliding

up the incline how can we calculate the

acceleration of the system how can we

derive an expression for it so it slide

in

in the positive x direction

f g is always going to be pulling the

block down

gravity pulls things down so this

component of the weight force

is going to cause it to go in the

negative x direction

based on the picture that's uh presented

and the frictional force will always be

opposite to direction of motion so

friction is also pointing in this way

so therefore for this example the net

force in the x direction is going to be

negative f

g minus f k

because they're both going in the

negative x direction to the left

so the formula is not going to change

the only thing that's changing

is the signs if it's positive or

negative

so therefore the acceleration is going

to be negative

g sine theta minus

mu k g cosine theta

so these two are working together to

slow down the block

now how would the situation change

if the block is sliding up the incline

but the incline is in the reverse

direction

so fg will still bring it down and fk

will still

oppose the block from sliding up but

this time

they're pointed in the positive x

direction

so therefore it's just going to be fg

plus fk

so in this direction acceleration is

positive

in the other example the acceleration

was negative and that's the only

difference

so just like before this is going to be

m a and

fg is going to be mg sine theta

and fk is mu k times mg

cosine theta

and so the acceleration of the system is

positive g

sine theta and this is supposed to be a

plus

so plus mu k g cosine theta

and so this is it

let's work on this physics problem as it

relates to inclined planes

a block slides down a 30 degree incline

starting from rest so let's draw an

incline

and here's the block

what is the acceleration of the block

so how can we find the answer to that

question

in order to find the acceleration we

need to find the net force

now we're going to define this as the

x-axis relative to the block

and this is going to be the y-axis

so what forces are acting on the block

in a horizontal direction that is

parallel to the incline

the only force that's acting on it it's

a component of gravity that brings it

down

i like to call this force fg

so whenever you have a typical incline

the main forces that you need to worry

about is the normal force

which is relevant if there's friction

fg the force the component of the

gravitational force that

accelerates it down the incline and

if you have any static or kinetic

friction which we don't have in this

problem

now if you wish to calculate fg it's

equal to

mg sine theta

fn the normal force is mg cosine theta

but we don't need that

in this problem

so the net force in the x direction is

therefore equal to this force because

that's the only force acting

in that direction now the net force is

going to be m a

mass times acceleration f g

is mg sine theta so notice that in this

problem

we don't need to know the value of m it

can be cancelled

so the acceleration in the x direction

parallel to the incline is simply g sine

theta

so in this problem sine theta

is going to be one half theta is 30 and

sine 30 is one half

so half of 9.8 is going to be

4.9 so the acceleration

is 4.9 meters per second squared

and so this is the answer

now let's move on to part b

what is the final speed of the block

after it travels 200 meters down the

incline

so the distance between these two points

is 200 meters

so how can we find the final speed for

kinematic problems like this

it's helpful if you make a list of what

you have and what you need to find

now the block starts from rest so the

initial speed is 0.

our goal is to calculate the final speed

we have the distance traveled as 200

meters and we know the acceleration

it's 4.9 meters per second squared

so what formula has these four variables

so if you look at the physics formula

sheet you'll see that it's v

final squared which is equal to v

initial squared plus

2ad the initial speed is zero a

is four point nine and d is two hundred

so it's two times four point nine which

is nine point eight

times two hundred so that's going to be

1960 and we need to take the square root

of both sides

so therefore the final speed

is 44.27 meters per second

so that's going to be the speed of the

block after traveled

a distance of 200 meters given

this constant acceleration number two

a block travels up a 25 degree incline

plane

with an initial speed of 14 meters per

second

part a what is the acceleration of the

block

well let's begin by drawing a picture

so let's say this is our incline plane

and it's 25 degrees above the horizontal

and let's draw a block here

so let's say this is the x-axis with

reference to the block

and this is the y-axis now the block

it's moving

with an initial speed of 14 meters per

second

now we have a component of the

gravitational force

that's going to slow the block down

as it goes up the incline plane

so let's write the net force where the

sum of the forces in the x direction

so the sum of the forces in the x

direction is only

this force that's the only force acting

on the block

in the x direction and it's in a

negative x direction so this is going to

be negative

f g net force according to newton's

second law

is mass times acceleration

fg we know it's mg sine theta

so we can cancel m and this will give us

the acceleration in the x direction

which is negative g sine theta

so now g is 9.8

and we're going to multiply that by sine

of 25 degrees

so the acceleration in the x direction

is negative 4.14166

meters per second squared

so that's the answer for part a now

let's move on to part b

how far up will it go

so how far up along the incline will

this block

go before it comes to a stop

well let's write that when we know we

know the initial speed in the x

direction

is 14 meters per second

what's the final speed when it comes to

a stop

when it comes to a stop the final speed

is going to be zero

we know the acceleration we have it here

so i'll just rewrite this

or missing is the distance

or the displacement along the x

direction

so what formula has these values

so this is the formula we need v final

squared

is equal to v initial squared plus 2

a d

v final is 0 v initial is 14

and the acceleration is negative 4

0.14 166 and then times d

14 squared is 14 times 14 so that's 196.

and then 2 times the acceleration

this is going to be negative 8.28

332 d now let's subtract

196 from both sides so moving this to

the other side

it's going to be negative 196 on the

left

and that's going to equal this

so now let's get d by itself

let's divide both sides by negative

eight point two eight

three three two

so d is going to be 23.662

meters