Introduction to Inclined Planes

Summary

This educational video script covers essential physics concepts for analyzing inclined planes. It explains the forces acting on a box resting on an incline, including the normal force and gravitational components. The script introduces SOHCAHTOA to relate these forces to the angle of the incline. It then derives formulas for acceleration on an incline with and without friction, emphasizing how to calculate these values. Practical examples, such as a block sliding down a 30-degree incline and another traveling up a 25-degree incline, illustrate the application of these principles, providing a comprehensive guide to inclined plane physics.

Takeaways

• 📏 The normal force on an inclined plane is calculated as \( mg \cos(\theta) \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline.
• 🔽 The component of gravitational force acting down the incline, often denoted as \( F_g \), is given by \( mg \sin(\theta) \) and is responsible for the block's acceleration down the slope.
• 📐 The sine and cosine of the incline angle are used to determine the perpendicular and parallel components of forces acting on a block on an incline.
• 🚀 The acceleration of a block sliding down a frictionless incline is \( g \sin(\theta) \), which is independent of the block's mass.
• 🛑 Friction opposes the motion of a block on an incline. The net force and thus the acceleration are affected by both gravitational and frictional forces.
• 📉 For a block moving up an incline, the acceleration is \( -g \sin(\theta) - \mu_k g \cos(\theta) \), where \( \mu_k \) is the coefficient of kinetic friction.
• 📈 When a block is pushed up an incline, the acceleration is \( g \sin(\theta) + \mu_k g \cos(\theta) \), showing that both gravity and friction contribute to the block's upward motion.
• 🔢 The final speed of a block after traveling a certain distance down an incline can be calculated using the kinematic equation \( v^2 = u^2 + 2ad \), where \( u \) is the initial speed, \( a \) is the acceleration, and \( d \) is the distance.
• 🕒 The time it takes for a block to come to a stop on an incline can be found using the formula \( t = \frac{v_f - v_i}{a} \), with \( v_f \) being the final speed (0 in this case), \( v_i \) the initial speed, and \( a \) the acceleration.
• 🔄 The direction of forces and their components is crucial in determining the motion of a block on an incline, with gravity and friction playing significant roles.

Q & A

What is the normal force experienced by a box resting on an inclined plane?

The normal force is the force that extends perpendicular to the surface of the incline, and it is equal to the weight of the object (mg) times the cosine of the angle of inclination (θ), expressed as F_n = mg * cos(θ).

What is the force component that causes a box to slide down an inclined plane?

The force component that causes a box to slide down an inclined plane is the gravitational force component parallel to the incline, known as fg, and it is equal to the weight of the object (mg) times the sine of the angle of inclination (θ), expressed as fg = mg * sin(θ).

How does friction affect the acceleration of a block sliding down an inclined plane?

Friction opposes the motion of the block and affects its acceleration. The net force in the direction of the incline includes the gravitational force component (fg) minus the frictional force (fk), leading to an acceleration given by a = g * (sin(θ) - μ_k * cos(θ)), where μ_k is the coefficient of kinetic friction.

What is the formula for calculating the acceleration of a block sliding down a frictionless incline?

On a frictionless incline, the acceleration of a block is independent of its mass and is given by the formula a = g * sin(θ), where g is the acceleration due to gravity and θ is the angle of the incline.

How can you determine the final speed of a block after it has traveled a certain distance down an incline?

The final speed of a block can be determined using the kinematic equation v^2 = u^2 + 2*a*d, where v is the final speed, u is the initial speed, a is the acceleration, and d is the distance traveled.

What is the significance of the term SOHCAHTOA in the context of inclined planes?

SOHCAHTOA is a mnemonic used in trigonometry to remember the primary trigonometric ratios: sine (opposite/hypotenuse), cosine (adjacent/hypotenuse), and tangent (opposite/adjacent). In the context of inclined planes, these ratios help in calculating the components of forces acting on an object.

How does the angle of inclination affect the acceleration of an object on an incline?

The angle of inclination affects the acceleration of an object on an incline through the sine function. The greater the angle, the greater the component of gravitational force acting down the incline, leading to a higher acceleration (a = g * sin(θ)).

What happens to the acceleration of a block on an incline if the angle of inclination increases?

If the angle of inclination increases, the acceleration of the block on the incline also increases because the component of gravitational force acting down the incline (and thus causing acceleration) becomes larger.

How can you calculate the distance a block will travel up an incline before coming to a stop?

The distance a block will travel up an incline before coming to a stop can be calculated using the kinematic equation d = (v^2 - u^2) / (2*a), where v is the final speed (0 in this case), u is the initial speed, and a is the deceleration (negative acceleration).

What is the role of the normal force in the context of an inclined plane with friction?

In the context of an inclined plane with friction, the normal force is crucial as it affects the magnitude of the frictional force, which in turn influences the net force and acceleration of the object. The frictional force is calculated as fk = μ_k * F_n, where F_n is the normal force.

Outlines

00:00

📚 Inclined Planes and Force Analysis

This paragraph introduces the concept of inclined planes and the forces acting on an object resting on one. It explains the normal force, which is perpendicular to the surface, and the component of gravitational force (mg) acting down the incline. The speaker uses trigonometric ratios (SOHCAHTOA) to derive equations for the forces parallel (x) and perpendicular (y) to the incline. The key equations discussed are the normal force (N = mg cos(theta)) and the force causing acceleration down the incline (Fg = mg sin(theta)). The speaker emphasizes the importance of understanding these forces for solving inclined plane problems.

05:00

🔍 Deriving Acceleration on Inclined Planes

The speaker continues by discussing how to calculate the acceleration of a block on an inclined plane. They explain that when there is no friction, the only force causing acceleration is the component of gravitational force down the incline (Fg). Using Newton's second law (F = ma), the acceleration down the incline is derived to be independent of the block's mass and solely dependent on the angle of the incline (a = g sin(theta)). The paragraph also addresses the scenario where friction is present, introducing the concept of kinetic friction force (Fk) and its effect on acceleration. The formula for acceleration with friction is given as a = g sin(theta) - μk g cos(theta).

10:00

📉 Calculating Acceleration with Friction

This paragraph delves into the calculation of acceleration when friction is present and the block is either sliding down or up an inclined plane. The speaker explains that friction opposes the motion, and its direction depends on the direction of the block's movement. For a block sliding down, the net force in the x-direction includes the force of gravity down the incline and the friction force opposing the motion. The acceleration is calculated as a = g sin(theta) - μk g cos(theta). Conversely, when the block is sliding up, the forces combine in the same direction, and the acceleration is a = -g sin(theta) + μk g cos(theta).

15:02

📐 Solving a Physics Problem with Inclined Planes

The speaker presents a specific physics problem involving a block sliding down a 30-degree incline from rest. They guide through the process of finding the block's acceleration using the previously derived formula (a = g sin(theta)). The sine of 30 degrees is calculated, and the acceleration is found to be 4.9 m/s². The paragraph then moves on to calculate the block's final speed after traveling 200 meters down the incline using the kinematic equation v² = u² + 2ad, where u is the initial speed, a is the acceleration, and d is the distance. The final speed is determined to be approximately 44.27 m/s.

20:02

🚀 Block's Motion Up a 25-Degree Incline

The final paragraph discusses a scenario where a block is traveling up a 25-degree incline with an initial speed. The speaker calculates the block's acceleration, considering the gravitational force component acting against its motion (a = -g sin(theta)). Using the kinematic equation, they determine how far the block will travel up the incline before coming to a stop. The distance is calculated by rearranging the kinematic equation to solve for displacement (d = v² / (2a)). The speaker also calculates the time it takes for the block to come to a complete stop using the formula t = (v - u) / a. The time is found to be approximately 3.38 seconds.

Keywords

💡Inclined Plane

An inclined plane is a simple machine that consists of a flat surface tilted at an angle. In the context of the video, it is used to demonstrate the physics of forces acting on an object, like a box, when it is placed on a slope. The video explains how the force of gravity can be broken down into components along and perpendicular to the incline, which is crucial for understanding motion and forces in such scenarios.

💡Normal Force

The normal force is the force exerted by a surface that supports the weight of an object resting on it. In the video, it is described as acting perpendicular to the inclined plane and is calculated as the product of the object's mass and the cosine of the angle of inclination. It's a key concept in analyzing the forces on an object on an incline, as it balances part of the gravitational pull.

💡Component of Force

In physics, a component of force refers to the part of a vector quantity that acts in a particular direction. The video script discusses how the force of gravity can be decomposed into components parallel and perpendicular to the inclined plane, which is essential for solving problems involving motion on an incline.

💡Trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. The video uses trigonometric functions (sine, cosine, and tangent) to relate the forces acting on an object to the angle of the incline. This is exemplified by the use of SOHCAHTOA, a mnemonic for sine, cosine, and tangent relationships in right triangles.

💡Net Force

Net force is the vector sum of all the forces acting on an object. The video explains that to find the acceleration of an object on an incline, one must calculate the net force along the incline, which is the difference between the gravitational force component pulling the object down the incline and any opposing forces, such as friction.

💡Acceleration

Acceleration is the rate of change of velocity of an object with respect to time. In the video, the concept is used to calculate how fast an object's speed changes as it moves down or up an incline. The script provides formulas for calculating acceleration on an incline, considering both scenarios without and with friction.

💡Friction

Friction is the force that opposes the relative motion or tendency of such motion of two surfaces in contact. The video discusses how friction affects the motion of a block on an incline, either slowing it down as it moves up or adding to the force pulling it down when moving down.

💡Kinetic Friction

Kinetic friction, also known as dynamic friction, is the frictional force that acts on a body when it is in motion relative to another surface. The video explains that kinetic friction opposes the motion of an object and must be considered when calculating the net force and acceleration on an inclined plane.

💡Static Friction

Static friction is the frictional force that must be overcome to set an object in motion. The video mentions static friction in the context of an object at rest on an incline, where it balances the component of gravitational force trying to move the object down the slope until the force exceeds the maximum static friction.

💡Newton's Second Law

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The video uses this law to derive formulas for acceleration on an incline, demonstrating how to calculate the motion of objects under various force conditions.

Highlights

Review of essential formulas for dealing with inclined planes.

Explanation of the normal force and its direction on an incline.

Introduction to the concept of resolving forces into components using trigonometry.

Use of SOHCAHTOA to relate the components of force to the angle of the incline.

Derivation of the equation for the normal force as mg cos(theta).

Identification of the component of gravitational force causing motion down the incline (fg = mg sin(theta)).

Formula for acceleration down an incline without friction: a = g sin(theta).

Incorporating friction into the analysis of inclined plane motion.

Derivation of acceleration with friction using the equation a = g(sin(theta) - μk cos(theta)).

Discussion on the direction of forces and acceleration when a block is moving up an incline.

Calculation of the acceleration of a block sliding down a 30-degree incline from rest.

Determination of the final speed of a block after traveling a certain distance down an incline.

Application of kinematic equations to solve for distance traveled and time taken.

Problem-solving approach for a block traveling up a 25-degree incline with initial speed.

Calculation of the distance a block will travel up an incline before stopping.

Estimation of the time it takes for a block to come to a stop on an incline.

Emphasis on the importance of understanding the direction of forces in inclined plane problems.

Transcripts

in this video

i want to review some formulas that you

need to know when dealing with inclined

planes

now 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 wave 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

so that's how far

up the incline it's going to go before

coming to a stop

now what about part c

how long

will it take before the block comes to a

stop

key phrase how long so what are we

looking for here

how long tells you

how you know the time how long it's

going to take

so we're looking for the time it takes

until it comes to a complete stop

so we got to calculate t

the kinematic formula that we could use

is this equation v final is equal to v

initial plus 18.

by the way if you need to brush up on

your kinematic formulas

just go to youtube type in kinematics

organic chemistry tutor and i have a

video dedicated to that topic

so it can give you a good review of how

to use those formulas so when you get a

chance

the full version of that video can be

found on my patreon page

which you can also access

in the description section below of this

video

so now let's move on with this equation

so the final speed when it comes to a

stop is zero

the initial speed is 14 the acceleration

well that hasn't changed it's negative

4.14166

so we just got to solve for t

let's move 14 to the other side

so this is going to be negative 14 is

equal to

what we have here

so t

is going to be

this number divided by that number

so negative 14 divided by

negative 4.14166

so we can round that to 3.38 seconds

so that's how long it's going to take

for the block to come to a complete stop

you