Introduction to Inclined Planes
Summary
TLDRThis 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
📚 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.
🔍 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).
📉 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).
📐 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.
🚀 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.
Mindmap
Keywords
💡Inclined Plane
💡Normal Force
💡Component of Force
💡Trigonometry
💡Net Force
💡Acceleration
💡Friction
💡Kinetic Friction
💡Static Friction
💡Newton's Second Law
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
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