Understanding Inclined Planes and Forces

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 an incline. The force that extends perpendicular to the surface, this is the normal force. The weight force is in the negative y direction, as always. So that's mg. Now 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 it 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, 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've heard of the term SOHCAHTOA in trigonometry. So the first part means sine is equal to the opposite side divided by 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 0, 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 the 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 inclined 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 as the y direction. The only force that's accelerating it in the x direction is Fg. So the net force in the x direction is Fg, 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 Fg 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 Fg plus negative Fk or simply minus Fk. Now always replace this with Me. Now we know Fg is Mg sine theta, and Fk, the kinetic frictional force, is mu k times the normal force. So make sure you know this equation. And the 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 ma 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 inclined 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's sliding in the positive x direction. Fg 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 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 Fg minus Fk, 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 sines, 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 pointing 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 Ma, 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. So let's work on this physics problem as it relates to inclined planes. A block slides down a 30 degree incline starting from res. 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 is 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 ma, mass times acceleration. Fg 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 1 half. Theta is 30, and sine 30 is 1 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, it's 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 your 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 0, A is 4.9, and D is 200. So it's 2 times 4.9, which is 9.8, times 200. 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 inclined 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 inclined 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 gets to move in 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 inclined plane. So let's write the net force, or 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. Fg. 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 what 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 0. 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 2AD. V final is 0. V initial is 14. And the acceleration is negative 4.14166 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 8.28332d 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 8.28332. 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 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 this video. So now let's move on with this equation. So the final speed when it comes to a stop is 0. 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.

Now 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 it 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, 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've heard of the term SOHCAHTOA in trigonometry. So the first part means sine is equal to the opposite side divided by 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 0, 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 the 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 inclined 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 as the y direction.

The only force that's accelerating it in the x direction is Fg. So the net force in the x direction is Fg, 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 Fg 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 Fg plus negative Fk or simply minus Fk. Now always replace this with Me. Now we know Fg is Mg sine theta, and Fk, the kinetic frictional force, is mu k times the normal force. So make sure you know this equation.

And the 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 ma 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 inclined 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's sliding in the positive x direction. Fg 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 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 Fg minus Fk, 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 sines, 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 pointing 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 Ma, 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. So let's work on this physics problem as it relates to inclined planes. A block slides down a 30 degree incline starting from res. 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 is 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 ma, mass times acceleration. Fg 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 1 half.

Theta is 30, and sine 30 is 1 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, it's 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 your 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 0, A is 4.9, and D is 200. So it's 2 times 4.9, which is 9.8, times 200. 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 inclined 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 inclined 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 gets to move in 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 inclined plane.

So let's write the net force, or 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. Fg.

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 what 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 0. 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 2AD. V final is 0. V initial is 14. And the acceleration is negative 4.14166 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 8.28332d 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 8.28332. 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 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 this video. So now let's move on with this equation.

So the final speed when it comes to a stop is 0. 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.

Transcript for:Understanding Inclined Planes and Forces

Transcript for:
Understanding Inclined Planes and Forces