High school physics
Inclined plane force components
Explore the various forces acting on a block sitting on an inclined plane. Learn how to break the force of gravity into two components - one perpendicular to the ramp and one parallel to the ramp. Finally, using geometry and trigonometry, learn how to calculate the magnitude of each component of force that is acting on the block. Created by Sal Khan.
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- How can something be perpendicular to the surface of the earth which is curved.5:20(54 votes)
- on a small scale it can be neglected.
if you stand on ground, you dont see it curved, you see it plane and straight. Thats because you are tiny comparatively(176 votes)
- Sal said normal force is perpendicular to the surface. But:
1. if normal force is reaction force (weight is 'action force') shouldn't it be positioned on a same line, that is, at same angle as weight of the body on a ramp? Third law states: "if one object exerts a force on another, the other exerts an equal and opposite force on the first." So, why is normal force 'normal' (perpendicular to the surface)?
2. In my experience, if a body is at inclined surface the part of the body closer to the angle of incline plane, (that is, left side of the box) pushes 'harder' to the ground than it's right side. If that is so, normal force shouldn't, in my opinion, be perpendicular to the surface because 'distribution of the pressure' on the inclined plane isn't even.
Thank you. Hope my question is clear...(19 votes)
- If there is no friction the surface of on abject can only impart a force on another object perpendicular to surface, any component of the force parallel to the surface requires friction. So the force is broken up into a Normal Force and a Friction Force.
With an incline that is frictionless and you have a block on it the block's weight is directed strait down but the normal force is perpendicular to the incline so when you add the force vectors you end up with a net force parallel to the incline pointing down and this is what causes the block to slide down the incline.
On a frictionless incline the force is the same along the entire surface of the box so there is no portion that pushes harder into the ground than another. Once you introduce friction you have gravity and the normal force essentially acting like it is all at the center of gravity of the box but the friction force is along the surface of contact between the box and the incline. This friction force not in line with the center of gravity of the box produces a torque causing the edge of the box away from the direction of the friction force to put a bit more pressure on the incline that the other.
But regardless of the amount of pressure at each point along the the surface of contact the force without friction is always perpendicular to the surface and is the normal force.(12 votes)
- how do i know what trig-ratio to use??
I'm confused on that..(7 votes)
- The way I remember is with SOHCAHTOA. The SOH means that the Sine is Opposite over Hypotenuse, C(osine) = A(djacent) / H(ypotenuse), T(angent) = O(pposite) / A(djacent)(23 votes)
- What are the arrows that Sal is drawing above the F and G?(8 votes)
- Those arrows show that they(F and G) are vector quantities(7 votes)
- what is force(1 vote)
- A force is a push or a pull...
So anytime you move you are pushing on somthing...
Walking- pushing on the ground
Biking- pushing the peddles etc(14 votes)
- i still dont understand why you break up the two forces of gravity. what's the point of doing this?(5 votes)
- We want to see how gravity affects motion on the inclined plane, so we break the force of gravity into components that are parallel and perpendicular to the plane, respectively. That way, we can see how gravity actually affects the sliding motion.(7 votes)
- How would you calculate normal force for a box on an incline? And what about if there is a force being exerted horizontally which keeps the object at rest neglecting friction? How would to force diagram for a scenario like this look and how would you solve for Fnormal and Fapplied?(5 votes)
- Breakout the gravity force vector into components which are parallel -mg*sin(Θ) and perpendicular -mg*cos(Θ) to the incline. The normal force will be equal and opposite to the perpendicular gravity component so N = +mg*cos(Θ). To keep the block from sliding, you would then need to apply a horizontal force equal and opposite to the parallel gravity component so that force would be F = +mg*sin(Θ).(6 votes)
- How come when we're trying to figure out the force of gravity parallel and force of gravity perpendicular to the inclined plane we use theta as the "focus point"?(3 votes)
- because trigonometry allows us to do that ; use angle as an identity to relate two lines(as in this case with two vectors) in a right angled triangle(4 votes)
- All of this is really complicated, and I'm having a hard time understanding how to solve this. Is there an equation or something that simplifies this whole process?(4 votes)
- The only quick tip I can give you is that the force of gravity will be multiplied by the sine of the incline and the normal force multiplied by the cosine of the incline.(3 votes)
- I want to know that if we find the resultant of the normal force and mg directly without resolving them by parallelogram method,then will the resultant be equal to mg sin θ???(3 votes)
Let's say I have some type of a block here. And let's say this block has a mass of m. So the mass of this block is equal to m. And it's sitting on this-- you could view this is an inclined plane, or a ramp, or some type of wedge. And we want to think about what might happen to this block. And we'll start thinking about the different forces that might keep it in place or not keep it in place and all of the rest. So the one thing we do know is if this whole set up is near the surface of the Earth-- and we'll assume that it is for the sake of this video-- that there will be the force of gravity trying to bring or attract this mass towards the center of the Earth, and vice versa, the center of the earth towards this mass. So we're going to have some force of gravity. Let me start right at the center of this mass right over here. And so you're going to have the force of gravity. The force due to gravity is going to be equal to the gravitational field near the surface of the Earth. And so we'll call that g. We'll call that g times the mass. Let me just write it. The mass times the gravitational field near the surface of the Earth. And it's going to be downwards, we know that, or at least towards the surface of the Earth. Now, what else is going to be happening here? Well, it gets a little bit confusing, because you can't really say that normal force is acting directly against this force right over here. Because remember, the normal force acts perpendicular to a surface. So over here, the surface is not perpendicular to the force of gravity. So we have to think about it a little bit differently than we do if this was sitting on level ground. Well, the one thing we can do, and frankly, that we should do, is maybe we can break up this force, the force due to gravity. We can break it up into components that are either perpendicular to the surface or that are parallel to the surface. And then we can use those to figure out what's likely to happen. What are potentially the netting forces, or balancing forces, over here? So let's see if we can do that. Let's see if we can break this force vector, the force due to gravity, into a component that is perpendicular to the surface of this ramp. And also another component that is parallel to the surface of this ramp. Let me do that in a different color. That is parallel to the surface of this ramp. And this is a little bit unconventional notation, but I'll call this one over here the force due to gravity that is perpendicular to the ramp. That little upside down t, I'm saying that's perpendicular. Because it shows a line that's perpendicular to, I guess, this bottom line, this horizontal line over there. And this blue thing over here, I'm going to call this the part of force due to gravity that is parallel. I'm just doing these two upward vertical bars to show something that is parallel to the surface. So this is the component of force due to gravity that's perpendicular, component of force that is parallel. So let's see if we can use a little bit a geometry and trigonometry, given that this wedge is at a theta degree incline relative to the horizontal. If you were to measure this angle right over here, you would get theta. So in future videos we'll make it more concrete, like 30 degrees or 45 degrees or whatever. But let's just keep in general. If this is theta, let's figure out what these components of the gravitational force are going to be. Well, we can break out our geometry over here. This, I'm assuming is a right angle. And so if this is a right angle, we know that the sum of the angles in a triangle add up to 180. So if this angle, and this 90 degrees-- right angle says 90 degrees-- add up to 180, then that means that this one and this one need to add up to 90 degrees. Or, if this is theta, this angle right over here is going to be 90 minus theta. Now, the other thing that you may or may not remember from geometry class is that if I have two parallel lines, and I have a transversal. So I'm going to assume this line is parallel to this line. And then I have a transversal. So let's say I have a line that goes like this. We know from basic geometry that this angle is going to be equal to this angle. It comes from alternate interior angles. And we prove it in the geometry module, or in the geometry videos. But hopefully this makes a little bit of intuitive sense, and you could even think about how these angles would changes as the transversal changes, and all of the rest. But the parallel lines makes this angle similar to that angle, or actually makes it identical, makes it congruent. This angle is going to be the same measure as that angle. So can we apply that anywhere over here? This line is perpendicular to the surface of the Earth. Right over here that I'm kind of shading in blue. And so is this force vector. It is also perpendicular to the surface of the Earth. So this line over here and this line over here in magenta are going to be parallel. I can even draw that. That line and that line are both parallel. When you look at it that way, you'll see that this big line over here can be viewed as a transversal. Or you could have this angle and this angle are going to be congruent. They're going to be alternate interior angles. So this angle and this angle, by the exact same idea here. It just looks a little bit more confusing here because I have all sorts of things. But this line and this line are parallel. You can view this right over here as a transversal. So this and this are congruent angles. So this is 90 minus theta degrees. This too will be 90 minus theta degrees. 90 minus theta degrees. Now, given that, can we figure out this angle? Well one thing, we're assuming that this yellow force vector right here is perpendicular to the surface of this plane or perpendicular to the surface of this ramp. So that's perpendicular. This right here is 90 minus theta. So what is this angle up here going to be equal to? This angle, let me do it in green. What is this angle up here going to be equal to? So this angle plus 90 minus theta plus 90 must be equal to 180, or this angle plus 90 minus theta must be equal to-- let me write this down. I don't want to do too much in your head. So let's call it x. So x plus 90 minus theta. Plus this 90 degrees right over here, plus this 90 degrees, needs to be equal to 180 degrees. Let's see, we can subtract 180 degrees from both sides. So we subtract 90 twice, you subtract 180 degrees and you get x minus theta is equal to 0, or x is equal to theta. So whatever the inclination of the plane is or of this ramp, that is also going to be this angle right over here. And the value to that is that now we can use our basic trigonometry to figure out this component and this component of the force of gravity. And to see that a little bit clearer, let me shift this force vector down over here. The parallel component, let me shift it over here. And you can see the perpendicular component plus the parallel component is equal to the total force due to gravity. And you should also see that this is a right triangle that I have set up over here. This is parallel to the plane. This is perpendicular to the plane. And so we can use basic trigonometry to figure out the magnitudes of the perpendicular force due to gravity and the parallel force due to gravity. Let's think about it a little bit. I'll do it over here. The magnitude of the perpendicular force due to gravity. Or I should say the component of gravity that's perpendicular to the ramp, the magnitude of that vector-- a lot of fancy notation but it's really just the length of this vector right over here. So the magnitude of this over the hypotenuse of this right triangle. Well, what the hypotenuse of this right triangle? Well, it's going to be the magnitude of the total gravitational force. I guess you could say that. And so you could say that is mg. We could write it like this. But that's really-- well, I could write it like that. And so this is going to be equal to what? We have the, if we're looking at this angle right here, we have the adjacent over the hypotenuse. Remember. We can do this in a new color. We can do this in a new color. SOH CAH TOA. Cosine is adjacent over hypotenuse. So this is equal to cosine of the angle. So cosine of theta is equal to the adjacent over the hypotenuse. So if you multiply both sides by the magnitude of the hypotenuse, you get the component of our vector that is perpendicular to the surface of the plane is equal to the magnitude of the force due to gravity times the cosine of theta. Times the cosine of theta. We'll apply this in the next video just so you can make the numbers a lot more concrete. Sometimes just the notation makes it confusing. You'll see it's really actually pretty straightforward. And then this second thing, we can use the same logic. If we think about the parallel vector right over here, the magnitude of the component of the force due to gravity that is parallel to the plane over the magnitude of the force due to gravity-- which is the magnitude of mg-- that is going to be equal to what? This is the opposite side to the angle. So the blue stuff is the opposite side, or at least its length, is the opposite side of the angle. And then right over here this magnitude of mg, that is the hypotenuse. So you have the opposite over the hypotenuse. Opposite over hypotenuse. Sine of an angle is opposite over hypotenuse. So this is going to be equal to the sine of theta. This is equal to the sine of theta. Or you multiply both sides times the magnitude of the force due to gravity and you get the component of the force due to gravity that is parallel to the ramp is going to be the force due to gravity total times sine of theta. Times sine of theta. And hopefully you should see where this came from. Because if you ever have to derive this again 30 years after you took a physics class, you should be able to do it. But if you know this right here, and this right here, we can all of a sudden start breaking down the forces into things that are useful to us. Because we could say, hey, look, this isn't moving down into this plane. So maybe there's some normal force that's completely netting it out in this example. And maybe if there's nothing to keep it up, and there's no friction, maybe this thing will start accelerating due to the parallel force. And we'll think a lot more about that. And if you ever forget these, think about them intuitively. You don't have to go through this whole parallel line and transversal and all of that. If this angle went down to 0, then we'll be talking about essentially a flat surface. There is no inclination there. And if this angle goes down to 0, then all of the force should be acting perpendicular to the surface of the plane. So if this going to 0, if the perpendicular force should be the same thing as the total gravitational force. And that's why it's cosine of theta. Because cosine of 0 right now is 1. And so these would equal each other. And if this is equal to 0, then the parallel component of gravity should go to 0. Because gravity will only be acting downwards, and once again, if sine of theta is 0. So the force of gravity that is parallel will go to 0. So if you ever forget, just do that little intuitive thought process and you'll remember which one is sine and which one is cosine.