WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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In this lesson we are going to talk about ramps and inclines.
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Our objectives include drawing and labeling free body diagram, showing all forces acting on an object on ramp.
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Drawing a pseudo free body diagram showing all components of forces acting on the object.
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Utilizing Newton’s laws of motion to solve problems involving objects on ramps.
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We have done a few of these already but I think it is worthwhile to really take some time and make sure we get these down because they come up so commonly.
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Drawing a free body diagrams for ramps.
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Choose the object of interest and draw these on a dot or a box.
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Draw and label all the external forces acting on the object and then scatter coordinate system choosing the direction of the objects motion as one of the axis.
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That means your X axis might be tilted in some direction and that is ok.
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For the case of an object on the ramp, the direction of the objects motion should most likely be up or down that ramp.
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If we have something like a ramp with a box on it, that is the same with the hammered with boxes on ramps for some reason.
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Some angle θ, there is a box what I would do to draw my free body diagram is draw my axis at the same angle as the incline of the ramp.
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I would call that my x axis then I will do my best to draw a line perpendicular to it for my y.
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There is x and y.
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We will draw a dot for our object.
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We will have our normal force perpendicular to the plane of the ramp.
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We will have our frictional force.
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In this case, assuming it is lying down the ramp will oppose that.
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We will call that our frictional force.
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There is our normal force straight down.
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That would be our free body diagram.
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When the forces do not line up with the axis, we can draw that pseudo free body diagram when we break some of those forces up and their components.
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We will redraw our diagram of all forces parallel to the axis.
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If we had something like let us draw it again.
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Our free body diagram for axis x and y.
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I'm just going to show how the angle of our plane is the same as the angle of our ramp there.
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We had a frictional force, we had a normal force, we had weight down.
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If that angle θ and this angle also with θ are going to break MG up in the components that are parallel to the y axis and parallel to the x axis.
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That is the way to do that is with pseudo free body diagram on a separate diagram.
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If we want all of our points in the AP test.
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We will draw our axis again and XY.
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We have our dot, we still have our frictional force, our normal force.
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We put in our components of weight.
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We have the component that is adjacent to our angle which I call MG perpendicular because it is perpendicular to the plane on the ramp is MG Cos θ.
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MG parallel the component of the weight that is parallel with the ramp was MG sin θ.
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Θ is the angle of the ramp itself or it is also that angle right there.
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Alright, so boxes on ramps.
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There are other forces the free body diagram in the pseudo free body diagram for box sitting on a ramp of angle θ.
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Then write Newton’s second law of equation for the X and Y directions.
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As we take a look here, let us start by drawing the forces on our box itself.
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Normal force will have a frictional force and MG.
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Cannot ever get too much practice doing this.
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We will draw our free body diagram over here.
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We would have normal force N, frictional force opposing motion and MG.
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Separate diagram can emphasize that enough our pseudo free body diagram.
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xy, we got our frictional force.
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Our normal force.
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Breaking up weight into a component parallel MG sin θ and perpendicular MG cos θ.
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If I were to write Newton’s second law of equations there in the x direction net force in the x direction.
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We are going to call down positive here we can write that as MG sin θ - frictional force = Ma X in the y direction
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net force in the y = N - MG cos θ is equal the Ma Y but the boxes sitting on the ramp
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it is not going to accelerate off the ramp nor is it going to fall through the ramp anytime soon.
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That is all equal to 0 and that would work.
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Let us do an example here.
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3 Forces act on the box and inclined plane as shown in the diagram below, weight, friction, and normal force.
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If the box is at rest the net force acting on it is equal to, if it is at rest the net force must be 0 equilibrium condition.
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Our answer there must be 0 because the box is at rest.
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Let us do another one.
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A 5kg mass is held at rest on the frictionless 30° incline by force F.
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What is the magnitude of force F?
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Let us start with their free body diagram.
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Put it up over here.
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There is X perpendicular to it.
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We will draw our Y so XY.
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We have our normal force, we have our applied force F and MG straight down.
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I'm going to draw our pseudo free body diagram.
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We put that over here on the left where we got some room.
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Our pseudo free body diagram, there is our x and y.
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We still have our normal force along the y, we have our applied force, but MG we break up in the components.
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We got the component parallel to the ramp MG sin θ and perpendicular to the ramp into the ramp MG cos θ.
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Let us call this our positive X and another Y direction when we write our Newton’s second law equation.
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Let us start in the x, net force in the x direction.
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It is going to be F - MG sin θ which is going to be equal to MA x
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Which, since acceleration is 0 because it is at rest it is all equal to 0.
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Therefore, F = MG sin θ which is going to be 5 kg × G 10 m / s² × the sin of 30° which is ½.
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The half of 50 is going to be 25 N.
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Take an example where we look at a truck on a hill.
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The diagram shows 100,000 N truck at rest on the hill that makes an angle of 8° with a horizontal.
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What is the component of the trucks weight parallel to the hill?
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MG parallel we just remember that equation.
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The component of the weight down the hill is going to be MG sin θ.
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It gives us the weight MG already which is 1 × 10⁵ × sin 8° which is right about 1.4 × 10⁴ N.
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Forces up the ramp.
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The block weighing 10 N is on a ramp inclined to 30° to the horizontal.
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A 3 N force of friction ff acts on the block as it is pulled up the ramp at constant velocity.
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It is always important to know constant velocity.
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The net force must be 0, acceleration is 0 as some force F parallel to the ramp.
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Find the magnitude of force F.
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Let us start with their free body diagram.
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I will draw over here on the left to begin with.
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The x and y and our object.
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A dot we have the normal force.
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We got some force F up the ramp, we have our frictional force.
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We have the weight of the box which it gives us is 10 N always down.
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We can draw our pseudo free body diagram.
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We will do that over here on the right YX.
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F still acting up the ramp and perpendicular to it for some friction down the plane on the ramp.
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We are breaking up MG into its components.
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We have the one perpendicular to the top of the ramp it is going to be 10 N cos 30° is 8.66.
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We will have 10 sin 30 the plane of the ramp parallel which is going to be 5 N.
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What is the magnitude of force F?
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Let us use Newton’s second law in the x direction.
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F net x = f - the force of friction -10 sin 30 as the equal 0 because it is not accelerating.
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It is moving at a constant velocity.
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Therefore, F equals force of friction + 10 sin 30 which is 5,
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Which is our force of friction that says is 3 N.
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That is 3 N + 5 N therefore F must be equal to 8 N.
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Let us take a look at some acceleration down a ramp.
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A 100 kg block slides down to frictionless 30° incline as shown, find the acceleration of the block.
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Just like we have been doing, let us draw our axis for free body diagram.
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Here is our X, there is our Y, there is our box, it is on the frictionless surface.
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We have normal force and we have its weight down.
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Friction and applied forces.
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Let us do the pseudo free body diagram.
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It looks like we have room beside it to do so.
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We will draw our pseudo free body diagram over here.
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We still have our normal force up the ramp, we are going to break MG in the components.
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We have MG sin θ parallel with the ramp and MG perpendicular which is MG cos θ perpendicular or in the plane ramp.
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There is our y, there is our x.
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If we want the acceleration of the block, it is going to accelerate in the x direction I am going to write that Newton’s equation first.
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F and x equals all the only thing we have there is MG sin θ which must equal M × acceleration in the x direction.
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Therefore, pretty straightforward acceleration in the x direction is just G sin θ is going to be 10 m / s² sin 30°.
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Sin 30 is half so that is just 5 m / s².
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Let us talk about what happens when we place the block on a ramp with an unknown coefficient of friction.
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The angle of the ramp is slowly increased until the block just begins to slide.
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To find the coefficient of static friction is a function of the ramps angle of elevation
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because the angle of the object just begins to slide is known as the angle of repose.
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Keep lifting it upward to starts to slide, you measure that angle and you can find up the coefficient of friction.
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Let us see how we can do all that.
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Here is a free body diagram of our block on our ramp.
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We have the force of static friction holding it in place.
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The normal force out of the plane in the ramp and its weight down and we are going to lift it up
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until that angle is right at the point where the box just begins to move.
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Right there and then we can draw our pseudo free body diagram just breaking up MG
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into components parallel and perpendicular to the ramp as shown.
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If we want to find the coefficient of friction we will start by writing Newton’s second law in the x direction.
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Net force in the x direction is going to be all we have that force of static friction - MG sin θ and all of that must be equal to MA x.
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Right at the point where it is just barely beginning to move acceleration is 0
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So therefore, the force of static friction = MG sin θ.
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Let us do the same basic analysis in the y direction.
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The net force in the y direction is going to be our normal force - MG cos θ all of the equals Ma Y.
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But ay= 0 the box is going to spontaneously fly up off the ramp or go through it.
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Therefore, we can write that the normal force is equal to MG cos θ.
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Putting all of this together then remember friction is fun.
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Force of friction equals μ × the normal force that means that μ are coefficient of friction must be the frictional force over the normal force
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which is MG sin θ / MG Cos θ.
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M /MG /G sin θ /cos θ is tan θ.
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If you want the coefficient of friction lifted up until the object just barely begins to slide measure that angle
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and the coefficient of friction is the tan of the angle.
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How slick is that?
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Let us take a look at one more simple problem.
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Jane rides a sled down the slope of angle θ at constant speed V.
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Constant speed right away a= 0.
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Net force =0.
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The term in the coefficient of kinetic friction between the sled and the slope neglect air resistance.
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Let us draw our free body diagram.
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There it is we will call this + Y in that + X direction and there is our object Jane and sled.
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We have the normal force and the plane of the hill.
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We have MG force of gravity down and some force of friction opposing motion.
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From then we can draw our pseudo free body diagram.
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Let us put that over here on the right.
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We have our positive y, positive x, we have our normal force.
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We still have our frictional force and MG we are going to break it in the components.
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We have MG sin θ parallel to the ramp and MG Cos θ perpendicular to it.
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We will go to Newton’s second law.
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Let us look in the x direction first F net x is equal MG sin θ - Force of friction and
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all of that is equal to Ma X which is equal to 0 because it is a constant speed.
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Therefore, force of friction equals MG sin θ.
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If we looked in the y direction net force in the y direction equals N - MG Cos θ equals 0.
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Therefore, N equals MG Cos θ.
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The coefficient of friction that the frictional force over the normal force or MG sin θ over MG Cos θ which is just tan θ.
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Almost the same thing you just get did, why is that?
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Before when it is just barely beginning to move on that previous problem define the coefficient of friction
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you have a net force of 0 that was right at the point where everything still balanced.
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Same idea here, you are a constant speed net forces 0 it is the same analysis just from a slightly different perspective.
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This is really an angle of repose question because it is moving constant speed.
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Alright, hopefully, that gets you a little more confident and feeling good about these ramps and incline problems.
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Thank you so much for watching www.educator.com.
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We will see you soon and make it a great day everybody.