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 friction.
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Our objectives include defining and identifying frictional forces.
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Explaining the factors that determine the amount of friction between two surfaces and
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determining the frictional force and the coefficient of friction between two surfaces.
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Let us start off by talking about the coefficient of friction.
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The ratio of the frictional force to the normal force provides the coefficient of friction.
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μ is the coefficient of friction.
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It is unit less and it is a force of friction divided by the normal force.
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Friction caused by the interaction of two objects and this coefficient of friction depends on the nature of the surfaces.
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Some approximate coefficients of friction for different types of friction.
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Kinetic friction is objects that are sliding against each other.
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Static friction is objects that are not sliding against each other.
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Rub around dry concrete for example has a kinetic coefficient of 0.68 and a static coefficient of 0.9.
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That means once it is moving or once it is sliding, it has less friction than it does and start trying to get it to start sliding.
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It is pretty common to see a static coefficient of friction larger than the kinetic coefficient of friction.
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Let us take a look at some situations.
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Are these situations static or kinetic?
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A sled sliding down a snowy hill that would be kinetic because we have objects that are sliding.
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How about a refrigerator at rest that you want to move?
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That is static because it is not sliding.
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That is why it takes more force to get something to start sliding to overcome static friction.
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Once you get it sliding you are dealing with kinetic friction which is less.
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A car with tires rolling freely, that is static because it is not sliding.
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If the tires are going down off the pavement in any given point, they are static with respect to the pavement.
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If you happen to skid across the pavement now you have sliding or kinetic friction.
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When you are using that coefficient of friction you will use either the kinetic (μ K) or the static (μ S) coefficient.
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Let us take the example of a car sliding.
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A car's performance is tested on various horizontal road surfaces.
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The brakes are applied causing the rubber tires of the car to slide along the road without rolling.
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Without rolling they are sliding and right away we are thinking kinetic.
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The tires encounter the greatest force of friction to stop the car on and you have a couple choices here
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which are dry concrete, dry asphalt, wet concrete, or wet asphalt.
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I am going to look for the largest coefficient of friction that would be right here on dry concrete and that must be our answer.
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Largest coefficient of friction, largest amount of frictional force.
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Here we have a block on an incline.
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It is sliding down a plane in angle θ.
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If the angle θ is increased, the coefficient of kinetic friction between the bottom surface of the block and the surface of the incline will?
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That is going to remain the same.
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Remember the coefficient of friction just depends on the nature of the surfaces and that does not change as you adjust the angle.
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How do we calculate the force of friction?
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The force of friction depends only upon the nature of the surfaces in contact, the μ and the magnitude of the normal force (fn).
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Or oftentimes I like to write normal force as just (n) so you can say the force of friction is μ × the normal force.
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Of course friction is fun.
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You can combine this with Newton’s 2nd law in the free body diagrams like we did in our last lesson to solve more involved problems.
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Here we go.
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The diagram below shows a 4kg object accelerating at 10m/s² on a rough, horizontal surface.
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What is the magnitude of the frictional force (ff) acting on the object?
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First thing that I am going to do is draw my free body diagram.
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Let us put our axis over here.
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As I label my forces we have the normal force.
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We have the weight of our object (mg).
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Those must be balanced because it is not accelerating up off the ground or into the ground because that would be goofy.
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We have a 15 N applied force and we have some frictional force which they are labeling (ff).
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Let us take a look.
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Let us apply Newton’s second law in the x direction, the net force in the x direction we will call it to the right positive.
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We have 15 N - the force of friction = mass × acceleration in the x direction which is 4 kg × 10 m/s².
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15 N - the force of friction = 40 N.
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Our force of friction must be 10 N.
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That is easy and straightforward.
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Let us take a look at a box on a wooden surface.
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A horizontal force of 8 N is used to pull a 20 N wooden box.
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A 20 N wooden box that is not its mass but 20 N is its weight.
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It is moving toward the right along a horizontal wood surface and it tells us the coefficient of kinetic friction is 0.3.
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Find the frictional force acting on the box.
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Remember friction is fun which is going to be 0.3 or coefficient of kinetic friction since it is sliding × the normal force or if we have 20 N as its weight,
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the normal force n must exactly balance that and be 20 N or otherwise it will accelerate off of the floor.
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Multiplied by 20 N is going to be 6 N.
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How about the net force acting on the box?
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The net force we got 8 N to the right and we got our frictional force to the left
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and that is going to be 8 N – our frictional force we just found is 6 N.
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Our net must be 2 N to the right.
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The mass of the box, all of its weight (mg) is 20 N.
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The net means its mass must be 20 N over the acceleration due to gravity (G) or 20 N / 10 m/s² is just going to be 2kg.
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Finally is the acceleration of the box.
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Net force in the x direction we said was 2 N and that is equal to (ma).
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A is equal to 2 N over our mass 2kg.
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It is going to accelerate it to 1 m/s² to the right.
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Static vs. Kinetic friction.
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Compared to the force needed to start sliding a crate across a rough level floor, the force needed to keep it sliding once it is moving.
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Well that is going to be less.
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You probably know that from practical experience but also our coefficient of kinetic friction is less than our coefficient of static friction.
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Drag forces.
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An airplane is moving with the constant velocity in a level flight.
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Compare the magnitude of the forward force provided by the engines to the magnitude of the backward frictional drag force.
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There is a key term in this question or one highlight, constant velocity.
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Whenever you say constant velocity, right away you should be thinking acceleration is 0 and the net force = 0.
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The forward force of the engines must exactly balance the frictional drag force.
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Frictional drag force they have to be equal otherwise it would be accelerating so they must be the same.
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Let us do a sled problem.
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Susie pulls the handle of a 20kg sled across the yard with the force of 100 N as shown.
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The yard exerts a frictional force of 25 N on the sled.
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We are asked to find the coefficient of friction and determine the distance the sled travels.
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If it starts from rest and Susie maintains her 100 N force for 5 seconds.
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Let us start by drawing our free body diagram.
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I am just going to sketch them up here.
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We have an applied force which is 100 N that is an angle of 30°.
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We have the normal force from the ground.
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We have the weight down.
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We have our frictional force (f) is 25 N.
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If we wanted to we could draw our pseudo free body diagram.
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We are going to do that down here in the bottom left.
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Let us see.
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Let us put it maybe right there.
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We still have our normal force up.
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We have our weight down.
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We have our frictional force to the left.
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We got to breakup our 100 N force and angle of 30° in the x and y components.
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The x component of that is going to be 100 cos 30.
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I will draw that in here 100 cos 30.
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The y component is 100 sin 30°.
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We can start our solutions.
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Let us start by looking at the net force in the y direction to find the coefficient of friction.
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Net force in the y direction is going to be (MA y) which we know is going to be 0.
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The net force in the y we got 100 sin 30 + our normal force - mg = 0.
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The normal force must equal mg -100 sin 30.
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Our normal force (mg) that is just going to be our 20kg 10 m/s² - 100 sin 30 that is just going to be 50.
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If 200 - 50 = 150 N for our normal force.
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Our force of friction then (μ) is our coefficient of friction is our frictional force divided by normal force.
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Our frictional force it gives us 25 N and our normal force we just said was 150 N.
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Our coefficient of friction must be 0.167.
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We want to see how far it travels given this information and that looks like a kinematics problem but I think first I need to find the acceleration.
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Let us take a look at the net force in the x direction that is going to be 100 cos 30 that is 86 .6 N - our frictional force which we just said was 25 N
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and that is going to be 86.6 -25 that 61.6 N.
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Our acceleration is our net force divided by the mass or 61.6 N / 20kg is about 3.1 m/s².
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I know our acceleration and our initial velocity and I know the time.
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I can use my kinematic equations now to find how far it goes.
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Δ x = V initial t + ½ (at)².
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It is a horizontal problem.
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Our initial velocity is 0 so this is just ½ × our acceleration 3.1 m/s² × our time 5s² or about 38.8 meters.
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Let us finish off by looking in an old AP free response problem.
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What we are going to do is have you go to this link to find the 2007 released mechanics exam free response number 1.
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I recommend you go to the site, look it over, print it out, even try it on your own first while you hit pause on the video and see how you do.
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Then come back here hit play on the video and check your answers as you go.
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If you get stuck use my solution to help get you going again through to the problem.
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I have that here the 2007 mechanics free response 1.
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We are given the box and we are applying a force at some angle.
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We are asked on the figure to draw a free body diagram showing all the forces on the block.
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We will start with the block and we should be pretty good at this by now.
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Our forces, we have the normal force.
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We have its weight.
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We have this force (f1).
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And we have our frictional force (ff).
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They are forces.
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Part B, just derive an expression for the normal force exerted by the surface on the block.
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If we want the normal force and I am going to look in the y direction with Newton’s 2nd law.
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Net force in the y direction must be n + we got to use the y component of (f1) that is going to be f1 sin our angle θ.
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And down we have - mg and all of that has to equal 0.
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Therefore, if I solve for the normal force that is going to be mg - f1 sin θ.
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Part C, derive an expression for the coefficient of kinetic friction between the block and the surface.
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To do that, I need to find first the force of friction.
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We will go to Newton’s 2nd law in the x direction.
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My x is going to be f1 cos θ - our force of friction.
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And because we know force of friction is μ × the normal force, we can write that as f1 cos θ - μ × our normal force = ma1 our acceleration.
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Therefore μ × the normal force must equal f1 cos θ –ma1 which implies then that μ must equal f1 cos θ –ma1/ the normal force.
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We already determined that the normal force is mg - f1 sin θ so this implies then that μ must equal (f1 cos θ - ma1) ÷ mg - f1 sin θ.
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Let see we now have got the part D.
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Let us go to the next page to give ourselves some room.
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It looks like we got some graphs to make.
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Let us see if we can draw these in here first.
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For part D, we are going to have a couple of graphs where we have a velocity graph and a position graph.
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For part D, we have velocity vs. time and we have x vs. time displacement.
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If we want to graph these from what we already know we must have a constantly increasing velocity because our object is accelerating.
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The T graph of that if we start from rest and tells us we do must look kind of like that.
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In our position time graph as it says, it starts at position 0 must have that shape.
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There are our graphs for part D.
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Part E, if the applied force is large enough the block will lose contact with the surface.
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Derive an expression for the magnitude of the greatest acceleration the block can have and still maintain contact with the ground.
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As we do that one, I am going to realize that when that first breaks contact with the ground, right at that point the normal force is 0.
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The ground is no longer pushing up on it and that is going to be my condition that is going to help me figure out where this occurs.
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If the normal force is 0 that implies them that f1 sin θ = mg or f1 is going to be equal to mg / sin (θ).
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We also know from our f net equation that force in the x direction is going to be f1 cos θ = ma.
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There is no friction because there is no normal force.
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Acceleration = f1 cos θ / m.
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We will call that equation 2 then equation 1 and as we combine them 1 and 2,
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that means that our acceleration (A) which is f1 cos θ / m must be equal to how we replace f1 with mg / sin (θ).
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We have mg / sin (θ).
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We still have a cos (θ) here as well as an (m).
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We can do a couple of simplifications m / m makes a ratio of 1.
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Sin / cos is tan (θ).
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I would write this as g / tan (θ).
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There is our acceleration right when it starts to leave ground.
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Right at the point where it is just barely touching the ground g / 10 θ.
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Hopefully that gets you a good start on friction.
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Thank you for watching educator.com.
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I look forward to seeing you again soon. Make it a great day everyone.