WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I am Dan Fullerton and in this lesson we are going to talk about Newton’s s and third laws of motion.
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Our objective is to understand the relation between the force that acts on an object and the objects acceleration.
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Perhaps, the most important part of the entire course.
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Write down and utilize vector equations resulting from the applications of Newton's second law.
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Applying it in third law and analyzing the forces between two objects.
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Solve problems in which application of Newton's laws leads to simultaneous linear equations.
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Let us start off by talking about Newton's s law of motion that is so useful in the course.
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It says the acceleration of an object is in the direction of indirectly proportional to the net force applied and inversely proportional to the objects mass.
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It is valid only in inertial reference frames.
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However, for the purposes of this course we are assuming everything we deal with is an inertial reference frame.
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The way we write it the net force or if you prefer the sum of all forces on an object is equal the mass × acceleration that = ma.
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How do we apply it?
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We start off with free body diagrams like we talk in our last lesson.
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Then, for any forces that do not line up with the x or y axis we break those in the components that do lie on the x or y axis with those pseudo free body diagrams.
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Then, we write expressions for the net force in the x and y directions and set the net force = mass × acceleration since Newton's law s law tells us f = ma.
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Finally, solve the resulting equations.
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We will do lots of examples.
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Let us start off with a block on the surface.
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Two forces f1 to the left and f2 to the right are applied to a block on the frictionless horizontal surface as shown.
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If the magnitude of the blocks acceleration is 2 m/s² find the mass of the block.
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Since you get a bigger force to the left let us call to the left our x direction here.
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If we look our net force in the x direction we have 12 in the positive direction - we have 2 in the negative x direction = mass × the acceleration in the x direction.
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12 -2 is 10 N must equal our mass × acceleration 2 m/s².
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Therefore, our mass must be 10 N / 2 m/s² or 5 kg.
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Nice and straightforward.
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How about if we have some concurrent forces?
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A 25 N horizontal force northward in the 35 N horizontal for southward at concurrently at a 15 kg object on a frictionless surface.
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What is the magnitude of the objects acceleration?
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Let us draw that.
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We have got 25 N to the north, we have 35 N that is our object.
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35 N south what is the magnitude of the objects acceleration?
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Why do not we call it down or south our y direction since we can see that is the bigger force.
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I am going to write that the net force in the y direction is 35 N -25 N because that points in the -y direction and
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that must be equal to mass × acceleration in the y direction or 10 N = 15 kg mass × acceleration.
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Or acceleration then is equal to 10 N / 15 kg which is 0.67 m/s².
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That is Newton’s second law.
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As we talk about Newton’s second law an opportune time to talk about mass vs. weight.
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If you recall mass is a measurement of an object’s inertia.
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It tells you how much stuff something is made up of.
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It is a constant.
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Your mass should be the same unless somebody comes over and starts chopping off limbs and other things that are unpleasant.
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Your mass is relatively constant.
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Weight is the force of gravity on an object.
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Weight which we write as mg can vary with gravitational field strength.
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Your weight on earth would be different than your weight on the moon.
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Your mass however would be the same.
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Let us do an example with mass vs. Weight.
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An astronaut weighs 1000 N on earth what is the weight of the astronaut on planet X where the gravitational field strength mg is 6 m/s².
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Let us figure out the astronauts mass first.
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The weight of the astronaut mg is 1000 N and therefore mass must be 1000 N / 10 m/s² here on earth or 100 kg.
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On this planet X, however, the weight is m × gravitational field strength.
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Let us call that g sub x for g on planet X which is 100 kg mass does not change × 6 m/s² or 600 N.
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An alien on planet X weighs 400 N what is the mass of the alien?
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The weight of the alien mg x is 400 N therefore the mass is 400 N ÷ gx which is 6 m/s² or 66.7 kg.
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What does this have to do with Newton’s second law?
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Here is the rule.
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We have been writing Newton’s second law f = MA.
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We have been writing the force of gravity the weight of an object as mg.
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Force of gravity is just a specific type of force is mass × we called g the acceleration due to gravity.
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Writing weight is mg is just an interpretation of Newton’s second law.
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To shorthand that and we have been using the same step.
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Already taking into account that we understand Newton’s second law.
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Let us look at a translational equilibrium problem.
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In the diagram a 20 N force due north and the 29 N due east act concurrently on an object in the same place of the same time.
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What additional force is required to bring the object and equilibrium?
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We need to have one force that is going to balance these two.
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The way I do this is the first thing I'm going to do is figure out what the resultant of these two are.
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Let us add them together to see what their net force would be.
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If we got 20 N to the right and 20 N up I am going to line those up tip to tail so I can add them.
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I am going to take this 20 N north and I'm going to move it here to the right to remember
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we can slide vectors as long as we do not change their direction or length.
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And then to find the resultant I will draw a line from the starting point in the first to the ending point of the last.
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And I can use the Pythagorean Theorem to figure out its magnitude.
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That will be the √20² + 20² which is about 28 N.
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Right now we have the equivalent of a 28 N force NE 45°.
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What force would be required to bring that in the equilibrium?
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We have the exact opposite force.
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We will start a force at the same point and have it go 28 N and down into the left.
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What we call this force the force that required to bring an object in the equilibrium is known as the equilibrant.
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Our answer would be 28 N southwest.
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Another translational equilibrium problem.
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A 3 N force in the 4 N force acting concurrently on a point.
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Which force could not produce equilibrium with these forces?
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Now let us see.
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The 1 N force produce equilibrium with these.
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We have a 3 N force, we have a 4 N force and have it go to the left.
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It looks like the addition of a 1 N force could completely balance those out.
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We get back to our starting point.
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1 would work that could produce equilibrium.
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I think 7 could work too.
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Let us take a look if we have 3 N and now we line up the 4 N force there.
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A 7 N force could bring this back to our starting point 7 N in the equilibrant that cannot be the answer.
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However, there is no way that we can have a 3 N force if 1 N force and no matter what we do with the 9 N force,
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no matter where we put it there is no way we get back to where we start where we get no net force.
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It is going to be 9 N force, that is not going to work.
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You may be wondering how you would do that with the 4 N force?
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That would work 3 N probably something like that.
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4 N and you can make a triangle out of it and you can get back to where you started.
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But 9 N cannot produce equilibrium with a 3 and 4 N force.
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A way you can test that is add up the 2 forces 3 + 4 = 7 and subtract the two forces.
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4 -3 is 1.
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Only forces that are between 1 and 7 could produce equilibrium with that.
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Anything outside of that range not going to work.
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Problem here we have got a 15 kg wagon pulled to the right across a surface by a tension of 100 N an angle of 30° above the horizontal.
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A frictional force of 20 N to the left axis simultaneously.
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What is the acceleration of the wagon?
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Let us start off by drawing a free body diagram.
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Let us get back to those.
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Especially as our problems start to get a little more complex.
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And as we look here it is pulled across the surface by a tension of 100 N and an angle of 30° above the horizontal.
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There is 100 N an angle of 30°.
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A frictional force of 20 N x to the left.
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There is our frictional force.
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What is the acceleration of the wagon?
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Let us do our next step and break that 100 N force up into its components with a pseudo free body diagram.
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We will draw our axis again.
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We still have force of friction to the left the x component of our 100 N is going to be 100 cos 30.
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That is going to be 100 cos 30 is 86.6 N.
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In the y direction, we are going to have 100 sin 30 the y component which would be 50 N.
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We can go to our Newton’s second law equation net force in the x direction is equal to
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we have 86.6 N to the right - our force of friction which it tells us is 20 N to the left
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and all of that must be equal to our mass × acceleration in the x direction.
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Acceleration in the x is 86.6 -20 or 66.6 N ÷ our mass 15 kg which is about 4.44 m/s².
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Alright, let us take a look at baseball problem.
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A 0.15 kg baseball moving at 20 m/s is stopped by a player in 0.01 s.
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What is the average force stopping the ball?
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We are going to learn some other ways we can solve this two here in a few lessons.
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But for now let us start by finding its acceleration.
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Its acceleration is its change in velocity ÷ time which is its final velocity - its initial velocity ÷ time.
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We can use our kinematic equations because it is a constant acceleration.
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It is going to be 0 - 20 m/s / 0.01 s or -2000 m/s².
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What is the negative mean?
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We are calling its initial velocity positive so this means it is in the opposite direction of its initial velocity.
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It is slowing down.
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If we want to know the force, the force is going to be mass × acceleration with Newton’s second law
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is going to be our mass 0.15 kg × acceleration -2000 m/s² or -300 N.
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And again, all that negative means it is in the direction opposite what we called positive the direction the ball is going initially.
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The force in the acceleration are always going to be in the same direction so they are both negative by Newton’s second law.
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Let us take a look at some steel beams.
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A steel beam of mass 100 kg is attached by cable to a 200 kg beam above it.
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The two were raised upward with an acceleration of 1 m/s² by another cable attached to the top beam.
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If we assume the cables are mass less find the tension in each cable.
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Let us start.
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We will call this a y direction and we will analyze beam 1 first.
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Here is our free body diagram for beam 1.
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We have tension 1 pulling up on it, we have tension 2 pulling down on it and we have its mass m1 g down.
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There is beam 1.
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We did the same thing with beam 2.
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We will draw a dot for our object.
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We have T2 pulling it up and only its weight m2 g pulling it down.
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Let us take a look at applying Newton’s second law here.
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For beam 1 write the net force in the y direction is equal to we have T1 - T2 – m1 g and all of that =m1 a or we could write that T1 = T2 + m1 g + m1 a.
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Let us go down and will start here on number 2.
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Net force in the y direction for 2 is going to be T2 – m2 g and that has to equal m2 a, Newton’s second law or T2 is going to be equal to m2 × g + a.
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Since we know those we can solve right away that is going to be equal to m2 which is 100 kg × g10 + a1 or 11 m/s².
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T2 = 1100 N.
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We can come back to beam 1 where T1 = T2 + m1 + m1 a
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T1 = T2 1100 N + we have m1 g 200 × 10 + m1 a 200 × 1
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So T1 therefore = 2000 + 200 + 1100 is 3300 N.
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There are the tensions in our cables.
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Let us take a look at another one here.
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The system below is accelerated by applying a tension T1 to the right most cable,
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Assuming the system is frictionless determine the tension T2 in terms of T1.
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You can do this in a very detailed way or we can use a little bit of systems thinking to try and make it nice and simple.
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The first thing I'm going to look at is I’m going to take a look and say you know what if all of these we consider one object for the moment.
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We have T1 applying a force on that entire object.
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We could write that T1 the net force on that big object is the total mass 7 × the acceleration a or a = T1 / 7.
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If we come over here and we look at just of this block is an object is a 2 kg block there is our free body diagram.
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We got a normal force, its weight, I will call this m2 I suppose and T2 acting on it.
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The only unbalanced force is going to be T2.
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Normal force and its weight have to exactly balance otherwise it spontaneously accelerate up off the table or go through it neither of which happen.
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Looking in the x direction at f net x = MA x or T2 = m2 which is 2 kg × ax
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If T2 = 2a and a= T1 / 7 T2 must equal 2 × a T1 / 7
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T2 = 2 / 7 T1.
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Taking a look at some banked curves.
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Curves can be banked at just the right angle that vehicles traveling a specific speed can stay on the road with no friction required.
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It is very important if you live somewhere like I do where there is lots of ice and the snow on the roads in the winter.
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Given a radius for the curve in a specific velocity at what angle should the bank of the curve be built.
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Let us start by drawing a free body diagram for our system.
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There is our y, there is our x.
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Our object are car even though it is going to be on an incline of some sort, is not moving in the direction that inclined.
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We are not going to tilt our axis here.
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Before we draw our object let us say we put our angle of our ramp.
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Let us make it really steep and dramatic here.
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Let us have it look something like that.
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There is the angle of our road just as a reference.
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We will call this angle θ which means that if we draw on normal to it, a normal force on our car this would also be θ.
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Here is our normal force.
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Here is the weight of our object of our car mg.
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Or if we want to do our pseudo free body diagram let us do that right beside it.
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If we break that up in the components we still have mg straight down.
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The normal force if we want the horizontal component that is going to be the adjacent.
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As I look at this angle the x component is going to be the opposite side in this case.
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That is going to be n sin θ here and the vertical piece is a side that right beside the angle that adjacent to it.
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This piece is going to be n cos θ.
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Let us write our Newton’s second law equations.
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Net force we will look at the x direction first is also going to be what is causing the centripetal force as it goes around a circle.
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That is what is causing the centripetal force to allow it to move in a circle to go around a curve of some
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radius is going to be the horizontal component of the normal force n sin θ.
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F net x which is also the net centripetal force is n sin θ which is ma by Newton’s second law.
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Since it is a centripetal acceleration moving in a curve, in a circle, we could write that as mv² / r.
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We will call that equation 1.
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Analyzing in the y direction, net force in the y direction we have n cos θ - mg and
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that has to be equal to 0 because there is no acceleration vertically.
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Therefore we can say that n must be equal to mg / cos θ.
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We will call that our second equation.
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Now to start to solve this let us see if we can combine equations 1 and 2 to write n sin θ = mv² / r we will start with 1.
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But we now know that n = mg / cos θ.
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This implies then that we still have a sin θ.
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We have an mg from our n up here ÷ cos θ must be equal to mv²/ r.
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On the left inside there, sin/ cos is the tan function so we have mg tan θ = mv² / r.
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I can divide a mass of both sides to say that g tan θ = V² / r.
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Or θ if we want the angle is going to be the inverse tan of V² / gr.
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If we know the speed we want the vehicles to travel on that curve.
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We know the radius of the curve.
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We can determine exactly what angle we should build that curve at so that the cars do not require a whole lot of friction on the road.
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Taking a look at tension in cords.
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Determine the tensions T1 and T2 in cables holding a 10 N weight stationary.
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As we look at this first thing I think I am going to do again is draw a free body diagram for my object.
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We will start to draw our axis here and I will draw the one for the pseudo free body diagram too
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because I can already see am going to have a force and angle.
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We will draw our x, that should do.
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Starting with the free body diagram there is our object.
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We have its weight which is given to us.
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It is not telling us it is mass when it says it is 10 N.
00:24:53.900 --> 00:24:57.300
That is its weight so we have 10 N down.
00:24:57.300 --> 00:25:04.700
We have tension T1 strings cords can only pull so that must be our T1.
00:25:04.700 --> 00:25:12.500
We have a T2 an angle of 60°.
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I'm going to do my pseudo free body diagram breaking up T2 into its components.
00:25:17.300 --> 00:25:23.000
I still have 10 N down.
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T1 to the left.
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The x component of T2 is going to be T2 cos 60°.
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We have T2 cos 60° here in the y component is going to be T2 sin 60°.
00:25:42.400 --> 00:25:45.400
I can start writing my Newton’s second law equations.
00:25:45.400 --> 00:25:47.200
We will start with the x.
00:25:47.200 --> 00:26:03.500
Net force in the x direction is going to be we have T2 cos 60 - T1 and all of that is going to have to be equal to 0.
00:26:03.500 --> 00:26:16.800
There is no acceleration it is sitting there in equilibrium which implies then that T1 is going to be equal to T2 cos 60°.
00:26:16.800 --> 00:26:20.200
Let us take a look at the y now.
00:26:20.200 --> 00:26:34.200
Net force in the y direction is going to be T2 sin 60° -10 N.
00:26:34.200 --> 00:26:37.700
Again, that has to be equal to 0.
00:26:37.700 --> 00:26:47.800
Therefore, T2 must equal 10 Newton’s / sin 60° which is 11.5 N.
00:26:47.800 --> 00:26:51.000
There is T2.
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I can put that into my equation for T1.
00:26:54.700 --> 00:27:08.000
Therefore T1 = 11.5 cos 60° or 5.77 N.
00:27:08.000 --> 00:27:12.900
We found the tension in our two cables there.
00:27:12.900 --> 00:27:16.800
Looking at this from a graphical perspective we will take an example where the velocity of
00:27:16.800 --> 00:27:20.800
an airplane in flight is shown as a function of time T in the graph.
00:27:20.800 --> 00:27:22.800
Velocity time.
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At which time is the force exerted by the plane's engines on the plane the greatest?
00:27:28.800 --> 00:27:34.800
The way I think about this one is we are going to have the greatest force when we have the greatest acceleration.
00:27:34.800 --> 00:27:44.700
We flee on the big force we need a big acceleration and since acceleration I should say force is proportional to acceleration.
00:27:44.700 --> 00:27:50.100
Acceleration is the slope or the derivative of the velocity time graph.
00:27:50.100 --> 00:27:52.100
Where do we have the biggest slope?
00:27:52.100 --> 00:28:02.800
It is probably occurring somewhere right around there which if I draw that down is right around 4.5 s.
00:28:02.800 --> 00:28:14.700
We have the greatest slope at around 4.5 s that must be the time that which we have the greatest force.
00:28:14.700 --> 00:28:21.400
Given the graph of velocity vs time for the particle shown below where V is proportional to T²
00:28:21.400 --> 00:28:27.400
sketch a graph of the net force exerted on the object as a function of time.
00:28:27.400 --> 00:28:32.500
If V is changing that way acceleration is the derivative of velocity.
00:28:32.500 --> 00:28:40.300
We start off with 0 slope if I wanted to put acceleration on the same graph in red we start off at 0 acceleration.
00:28:40.300 --> 00:28:46.500
As we go along we are going to have higher and higher, larger and larger amounts of acceleration.
00:28:46.500 --> 00:28:56.400
It looks like our acceleration graph is probably going to be linear something of this shape.
00:28:56.400 --> 00:29:05.700
If we have that shape for acceleration we have to have the same shape for our force time graph since force = mass × acceleration.
00:29:05.700 --> 00:29:16.900
So I would draw our force graph with the shape like that when you are increasing up to the right.
00:29:16.900 --> 00:29:20.400
Newton’s third law this is a pretty cool one as well.
00:29:20.400 --> 00:29:25.100
All forces come in pairs, you never have just a single force, they are always pairs.
00:29:25.100 --> 00:29:35.200
If object 1 exerts a force on object 2, object 2 exerts a force of equal magnitude but opposite in direction back on object 1.
00:29:35.200 --> 00:29:45.000
Or if you wrote it mathematically the force of object 1 on object 2 is equal in magnitude but opposite in direction to the force of 2 on 1.
00:29:45.000 --> 00:29:50.600
I should note that those are vectors of course, Newton’s third law.
00:29:50.600 --> 00:30:01.800
If I punch you in the nose with a force of 100 N, your nose punches my fist with a force of 100 N back in the opposite direction.
00:30:01.800 --> 00:30:03.600
Alright, some examples.
00:30:03.600 --> 00:30:09.700
How does a cat run forward if the cat runs forward does not it push backwards on the ground?
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It is the ground actually propels the cat forward?
00:30:13.400 --> 00:30:16.900
Or if you want to swim forward which way do you push the water?
00:30:16.900 --> 00:30:25.200
You push backward on the water applying a force on the water and the water’s reactionary force by Newton’s third law is what pushes you forward.
00:30:25.200 --> 00:30:28.200
Or if you want to jump in the air which way do you push with your legs?
00:30:28.200 --> 00:30:32.300
Do not you push down on the ground as hard as you can as quickly as you can?
00:30:32.300 --> 00:30:42.800
You push down on the ground it is the opposite force of the ground pushing up on you that propels you up into the air.
00:30:42.800 --> 00:30:47.400
When we have these two forces from different objects we call those action reaction pairs.
00:30:47.400 --> 00:30:51.600
If a girl is kicking a soccer ball let us identify some of those pairs.
00:30:51.600 --> 00:31:06.000
We have the force of the girl’s foot on the ball and the ball is actually applying a force back on the girl’s foot or a rocket ship in space.
00:31:06.000 --> 00:31:14.000
The rocket ship is propelled by pushing gases out.
00:31:14.000 --> 00:31:23.700
The reactionary force is that those gases push the rocket causing the acceleration.
00:31:23.700 --> 00:31:27.100
How about gravity on you? gravity on me?
00:31:27.100 --> 00:31:34.600
Right now the Earth's gravity is pulling me down.
00:31:34.600 --> 00:31:44.100
Guess what though, I'm pulling the earth upward with the exact same force.
00:31:44.100 --> 00:31:46.200
How come I have such a more dramatic effect?
00:31:46.200 --> 00:31:53.100
If I jump off a cliff, I'm being pulled down toward the earth with the same force I'm pulling the earth up toward me,
00:31:53.100 --> 00:31:56.300
why does it look like it is going to hurt the earth nearly as much as me?
00:31:56.300 --> 00:32:01.800
I'm going to accelerate a lot more because for the same force I have a much smaller mass.
00:32:01.800 --> 00:32:13.200
That same force applied to the Earth's huge mass gives a very little almost an immeasurable acceleration.
00:32:13.200 --> 00:32:17.600
Earth's mass is approximately 81 × the mass of the moon.
00:32:17.600 --> 00:32:24.500
If Earth exerts a gravitational force of magnitude f on the moon, the magnitude of the gravitational force of the moon on Earth is.
00:32:24.500 --> 00:32:27.900
It is a straightforward application of Newton’s third law.
00:32:27.900 --> 00:32:33.300
If the Earth is pulling on the moon with force f, the moon must pull on the earth with force f.
00:32:33.300 --> 00:32:38.200
Forces come in pairs that are equal and opposite.
00:32:38.200 --> 00:32:40.700
Or the sailboat question.
00:32:40.700 --> 00:32:49.700
A 400 little girl standing on a dock exerts a force of 100 N on the 10,000 N sailboat as she pushes it away from the dock.
00:32:49.700 --> 00:32:53.100
How much force does the sailboat exert on the girl?
00:32:53.100 --> 00:32:59.800
In order to know the force the sailboat exerts on the girl we need to find the force that the girl exert on the sailboat.
00:32:59.800 --> 00:33:07.300
She exerts a force of 100 N on the sailboat, the sailboat must have that exact same magnitude of force back on her.
00:33:07.300 --> 00:33:17.400
This 10,000 N here that is just describing the weight of the sailboat not talking about the interaction of the girl and the sailboat.
00:33:17.400 --> 00:33:20.900
Alright and another one, hammer and nail.
00:33:20.900 --> 00:33:26.300
A carpenter hits the nail with a hammer, compare to the magnitude of the force the hammer exerts on the nail.
00:33:26.300 --> 00:33:35.000
The magnitude of the force the nail inserts on the hammer during contact is the exact same.
00:33:35.000 --> 00:33:39.200
Why does the nail get propelled forward and the hammer not so much different?
00:33:39.200 --> 00:33:46.100
The nail has a much smaller mass.
00:33:46.100 --> 00:33:47.700
And our last question here.
00:33:47.700 --> 00:33:54.400
If force is only come in pairs that are equal and opposite why do not all forces cancel each other?
00:33:54.400 --> 00:34:02.400
The trick is these forces act on different objects.
00:34:02.400 --> 00:34:07.700
They all act on the same object they would cancel out but if they do not they are acting on different objects.
00:34:07.700 --> 00:34:13.100
Think about it in terms of free body diagram to be doing a free body diagram for each of these objects.
00:34:13.100 --> 00:34:24.900
You are not going to have a free body diagram that has the initial force in the reactionary force or the paired force all on the same diagram.
00:34:24.900 --> 00:34:30.400
You may have heard Newton’s third law stated as for every action there is an equal and opposite reaction.
00:34:30.400 --> 00:34:34.800
Which is something of a restatement of the law but it also can be somewhat confusing
00:34:34.800 --> 00:34:37.800
because you are not really talking about what actions and reactions are.
00:34:37.800 --> 00:34:47.800
Instead, Newton’s third law the force of object 1 and object 2 is equal in magnitude and opposite direction to the force of object 2 on object 1.
00:34:47.800 --> 00:34:52.900
Forces only come in pairs, you can never have a singular force.
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Thank you for watching www.educator.com.
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We will see you again real soon and make it a great day everybody.