PK {iA@ @ + Problem_Set_3/01-The_Feather_Problem.en.srt1
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Let's start this problem by talking about the feather problem.
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Remember back in unit 2 that Aristotle said that "objects fall at a constant speed,"
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and he said that speed is proportional to the mass.
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Galileo on the other hand said that "all objects fall with constant acceleration."
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We've seem to have taken Galileo's side in this debate, but is that the right thing to do?
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After all the feather or a sheet of paper seems to fall down and what seems to be constant rate.
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Let's consider something called air resistance.
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Now, if you're driving in a car on a highway and you stick your hand out the window,
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you don't need to have a ridiculously long arm but it helps.
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You can feel this force of air resistance, and this is the force of a quickly moving object.
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For example, your hand feeling the actual individual molecules of air as they slam up against your hand,
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and that provides an actual force and its always opposite the direction of motion.
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So, if the car is going this way the force is that way.
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The size of the air resistance force is more or less approximated by an equation like this.
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Here v is the speed your traveling and b is some number that depends on
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maybe the size and shape of your hand or the feather.
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As you can see, this force goes bigger as your velocity increases.
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So, let's see how this feather and it's being pulled down by gravity.
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As its speed increases, the air resistance force it feels will also increase.
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What eventually happens is the feather reaches what's called terminal velocity
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and that is the maximum speed that the feather reaches.
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So, let's say the feather with mass 0.001 Kg so it would be 1 gram
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and this b, this constant up here of 0.04.
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I'm not going to tell you the units, but I bet you can tell me what they are on the form.
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I want you to tell me knowing what you know about motion at a constant speed.
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What will be the feather's terminal velocity in m/sec.
PK {iA0/b 4 Problem_Set_3/02-The_Feather_Problem_Solution.en.srt1
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When we know the constant speed is the same as being at rest from the point of view of forces,
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which means the air resistance force must equal to weight then they balance.
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When I carry out this Algebra, I find the terminal velocity or the maximum speed
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that this feather will reach is 0.5 m/s².
PK |iAK; , Problem_Set_3/03-Elevator_Weight_Loss.en.srt1
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Here's a man in elevator. This man wants to lose weight.
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And he is very happy because he thinks he has found a pretty clever way to lose his weight.
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This man is standing on a scale, and as a quick reminder,
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let's talk about the forces acting on this man when the elevator is completely still.
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We can represent him as a box of mass m. Gravity is pulling down with the strength equal to mg.
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And a normal force is pushing up with the exact same strength, because he is motionless.
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There is no acceleration. Now remember, the scale doesn't read mg.
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It doesn't tell you exactly what your weight is.
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It tells you how hard it has to push to keep you from falling through the scale.
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So this man's realized that if he can just reduce Fn, then he can reduce his perceived weight.
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Of course, this will lead to unbalanced forces meaning he'll have to accelerate downwards.
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In order to accelerate downwards, he's going to need an elevator.
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This man finds that when he is at rest in the elevator, the scale says 800 N.
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He wants the scale to say 600 N. Now, I have two questions for you.
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First, what must his downward acceleration be
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to actually see this target weight of 600 N on the scale?
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Next, if he travels at acceleration for 5 seconds
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and assuming the elevator started at rest, how far will he travel?
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And this is more of a review of unit 2. Enter your answers here and here.
PK }iA 5 Problem_Set_3/04-Elevator_Weight_Loss_Solution.en.srt1
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If his measured weight is 800 N at rest, that must be his true weight,
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because at rest, the normal force is equal to the weight and so both are equal to 800 N.
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If his target weight is 600 N, that's what he wants the scale to read.
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That's how hard he wants the scale to be pushing up on him that should be the normal force.
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Well, now we can do F=ma.
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For F, we'll plug in where 800 is 200 bigger than 600, so we have an unbalanced force of 200 N.
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M is going to be 80 and that's because this 800 N. That's equal to mg and g is 10.
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If I divide both sides by 10, m is 80, so if 200=80a. a=2.5 m/s².
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And how far will this elevator travel in 5 seconds if this is its acceleration?
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Well, we can use this equation, which we know from unit 2.
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We assumed it started from rest as soon as your velocity is 0.
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And that gives us a Δy of about 31 meters. Good job.
PK }iA0
RC C , Problem_Set_3/05-Is_g_Really_Constant.en.srt1
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In this problem, we're going to probe the acceleration due to gravity.
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Up until now, we've been saying that this acceleration is just 10 m/s² anywhere on the earth.
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We've been assuming it didn't matter if you are here, or here, or here, or there.
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Little g was equal to 10 m/s².
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Now, one thing I have to confess that it's actually closer to 9.81 m/s²
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but that's something we're not going to worry about.
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What I really want to get in this problem is the underlying cause of this acceleration
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and examine any sort of differences in this acceleration due to, well, let's see.
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We know this gravitational acceleration is due to a gravitational force, which we can call FG.
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And we've talked about how that force is proportional to 1/r².
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Now, when we stand on the earth, the r that we're talking about is actually the distance
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between the center of the earth and the center of our bodies, and the distance is about 6,400 km.
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That's the radius of the earth.
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Now, since the force of gravity has this strange dependence, 1/r²,
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and remember, this means if I get twice as far away from the earth,
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so maybe if I were out over here somewhere, the force of gravity would be 1/4 as strong
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because 1/4 is 1/2².
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Now, we know that there's this strange dependence on the force of gravity, it goes like 1/r².
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So that means if I get twice as far away, the force gets actually four times weaker.
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Now, remember it's this force of gravity which is causing this acceleration due to gravity.
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And since force is proportional to acceleration,
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well that means that acceleration like gravity is proportional to 1/r².
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By what percentage does the acceleration due to gravity change when we stand on Mount Everest?
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Now, Mount Everest would be somewhere over here.
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And the height of Mount Everest is about equal to 9 km. This is a tricky question.
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But try to think back about what you know about proportionality to give us an answer.
PK iAEC , Problem_Set_3/06-Is_g_Really_Constant.en.srt1
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If acceleration is proportional to 1/r², I can write an equation like this a₁ /a₂ = (r₁/r₂)⁻²
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and the negative is because this is in the denominator and the 2 is because r².
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Well let's take a₁ to be our normal acceleration due to gravity g, a₂ would be our acceleration
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on the top of the mountain, we call that a₂ still,
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r₁ is our distance from the distance center of the earth when we're at sea level
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so that's 6,400 km and r₂ is 9 bigger than that because we're on the top of Mt. Everest.
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And when you do this math out, you find that the acceleration on the top of Mt. Everest
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is equal to the normal acceleration times 0.997.
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Well that's exactly equal to 99.7% of the normal acceleration due to gravity,
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which means we've decreased by about 0.3% and if you had made this negative, that was okay too.
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So going from sea level to the highest place on earth, the acceleration due to gravity
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only changes by <1%.
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So g really is pretty close to being constant when you're on the surface of the earth.
PK iAUV! ! ' Problem_Set_3/07-Inclined_Planes.en.srt1
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Before we do more problems, let's do a quick lesson on inclined planes
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because they're really a fascinating tool for making nature work for us.
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Now we already know that Galileo had a lot fun rolling balls down inclined planes.
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And if you remember the reason he did this was because it slowed down motion and allowed him
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to make some measurements and observations that he couldn't have made with just objects in free fall.
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The question though is why did rolling balls down an inclined plane slowed down their motion.
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Sure it seems sort of obvious but let's analyze it.
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Instead of rolling balls, let's imagine sliding blocks because the physics of rolling
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is slightly more complicated than I want to get into right now.
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We're zoomed in now, and here's our block and it has mass m.
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And we're going to let it slide down this inclined plane.
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For now, we're pretending this plane has no friction.
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Remember friction is the force associated with things rubbing.
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Well, let's do what we always do and draw a force diagram.
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As always, weight points straight downwards and then normal force
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and remember normal means perpendicular, actually points from this direction
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and that's because it has to be perpendicular.
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We're not allowing for any friction. Friction would point up or down the inclined plane.
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We're only allowing for this normal force.
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We'll up to now what we would have done was take this force
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and break it into vertical and horizontal components.
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It turns out that's not the best way to solve this problem because we know from experience
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that this block is going to slide down the ramp.
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It's not going to slide straight to the right or straight up or straight down.
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It's going to slide in this direction, so that's the direction we should care about.
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And how do we care about a direction?
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Well, instead of breaking this force down, we break the weight down.
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We replace this downwards force with a force perpendicular to the surface,
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which I've indicated with this little perpendicular symbol here and a force parallel,
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which I've indicated with these two parallel lines.
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Now, the thing that's really cool is that if this plane was inclined at an angle α,
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we can do some geometry to show that this angle here inside this triangle is also equal to α.
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And since this here is a right angle, we can do trigonometry.
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In fact, if we want to look at the sine of this angle for example,
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well that's going to equal opposite over adjacent so F parallel over mg and when we solve that
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we find that F parallel is just mg times the sine of this angle.
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I mean, just a similar thinking to show that F perpendicular is mg times the cosine of the angle.
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Now let's think about what's going on here.
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Well I know that a block on a plane like this is going to slide this way.
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It won't jump off the plane, and it won't accelerate down into the plane. Okay.
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So that means that in this direction the forces are all set. They're balanced.
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Fn must equal F perpendicular. In fact, the only unbalanced force here is this parallel force.
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And this parallel force since it's unbalanced is what's causing the acceleration.
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It must be equal to ma. Well we can plug in what we know. F parallel is mg sin θ.
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So this must be true and this tells me that the acceleration of an object
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sliding down an inclined plane is equal to g times the sin of θ.
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Now that's very amazing. This number sine of θ is always between 0 and 1.
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It's always going to make g smaller.
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So this is exactly why Galileo chose inclined planes, they slow down acceleration.
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They make it smaller than g. Now, this was a lot of material.
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If you have any questions, go to the forums and ask.
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Now we're going to move on to a couple more problems as I have promised.
PK iA( - Problem_Set_3/08-San_Francisco_Parking.en.srt1
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Now the Udacity offices are in Palo Alto, CA, which is very close to San Francisco.
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San Francisco is notorious for its hills and the difficulty of parking.
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When you pull up to a spot in San Francisco, you have to pull your emergency brake
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and that locks the tires and prevents them from rotating.
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The force that keeps you from sliding down the hill though is actually what's called the force of friction.
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And that exists between the tires and the road.
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I didn't draw the arrow coming from down there, but that's really where the force exists.
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Now obviously friction can only do so much.
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Obviously if you park your car on a hill that's completely vertical, well, you're in trouble.
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Friction is not going to provide enough force to prevent you from falling down the hill.
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On the other hand, if you park in a pretty gentle hill you should be fine.
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I want to know what's the angle where your car is going to slip.
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There must be one. Let's figure out what it is.
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So let's say for your tires and this road surface
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the force of friction has a maximum value of 80% of the car's weight.
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I want to know what is the steepest angle hill, and this is what angle I'm talking about here,
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that you can safely park on, so if the hill got any steeper you're in trouble when the car falls down.
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Enter your answer here in degrees.
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If you're not sure what to do go back to the previous lesson
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and learn some more about inclined planes.
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And if you're still confused, go to the forums. I understand this is quite a tricky question.
PK iAy2Ƥ 6 Problem_Set_3/09-San_Francisco_Parking_Solution.en.srt1
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Well, let's take a look at the forces acting on this car. Well, we know there's the normal force.
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And we know there's gravity but we don't want to write that as a downwards arrow.
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But rather as a perpendicular arrow and a parallel one.
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And in the last video, we showed that the parallel force is equal to
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the weight of the car in mg times the sin of α.
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Now the steepest angle hill, well that occurs when friction force is totally maxed out.
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When it is at exactly its maximum value, which is 80% of the car's weight
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or put in another way Ff = 0.8 mg.
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Well in this situation where the car is just barely holding on, we know the friction force
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is balancing the parallel force so we can set these two equal to each other.
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Notice something amazing that happens. The m's and g's, the whole weight cancels out.
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This doesn't depend at all on the mass of your vehicle. Amazing!
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Well now, I have to take the inverse sine of 0.8 to solve for the sin of α.
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But still, I take this inverse sine of 0.8 and I get an answer of 53°.
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I'm very impress if you're able to get this on your first try or even your second or third.
PK iA",˄ &