Subjects

Sections | |||
---|---|---|---|

Intro to Forces + Newton's Laws | 16 mins | 0 completed | Learn |

Force Problems with Motion | 28 mins | 0 completed | Learn |

Forces with Multiple Objects | 26 mins | 0 completed | Learn |

Vertical Forces & Equilibrium | 34 mins | 0 completed | Learn |

More 1D Equilibrium | 16 mins | 0 completed | Learn |

Vertical Forces & Acceleration | 26 mins | 0 completed | Learn |

Landing & Jumping Problems | 19 mins | 0 completed | Learn |

Forces in 2D | 35 mins | 0 completed | Learn |

2D Equilibrium | 25 mins | 0 completed | Learn |

Concept #1: Landing & Jumping Problems

**Transcript**

Hey guys so in this video I want to show you a few more problems of acceleration in the Y. axis Let's get started. So I'm going to go over three different examples here the first one I have a five thousand kilograms spaceship so Massy Cause I'm sorry seven thousand that comes to rest at a constant acceleration from eight hundred meters per second down so it was originally moving with eight hundred meters per second so this is my initial velocity. Going down.And it's eventually going to come to a rest over here so v final equals zero right so that's the set up and it does this in 90 seconds so that's the delta T. between these two points. And I want to know.What braking force must that must the rockets provide So if you have a spaceship The idea is that it's moving down but to stop it it has to accelerate up right because my velocity is going from going down to No velocity at all so I have to accelerate up you can always think of acceleration as tugging on the velocity vector so my velocity vector looks like this and as I start tugging up on it it gets smaller and smaller and eventually it's zero take so this is what's happening with my velocity so the acceleration has to be up its absolute velocity because I'm slowing down all right so the forces on the object if you were to draw free body diagram would be mg There's always mg down if you're a rocket right obviously so in an up I have a braking force OK and if I want to compare the suit I want the acceleration to be up so this has to be the stronger force so the braking force has to be greater than the mg so that accelerates up that way so this is my free body diagram right so if I want to find the force I use that for. So some of our forces on a rocket is the mass of the mass of the rocket and the acceleration of the rocket and there are two forces there is fb going up. and there is mg going down, Now I'm looking for Fb So I have to have everything else I have mass I have gravity I have mass i forgot to put in a here but I don't have the acceleration so I have to find a acceleration to be able to find have be I'm stuck on f=ma I'm going to go to my motion equations there's enough information here those I already have one two three pieces of information there's enough information for me to find a K. so the if you look around the variable you're missing here is your Delta Y. So this is my ignore variable and that allows me to figure out which equation to use I'm going to use the first equation to find the acceleration so I am gonna go over here on side and find what the acceleration is. OK and i can use V. equals relational plus eighty the final velocity is zero the initial velocity careful this is going down so it's negative eight hundred. The acceleration is not nine point eight because the rockets are pushing me up it's actually made positive acceleration in the time as ninety seconds so if I move this around I have the acceleration is eight hundred by ninety. And that gives me. Seventy seven point eight. meters per second. Squared pretty intense right no with this acceleration I can just plug it back in here so FB is going to be just ma + mg So Fb is the masses seven thousand the acceleration is seventy seven point eight plus seven thousand in gravity nine point eight OK remember. There's gravity here G. is not going to be negative nine point eight The negative is already here so remember G.'s always 9.8 if you want to make it negative because the acceleration Y. axis happens to be negative G. for freefall you do that by putting negative in front of it not by making the G. negative itself so that's important because when I'm here I need to know not to plug in negative nine point eight for G. but to plug in just nine point eight OK so if you do all of this you get the Fb equals six one four six zero zero Newtons.

In that is the final answer for this one so here i have a two kilograms purse mass equals two kilograms it's dropped from the top of the Leaning Tower of Pisa as matter were dropped from but it would matter is that it drops fifty six meters before hitting the ground so you drop it so we can assume that the initial velocity is zero it falls for a distance of fifty six so Delta y is fifty six but that fifty six is negative because I'm fall I'm going down. And I was always going up it is positive when I was trying to get to the bottom with the final velocity of twenty six that is also negative because it's going down and the idea here is that the acceleration is not gravity it's not nine point eight because there is air resistance and I want to know how much air resistance we have now we haven't formally introduced sort of the full. Look into air resistance but we can still solve this because this is just the basic forces with motion problem right so the idea is that as you're falling obviously mG.'s pull pull you down but there's also a force that resists that F. air so that you don't accelerate with nine point eight down you accelerate it a little bit less so your final velocity down here is a little bit less than what you would normally be. So let's say with no in no air resisting to get to the ground with thirty or whatever you could calculate it we don't have to but instead you get there with twenty six OK this is still mg is still bigger than this resistance force because you have to you're falling right into a tug of war and mg pull harder so that you accelerate down but not as fast as you would if there was no resistance at all so this is your free body diagram right there and if I want to find the force of air resistance a force I'm going to find a force by writing f=ma some of all the forces on purse is the mass of the purse acceleration of the purse the forces are F air going up and mg going down.

And I want this guy here and if you look around I have everything except acceleration and I'm stuck here I mean I have to get acceleration from motion I hear and if you look around here I have three out of five I don't have time but I don't need it so time is my ignore variable and this tells me I can use a second equation to find the acceleration. So I can use the second equation. The final velocity is negative twenty-six to get squared this is zero two is what I'm looking for Delta wise negative fifty six this is a twenty six right here so this becomes six seven six positive and I have to divide by two and negative fifty six over here and if you put this in a calculator you get that a equals negative six point zero four. meters per second squared. OK so that's a and I can plug this back in here to find F. Air F. air is going to be ma plus mg OK so. Mass is two. The acceleration is made of six point zero four mass is two gravity is nine point eight Remember gravity is not negative nine point eight that's the acceleration if you're free falling with gravity when you write in mG. is the G. just nine point eight yes it's pulling down that's why I had a negative here but the negative went away when I moved to the other side so when you just plugging in G. it's nine point eight positive OK if you do all this the answer is seven point five two. Newtons right that said now letÕs see if we can make sense out of the SR your mg for two kilograms if you mass is 2 2 your M G is nineteen point six look at what's happening this object being pulled down by an M.G. of nineteen point six and it's being pulled up by an F. air of five point fifty two as we expected the mg is bigger so we've got a number that's reasonable and that gives us greater confidence in its correctness our We'll have a big article jumper jumping up which is a two part problem so lets see when you jump your body accelerates up while you're pushing against the floor so the way you jump is you push down against the floor and because of action reaction actually action for pushes up with the force of normal in if that force is stronger than your and G. then you go up you can walk the floor so again you push down there's a force of you on the floor and the reaction of the floor pushes back if that's the force of the floor on new which is normal OK force of floor.

On you which is normal and as long as normal is bigger than mg.

You will come off the floor you will jump you actually lift off the floor OK let's the idea and what I want to know is what average force to the floor apply on you so i am looking for m Or the force of you on the floor Those are the same or even the force of you on the floor because these are the same magnitude as well in this case obviously when I write f=ma when I write f=ma for the object or the thing that has the forces applied on it so if the forces are applied on you I'm going to write some of our forces on you is your mass. And your acceleration OK So we go look for all the forces on you so you look for this guy which is the same as normal. OK so free body diagram would look like this.

Normal and mG. and I want normal to be bigger than the M.G. so that I'm actually accelerating up this is free body diagram right here complete free body diagram so it says when you jump a this is a two part problem and I merely motion equations to figure this out so let me draw this real quick when you jump it causes your legs stretch out your legs stretch for fifty centimeters before you come off the floor so here's the idea and we're going to be treating the body as a point mass all these problems are like this so the idea is here's a little point mass it stretches up while jumping it stretches up the distance of.

Fifty centimeters that's point five positive because it's going up right so this is you stretching up before you come off the floor so why are you doing this you have an acceleration that is going up.

okay your initial velocity here is zero in your velocity at this point we don't know.

Actually I'm going to call this a B. so va is zero vB. I don't know. In then what happens at point B. You come off the floor you go all the way up and then you reach your max might and you fall. And it says that you come off the floor for a maximum height of forty centimeters going to go up a little bit more to point C. where your velocity is zero because that's the maximum height in the here you're delta y is point four also positive because you're still going up OK but now from B. to C. you are in the air.

Here this is where you are actually jumping right so since you're in the air you're floating your free fall fall in your acceleration is gravity so it's negative G. negative nine point eight OK so let's see I have this is my two part motion diagram what am I missing here missing time here and I'm missing time here OK So this is what the setup were looks like the reason I'm doing this is because I know I'm going to need a little bit later but if you weren't sure that you needed that you would keep going with that because I'm a cause i'm after all I'm asking for force so that's how to start force problems with f=ma and then you would get stuck needing the acceleration Let me show you so the force on you.

Is the sum of all forces on you is normal going up M.G. going down. And then your mass and your acceleration and if I'm looking for n n equals to ma+mg and I have all of these except a I don't have a.

So I have to go find a right but which a well I want the forces on you while you're going up so I need this acceleration for the first part here so I'm going to call this a one and call this a2. and what I want is I want a1 so let's find a 1

So I have to play with this bottom part of the diagram here with the. Bottom part of my motion so go here I just need three variables and you're looking for a one.

if You count I have one two I only have two variables the third one is going to have to be either V.B. or delta T. and hopefully you can see right away that since I have two intervals and vB.'s is in the middle of them. I'm going to be able to get V B from the second interval because they have that as a shared piece of information right and I'm going to be able to use it over here I can't find T S O T's going to be my ignored verb because I can only where else to get right but I can go up here to get V.B. and the fight if you count up here I know one two three things so I will be able to find V.B. from here and not be able to use V.B. to find this a all right so I'm actually not right away looking for acceleration I'm actually going to be looking for.

V B.

If you look at the second set here my ignore variable is time.

Some going to be able to use the second equation.

There's a lot of steps here and the final velocity is zero the initial velocity is v B And that's what I'm looking for the acceleration is gravity. And Delta y is positive point for oKay. and I have this.

If you move everything you get that V B equals two point eight I get a positive and that's good because I'm going up so this Vb is a positive. Two point eight because at this point you're still growing up all right so now I can come back here.

And find a way because now I have. Three pieces of information for the bottom part as walk in to find a I can use so not as hard start here then here and here I'm here and I'm back looking for a so dont get lost

One more.

Cool so to find a I have three variables and delta T. again is my ignored variable some going to use again equation number two we kind of make some room here.

All right so v squared equals v initial square plus 2a Delta Y. and I'm looking for. a

Your final velocity from a to B. is two point eight.

The initial velocity you start at the ground right with zero and then you went up so this is zero + 2A is what I'm looking for Delta y how far up I went which here is point five right I might stretch point five coming up that's my acceleration distance and if you solve for everything here your a which is your a one is going to be seven point eighty four comes out as a positive which is good because it is a positive it is blown up and that's what I get for a one. So I can start on the left made my way to the right came back to the left pretty crazy and now I'm going to be able to find n OK. So n equals let me just kind of number the steps one.Two. Three.

And then back here at four and n=m which is your mass is eighty so eighteen times acceleration which is seven point eight four plus eighty times gravity which is nine point eight positive right be careful and if you do this it Rounds up to three if you used to significant figures for 1410

Newtons is the amount of normal force you need if you're going if you're eighty kilograms and you're going to pull you're going to jump the maximum height of forty centemeters OK if you're going to come off the floor at the bottom of your feet I guess are forty centimeters above the ground so that's it for this one obviously sort of a more advanced vertical motion problem with a jump a similar one would be if you're landing the same thing happens when you're landing where you hit the ground and then your knees bend so that you sort of decelerate while compressing your legs. OK So hopefully this makes sense let me know if you guys have any questions

0 of 1 completed

Concept #1: Landing & Jumping Problems

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.