1

concept

## Weight Force & Gravitational Acceleration

7m

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Hey, guys. So up until now we've been dealing with forces that act only on the horizontal axis like this. We're gonna start to look at some forces now to act on the vertical axis. And one of the main ones you're going to use in almost all of your problems is the gravitational or the weight force. So we'll be talking about this weight force and gravitational acceleration in this video. Let's check it out. So, guys, the main idea here is that all objects that are near the earth are affected by a phenomenon called gravity and gravity is going to produce a force. And one of the things that we've seen this chapter is that if you have forces that act on objects and there's no other forces, the next going to produce an acceleration because of F equals M A. If you have a net force, then there's gonna be some acceleration. So these these three related ideas here gravity the force due to gravity and the acceleration due to gravity. Let's talk about them in more detail. So, guys, whenever your textbooks measure this effect or phenomenon called gravity, really, all that all that is is It's a conceptual phenomenon that tells us that objects that have mass, for instance, like this asteroid and our earth over here are going to attract each other. So these two objects, because they have mass, they are going to attract each other like this. So how do they actually attract each other? Well, they do it by producing forces. So this force that is due to gravity basically what happens is you have this earth that is pulling on this asteroid, and I'm just gonna look at the asteroid for a second. There is a force due to gravity. So we actually have a name for that. It's called the Weight Force or the Gravitational Force. So really, this arrow is actually a vector. We're gonna call this W. But the other symbol that you might see for it is also equal to F. G. So you might also see f g. The equation is very straightforward. The weight force is just equal to little mg mass times, gravity, mass times, gravitational acceleration. This little G here and what you need to know about this weight forces that the units, just like all other forces, are in Newtons and what's special about this force is that it's always going to point towards the earth's center. So in this case, the asteroid, the weight forces going to point towards Earth Center like this now almost always in your problems, that's going to be pointing down towards the ground unless you explicitly told that the ground is somewhere else. All right, so now this force here is going to produce an acceleration. It's going to cause this asteroid to fall towards the earth. How do we actually calculate that acceleration? We can just use F equals Emma. So if this weight force here this F g, which is equal to MG here is the only force that's acting on this object, then we can use F equals to figure out this acceleration. The only force that's acting on this thing is little mg, and this is going to be equal to little m A. So what happens is our EMS are going to cancel, and your acceleration is just going to be little G. So this acceleration here, that's due to gravity, which has a name. It's also called The gravitational acceleration is really just a which is just the little G that we've been using for a bunch of chapters so far, all the way back to our motion chapters. We know that this little G near the Earth has a value. It's 9.8 near the Earth, and its units are not Newton's. Its units are going to be meters per second squared. Now when you need to know about this acceleration here. So we know that this object is going to accelerate down like this is that this acceleration is not constant. This acceleration does vary by location. For example, we know that the gravitational acceleration on the earth is that 9.8 that we've been using for some time now. But if you were to go to the moon or Jupiter or somewhere else, that gravitational acceleration is gonna have a different value on the moon. That's equal to 1.62. You might have seen those videos of astronauts who are bouncing around on the surface of the moon, and that's because the gravitational acceleration is weaker there so they can jump higher. They can balance all that kind of stuff. All right. So, again, just to recap the effect of gravity produces a force and that force always gonna point where the Earth Center that forces the weight force and it's given in terms of Newton's and that force, if it's the only force that's acting on an object, produces an acceleration called the gravitational acceleration. That's our little G, and that is equal to 9.8 m per second squared. That's the units. All right, so let's move on now. One thing you need to know about this weight force is that that term weight is actually used incorrectly in everyday language. What do I mean by that? Let's check out this example. Is this that you step on a bathroom scale and it measures your weight to be kg? So what is your real weight on the Earth's surface? Well, the reason that weight is in quotation marks like this is because the weight force is not supposed to be given in units of kilograms. Remember your weight force, just like all other forces has to be given in Newton's. So really, what's going on here, guys, is that scales don't measure weight. Instead, what scales measure as they measure your quantity of mass so mass, which is given in terms of kilograms is really just the quantity of matter. It's the amount of matter that makes up you or another an object or something like that, and this mass is not going to change at different locations. So, for instance, if you a 70 if you have 70 kg of mass on the earth's surface, if you go to the moon or Jupiter, it's still gonna be 70 kg, because the quantity of matter, the amount of stuff that makes up you hasn't changed. Wait, on the other hand, we know is given in terms of Newton's, and that's a force that's due to gravity. And unlike the mass, your force is going to change. This force does change depending on different locations. We can actually just see that by the equation, w equals mg. So just as the gravitational acceleration little G varies by location, then so does your weight force, because that's part of that equation. So your MG service or your weight force, just like little G will change based on different locations. So what's your real weight on the earth's surface? Well, if your mass is equal to 70 kg, then we can just use 70 kg times 9.8 and you're gonna get 686 Newtons. So that's your real weight on the Earth's surface. 70 kg is just your mass. Let's check out this next example here. If an object has 10 kg on the Earth, what's its mass on the moon? So basically what they're saying here is if M Earth is equal to 10, then what is M moon? Well, remember, guys that your mass, which is the quantity of matter, does not change debate based on your location. So if you have 10 kg on the moon, if that's your mass and that means that this object is going to be 10 kg. Sorry. If you're 10 kg on the earth's surface, then that means on the moon. It's also going to be 10 kg. So what's its weight on the earth? Well, how do we get Wait, We just use the equation. W equals mg. So if 10 kg, then that means the weight on the earth is going to be 10 times little G Earth, which is 9.8. So this is just gonna be 98. And that's Newton's So if the same object 10 kg will be put on the moon, its weight on the moon would be the 10 kg times the gravitational acceleration of the moon, which is 1.62. And so therefore be 16.2 Newtons. So your weight would be less, but your mass would be the same. That's it for this one. Guys, let me know if you have any questions.

2

Problem

The Mars Rover Perseverance weighed about 10,000 N while on Earth. After it reached the surface of Mars, it weighed about 3790 N. What is the gravitational acceleration on Mars?

A

0.039 m/s

^{2}B

0.27 m/s

^{2}C

3.7 m/s

^{2}D

26 m/s

^{2}3

concept

## Vertical Forces And Acceleration in the Y-axis

6m

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Hey, guys. So up until now we've seen lots of problems. We have forces in the vertical axis that cancel out, and therefore the object was at equilibrium. Well, you're gonna need to know how to solve problems where you have vertical forces that don't cancel. And so this actually gonna cause objects to accelerate in the Y axis. But really, we're just gonna use the same list of steps that we use for any forces problems to solve these kinds of problems. So this is really straightforward. We're just gonna get right to the example. So you've got a 5.1 kg block, it's in the air and we're pulling it using a vertical massless string. So we want to do is we want to find the acceleration for each of the following varying tension forces. So we got the first one here, which is tension equals 70. So the first we have to do is we have to draw the free body diagram. So we've got a block like this, Remember, we check for the weight force first. This is our weight force, and this is equal to negative MG because it's downward, Remember, up is positive down is negative. This is gonna be negative. 5.1 times 9.8 equals negative 50. So that's our weight force. And then there's no applied forces, right? There's nothing pushing or pulling this thing. But we do have a tension force because we have some string that tension forces going to act upwards, right, because we're basically just hanging this block in the string. And so we know that this tension force is equal to 70 and it's upwards, so there's no contact forces, right? There's a normal or friction because this thing is basically suspended in the air. So we just have the weight and the tension. So now we just go ahead and write ethical dilemma, right? So we want to find the acceleration so we want f equals m A. And so there's really only two forces to consider our attention and our weight force. Remember, you just add the forces tension and weight, and that equals m A. And now we can just replace the values. So our attention is 70. Our weight forces negative 50. Don't forget that negative sign and this is equal to 5.1 times a So you got 20 equals 5.1 a. And so therefore, your acceleration is equal to 3.92 m per second squared. So we've got two things here. We know that this acceleration is not going to be zero, right, because the forces are not going to cancel. But the fact that we get a positive sign also means that we know the direction of this acceleration. So if this 70 Newton tension force is greater than your Newton force of gravity that acts downwards, If you think about this like a tug of war, then that means your tension force upwards is going to win. And so therefore, you would expect that the acceleration is going to be upwards, and that's what that positive sign tells us. All right, let's move on to the next one here. So now we've got 30 Newtons instead of 70. But really, we're just gonna do the same exact thing. So we've got our box like this. We've gotta wait for us. We know that this is W. We know the Sequels to negative 50. And then we know that this tension force is actually now 30 instead of 17. We're gonna write a little bit smaller, that arrow. So here's the thing. So if we have a 30 Newton tension force we have what we have in part a is that if this 70 Newton force was bigger than our, uh, 50 Newton force downwards, then we had an acceleration that was up. Well, here we've got this 30 that's actually smaller than our downward wait for us. So we expect that the acceleration is gonna point downwards. Alright, so now we're just gonna do f equals m A. So we've got our attention. Plus our weight equals mass times acceleration. So now we've got our 30 plus negative 50 equals 5.1 a. And so this is negative. 20 equals 5.1 a. And so here the acceleration is negative. 3.92 m per second squared. This should make some sense, right? Because basically, now we know that the weight force was bigger or attention forced upwards was smaller than our weight force. And so here what happens is you have an acceleration that points downwards. And so even though you're trying to pull this block up, the weight forces still bigger than how hard you're pulling upwards. And so this blocks is still going to accelerate. Downwards. Alright, so now we've got here is we've got 50 Newtons and we're actually just gonna fill these out. We're gonna fill these out in just a minute here. So we've got 50 Newtons. So we've got our block like this. We know our weight force. It was negative. Mg. We know that's negative. 50. But now what happens is we're pulling upwards with attention. Force of 50. So when you write out your ethical dilemma, we know that we have tension. Plus, weight equals Emma. So you've got 50 plus our negative 50 and that equals 5.1 a. So are 15 negative 50? Just cancel out to zero. And so what that means is that this block is in equilibrium because your upward forces and downward forces perfectly cancel each other out. So we know that the acceleration is equal to zero in this case because your forces are perfectly balanced. Remember, this is the equilibrium means Now, finally, we have attention of zero Newton's. That's not a typo. So here, what we've got is our block like this. And now we've got our weight force we know are w equals negative 50. But now we have our tension of zero which basically means that we don't even draw the tension force. Right? Really? The only forces acting on this thing is the weight force. So we do our f equals m A. And there's only one force acting on it. This time our weights equals m A. So we have negative m g equals m A. So negative 50 equals 5.1 a. And so you work this out and you're gonna get a is equal to negative 9.8 m per second squared. This shouldn't come as a surprise because if there's only force that's acting on, this is the weight force. Then we know from our video on the weight force that all objects are gonna accelerate downwards with an acceleration of negative G, which is negative 9.8. There's no other forces that are acting on this, right? So let's go ahead and summarize real quick if you're upward forces the magnitudes right? We're just thinking of the numbers Here are greater than your downward forces like we saw in this example. Here are 70 was bigger than our 50 downwards of weights. Then that means your acceleration is going to be positive, which means it points upwards now. In this case, when we had our upward forces less than our downward forces are, acceleration turned out to be negative, right? Basically, our weight forces in our downward forces were bigger or 50 was bigger than the 30. And so therefore the acceleration was downwards. And now if your upward forces perfectly equal your downward forces, then we know that to be equilibrium in your acceleration is equal to zero in this case. And then finally, if your upward forces are equal to zero, then that means that your acceleration is going to be exactly at negative G. Assuming that there's no other forces involved. Right? So that takes care of this video. Let me know if you have any questions.

4

Problem

A 3-kg bucket is being pulled upwards by a cord. The tension in the cord is 35 N. What is the acceleration of the bucket? (The mass of the cord is negligible, which means you can assume m_cord = 0.)

A

25 m/s

^{2}B

1.9 m/s

^{2}C

21 m/s

^{2}D

8.4 m/s

^{2}5

example

## Lowering a Load of Bricks

5m

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Hey, guys, let's check out this problem. We've got 100 kg load of bricks that's being lowered on a cable. So basically, I've got this load of bricks that's 100 kg just gonna draw as a box. It's being suspended by a cable, But that cable is actually lowering This load of bricks into the velocity here is equal to 5 m per second and eventually what happens is over some period of two seconds, it's going to slow to a stop, which means that there are some time period here. I'm gonna call this tea, which equals two seconds. Eventually, this load of bricks will come to a stop, which means the velocity is going to equal zero. So I'm gonna do is I'm gonna call this the initial velocity, which is my 5 m per second Final velocity is zero. What I want to do is I want to figure out what is the tension in the cable so that this, uh, this load of bricks actually comes to a stop. Right? So the first we have to do is going to draw a free body diagram. So I'm gonna do that over here is gonna be my free body diagram. So basically, I'm gonna have this little dot like this. I have the weight force. The weight force is going to be downwards. This is my w equals mg. And that also got attention force like this. And this is basically what I'm trying to solve for All right, So if I'm basically trying to solve for this tension force again, we look at any other forces. There's no applied forces because I don't have anything pushing or pulling. You also have a normal or friction because this thing is in the air. Right? So our free body diagram is pretty straightforward. Only these two forces and now we're going to f r f equals M A. So if our ethical dilemma we have to expand our forces, we need to know the direction of positive. So we're just going to choose the upward direction to be positive. All right, so now we've got our forces. We've got tension that's upwards. And then RMG is downwards, so our equation becomes t minus. MG equals m a. Now we want to figure out what this tension is, so I need to figure out everything else in the problem. I have to know mass. I have to know G have to know mass and the acceleration. But do we actually know the acceleration? We actually don't If you look at the problem, all we're told is that this thing is going downwards at 5 m per second and then over two seconds, it stops. That doesn't tell us what the acceleration is, and so we're kind of stuck here. How do we figure this out if we don't know the acceleration? Well, remember, what happens is that if you ever get stuck solving for a, you can always try to solve it using a motion equation. And that's exactly what I'm gonna do here. So if you want to figure out the acceleration, I'm gonna need to write out my five cinematics variables, right? I'm giving stuff like initial velocity, final velocity time. So these are all motion variables. So I need to know the Delta y this is my V, not V a n t. So I just need to know three out of five variables and then I can pick an equation to solve. So I don't know what my delta Y is. I don't know the distance that this thing is stopping through. But I do know my initial velocity is five. Is it five, though, or is it negative? Five. Remember, what happens is that the direction of velocities and accelerations depend on which direction you choose to be positive. We chose up to be positive. So this V not actually points down, even though we write it as a positive in the diagram we're doing math. We actually have to write it with the correct sign. So it's negative. Five. The final velocity zero. The acceleration is actually what we're trying to find here, and we know the time is equal to two seconds. So fortunately, what happens is we know 3 to 5 variables 123 So we can just pick the equation that ignores my delta. Y right, I'm just gonna ignore that one. And that equation is going to be the simplest one Number one which says that the initial velocity Final velocity is going to be initial velocity plus a times T, so we can use this to find a. So our final velocity is zero initial velocity is negative. Five plus and then we got eight times two. So if you bring this negative five words to the other side, it becomes positive. So five equals two a and so your acceleration is going to be 2.5 m per second squared. So let's talk about the sign. Remember that when you saw for the acceleration, as long as you've plugged in everything correctly, you should get the correct sign and it should indicate that the accelerations direction so which means, which means that we've got a positive number. That just means that we got an upward acceleration. So the upward acceleration is 2.5. This should make some sense, right? So if the load of bricks is going downwards with 5 m per second in order for it to come to a stop, the acceleration has to point upwards. Right, So we got an acceleration of 2.5. Now, you can just plug this number back into our F equals m a and then solve the tension. So what happens is we're gonna move this mg to the other side. Intention becomes mg. Plus, you could also just move. You can merge this into under under parentheses. That's up to you. So the tension is equal to, and now we just plug everything in 100 times 9.8 plus 100 times the acceleration, which is 2.5. We plug it in as a positive, remember? So if you work this out, you're gonna get is 1230 Newton's. So we look to our answer choices 12. 30 and that's going to be answered. Choice A. So we've got our attention is 12. 30. This should make some sense that we got a 12 30. Because in order for the acceleration to be up right, which to which we want for the load of bricks to slow to a stop, it has to be bigger than your weight force, which is 980. So even though the load of bricks is moving downwards, the tension force still has to be bigger to produce an upward acceleration so that the load of bricks stops. All right, so that's if this one guys

Additional resources for Vertical Forces & Acceleration

PRACTICE PROBLEMS AND ACTIVITIES (11)

- You walk into an elevator, step onto a scale, and push the “up” button. You recall that your normal weight is ...
- At the surface of Jupiter’s moon Io, the acceleration due to gravity is g = 1.81 m/s2. A watermelon weighs 44....
- At the surface of Jupiter’s moon Io, the acceleration due to gravity is g = 1.81 m/s2. A watermelon weighs 44....
- An astronaut’s pack weighs 17.5 N when she is on the earth but only 3.24 N when she is at the surface of a moo...
- A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The ho...
- Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E5.14). ...
- Two 25.0-N weights are suspended at opposite ends of a rope that passes over a light, frictionless pulley. The...
- Two 25.0-N weights are suspended at opposite ends of a rope that passes over a light, frictionless pulley. The...
- A 550-N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined m...
- A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The ho...
- Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E5.14). ...