Hey, guys. So up until now, we've been dealing with forces that act only on the horizontal axis like this. We're going to start to look at some forces now that act on the vertical axis. 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 in this chapter is that if you have forces that act on objects and there's no other forces, then that's going to produce an acceleration because of tafema. If you have a net force, then there's going to be some acceleration. So 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 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 going to 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 going to call this w, but the other symbol that you might see for it is also equal to fg. The equation is very straightforward. The weight force is just equal to little mg, mass times gravitational acceleration, this little g here. And what you need to know about this weight force is 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 force is going to point towards Earth's center like this. Now almost always in your problems, it's going to be pointing down towards the ground unless you're explicitly told that the ground is somewhere else. Alright. 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=ma. So if this weight force here, this fg, which is equal to mg here, is the only force that's acting on this object, then we can use f=ma 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 ma. So what happens is our m's are going to cancel, and your acceleration is just going to be little g. So this acceleration here that's due to gravity, which just 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 newtons. Its units are going to be meters per second squared. Now what 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 going to 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 the surface of the moon, and that's because the gravitational acceleration is weaker there. So they can jump higher, they can bounce all that kind of stuff. Alright. So, again, just to recap, the effect of gravity produces a force, and that force always going to point toward the Earth's center. That force is the weight force and it's given in terms of newtons. 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 meters per second squared. That's the units. Alright. 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. It says that you step on a bathroom scale, and it measures your weight to be 70 kilograms. 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 that your weight force, just like all other forces, has to be given in newtons. So really what's going on here, guys, is that scales don't measure weight. Instead, what scales measure is 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 object or something like that. And this mass is not going to change at different locations. So for instance, if you weigh 70 if you have 70 kilograms of mass on the Earth's surface, if you go to the moon or Jupiter, it's still going to be 70 kilograms because the quantity of matter, the amount of stuff that makes up you hasn't changed. Weight, on the other hand, we know is given in terms of newtons, and that's a force that's due to gravity. And unlike the mass, your force is going to change if this force does change depending on different locations. We can actually just see that by the equation w=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 so sorry. 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 kilograms, then we can just use 70 kilograms times 9.8, and you're going to get 686 newtons. So that's your real weight on the Earth's surface. Seventy kilograms is just your mass. Let's check out this next example here. If an object has 10 kilograms 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 based based on your location. So if you have 10 kilograms on the Moon, if that's your mass, then that means that this object is going to be 10 kilograms sorry. If you're 10 kilograms on the Earth's surface, then that means on the moon, it's also going to be 10 kilograms. So what's its weight on the Earth? Well, how do we get weight? We just use the equation w=mg. So if 10 kilograms, 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 going to be 98, and that's newtons. So if the same object, 10 kilograms, would be put on the moon, its weight on the moon would be the 10 kilograms times the gravitational acceleration of the moon, which is 1.62. And so, therefore, it would 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.

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# Vertical Forces & Acceleration - Online Tutor, Practice Problems & Exam Prep

Gravity exerts a downward force known as weight (W or F_{g}), calculated using the equation W = mg, where m is mass and g is the acceleration due to gravity (approximately 9.8 m/s² on Earth). This force influences an object's motion, leading to acceleration (a). If upward forces exceed downward forces, the object accelerates upward; if downward forces are greater, it accelerates downward. In equilibrium, forces balance, resulting in zero acceleration. Understanding these principles is crucial for solving problems involving vertical forces and motion.

### Weight Force & Gravitational Acceleration

#### Video transcript

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?

$g_{Mars}=0.039m/s^2$

$g_{Mars}=0.27m/s^2$

$g_{Mars}=3.7m/s^2$

$g_{Mars}=26m/s^2$

### Vertical Forces And Acceleration in the Y-axis

#### Video transcript

Hey, guys. So up until now, we've seen lots of problems. We have forces in the vertical axis that canceled out, and therefore, the object was at equilibrium. Well, you're going to need to know how to solve problems where you have vertical forces that don't cancel, and so this is actually going to cause objects to accelerate in the y-axis. But, really, we're just going to 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 going to get right to the example. So you've got a 5.1 kilogram block. It's in the air, and we're pulling it using a vertical massless string. So what 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 thing we have to do is we have to just draw the free body diagram. So we've got our block like this. And 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. So this is going to 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 nothing pushing or pulling this thing, but we do have a tension force because we have some string. That tension force is going to act upwards, right, because we're basically just hanging this block from 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 no 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 s=ma. Right? So we want to find the acceleration. So we want s=ma. And so there's really only 2 forces to consider, our tension and our weight force. So remember, you just add the forces, tension and weight, and that equals ma. And now we can just replace the values. So our tension is 70. Our weight force is negative 50. Don't forget that negative sign. And this is equal to 5.1 times a. So you got 20 equals 5.1a. And so, therefore, your acceleration is equal to 3.92 meters per second squared. So we've got 2 things here. We know that this acceleration is not going to be 0, 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 50 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 the acceleration is going to be upwards, and that's what that positive sign tells us. Alright. Let's move on to the next one here. So now we've got 30 newtons instead of 70. But, really, we're just going to do the same exact thing. So we've got our box like this. We've got our weight force. We know that this is w. We know this equals to negative 50. And then we know that this tension force is actually now 30 instead of 70. So we're going to write a little bit smaller, the arrow. So here's the thing. So if we have a 30 newton tension force, what we have, what we had in part a is that if this 70 newton force was bigger than our, 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 weight force. So we'd expect that the acceleration is going to point downwards. Alright? So now we're just going to dof=ma. So we've got our tension plus our weight equals mass times acceleration. So now we've got our 30 plus negative fifty equals 5.1a and so this is negative twenty equals 5.1a. And so here, the acceleration is negative 3.92 meters per second squared. This should make some sense, right, because, basically, now we know that the weight force was bigger. Our tension force 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 force is still bigger than how hard you're pulling upwards, and so this block is still going to accelerate downwards. Alright? So now, what we've got here is we've got 50 Newtons. And we're actually just going to fill these out. We're going to 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 this negative 50. But now what happens is we're pulling upw

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.)

^{2}

^{2}

^{2}

^{2}

### Lowering a Load of Bricks

#### Video transcript

Hey, guys. Let's check out this problem. We've got a 100 kilogram load of bricks that's being lowered on a cable. So basically, I've got this load of bricks that's 100 kilograms. I'm just going to draw it as a box. It's being suspended by a cable, but that cable is actually lowering this load of bricks. So we know the velocity here is equal to 5 meters per second. And eventually, what happens is over some period of 2 seconds, it's going to slow to a stop, which means that there are some time period here, I'm going to call this t, which equals 2 seconds. Eventually, this load of bricks will come to a stop, which means the velocity is going to equal 0. So what I'm going to do is I'm going to call this the initial velocity, which is my 5 meters per second. The final velocity is 0. What I want to do is I want to figure out what is the tension in the cable so that this load of bricks actually comes to a stop. Right? So the first thing we have to do is we're going to draw a free-body diagram. So I'm going to do that over here. This is going to be my free-body diagram. So basically, I'm going to have this little dot like this. I have the weight force, the weight force is going to be downwards. This is my W=mg, and that also got a tension force like this. And this is basically what I'm trying to solve for. Alright? 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 you don't have anything pushing or pulling. You also don't have a normal or friction because this thing is in the air. Right? So our free-body diagram is pretty straightforward. It's only these two forces, and now we're getting into our F=ma. So if our F=ma, 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. Alright? So now we got our forces. We've got tension that's upwards and then our mg is downwards. So our equation becomes t-mg=ma. 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. I 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 meters per second, and then over 2 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 we're going to do here. So if we want to figure out the acceleration, I'm going to need to write out my 5 kinematics variables. Right? I'm given 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 naught, v, a, and t I just need to know 3 out of 5 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 5. Is it 5, though, or is it negative 5? Remember what happens is that you the, direction of velocities and accelerations depend on which direction you choose to be positive. We chose up to be positive, so this v naught actually points down even though we write it as a positive in the diagram. When we're doing math, we actually have to write it with the correct sign. So it's negative 5. The final velocity is 0. The acceleration is actually what we're trying to find here. And we know the time is equal to 2 seconds. So fortunately, And that equation is going to be the simplest one, number 1, which says that the initial, sorry, 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 0. Initial velocity is negative 5 plus, and then we've got a times 2. So if you bring this negative 5 over to the other side, it becomes positive So 5 equals 2a and so your acceleration is going to be 2.5 meters per second squared. So let's talk about the sign. Remember that when you solve for the acceleration, as long as you've plugged in everything correctly, you should get the correct sign, and it should indicate the acceleration's direction. So which means which means that we 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 meters 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 we can just plug this number back into our F=ma and then solve for the tension. So what happens is we're going to move this mg to the other side and tension becomes mg+ma. You could also just move you can merge those into 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, and we plug it in as a positive, remember. So if you work this out what you're going to get is 1230 Newtons. So we looked through our answer choices 1230, and that's going to be answer choice a. So we've got our attention is 1230. This should make some sense that we got a 1230 because in order for the acceleration to be up, right, 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. Alright? So that's it for this one, guys.

## Do you want more practice?

More sets### Here’s what students ask on this topic:

What is the difference between weight and mass?

Weight and mass are often confused but are fundamentally different. Mass is a measure of the amount of matter in an object and is measured in kilograms (kg). It remains constant regardless of location. Weight, on the other hand, is the force exerted by gravity on that mass. It is calculated using the equation $W=mg$, where $W$ is weight, $m$ is mass, and $g$ is the acceleration due to gravity (approximately 9.8 m/s² on Earth). Weight is measured in newtons (N) and varies with the gravitational field strength of the location.

How do you calculate the weight of an object on Earth?

To calculate the weight of an object on Earth, you use the equation $W=mg$, where $W$ is the weight, $m$ is the mass of the object in kilograms (kg), and $g$ is the acceleration due to gravity, which is approximately 9.8 m/s² on Earth. For example, if an object has a mass of 10 kg, its weight would be $10\times 9.8=98$ newtons (N).

What happens to the weight of an object if it is taken to the moon?

The weight of an object decreases if it is taken to the moon because the moon's gravitational acceleration is weaker than Earth's. The gravitational acceleration on the moon is approximately 1.62 m/s². Using the weight equation $W=mg$, if an object has a mass of 10 kg, its weight on the moon would be $10\times 1.62=16.2$ newtons (N), compared to $98$ newtons on Earth.

How do you determine the acceleration of an object when vertical forces do not cancel out?

To determine the acceleration of an object when vertical forces do not cancel out, use Newton's second law, $F=ma$. First, draw a free-body diagram to identify all forces acting on the object. Sum the forces to find the net force $F$_{net}. Then, use the equation $F$_{net}

What is gravitational acceleration and how does it vary by location?

Gravitational acceleration, denoted as $g$, is the acceleration due to the force of gravity acting on an object. On Earth, it is approximately 9.8 m/s². However, this value varies depending on the location. For example, on the moon, gravitational acceleration is about 1.62 m/s², and on Jupiter, it is approximately 24.79 m/s². These variations occur because different celestial bodies have different masses and radii, affecting the strength of their gravitational fields.

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- (II) A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale ...
- (II) What will a spring scale read for the weight of a 58.0‑kg woman in an elevator that moves(a) upward with ...
- (II) What will a spring scale read for the weight of a 58.0‑kg woman in an elevator that moves(b) downward wit...
- (II) What will a spring scale read for the weight of a 58.0‑kg woman in an elevator that moves(d) with a downw...
- (II) An exceptional standing jump would raise a person 0.80 m off the ground. To do this, what force must a 68...
- (II) What will a spring scale read for the weight of a 58.0‑kg woman in an elevator that moves(e) in free fall...
- (II) An 18.0-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a t...
- (III) Determine a formula for the acceleration of the system shown in Fig. 4–49 (see Problem 55) if the cord h...
- "(II) High-speed elevators function under two limitations: (1) the maximum magnitude of vertical acceleration ...
- (III) An inclined plane, fixed to the inside of an elevator, makes a 38° angle with the floor. A mass m slides...
- (II) A narrow but solid spool of thread has radius R and mass M. If you pull up on the thread so that the cm o...
- (II) How much tension must a cable withstand if it is used to accelerate a 1400-kg car vertically upward at 0....
- (III) An object moving vertically has v→ =v₀ → at t = 0 . Determine a formula for its velocity as a function o...
- (II) As shown in Fig. 4–48, five balls (masses 2.00, 2.05, 2.10, 2.15, 2.20 kg) hang from a crossbar. Each mas...
- (III) An inclined plane, fixed to the inside of an elevator, makes a 38° angle with the floor. A mass m slides...
- A 6750-kg helicopter accelerates upward at 0.80m/s² while lifting a 1080-kg frame at a construction site, Fig....
- A 6750-kg helicopter accelerates upward at 0.80m/s² while lifting a 1080-kg frame at a construction site, Fig....
- A wet bar of soap (m= 150g) slides freely down a ramp 3.0 m long inclined at 8.5°. How long does it take to re...
- (II) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.140 g by a single cable. Determine(a...