# Video: Determining the Apparent Weights of Objects Applying a Force to Another Object

A hot-air balloon has a mass of 350 kg. The balloon is near the ground, and two ropes were thrown out for people on the ground to grab onto in order to help pull the balloon to the ground. Before the ropes are pulled, the balloon accelerates vertically downward at 0.25 m/s². And when the ropes are pulled, it accelerates downward at 1.5 m/s². Each rope has the same force applied to it, and each person pulling on the rope has a mass of 75 kg. How many newtons does each person who holds a rope apparently weigh while holding onto it?

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### Video Transcript

A hot-air balloon has a mass of 350 kilograms. The balloon is near the ground, and two ropes were thrown out for people on the ground to grab onto in order to help pull the balloon to the ground. Before the ropes are pulled, the balloon accelerates vertically downward at 0.25 meters per second squared. And when the ropes are pulled, it accelerates downward at 1.5 meters per second squared. Each rope has the same force applied to it, and each person pulling on the rope has a mass of 75 kilograms. How many newtons does each person who holds a rope apparently weigh while holding onto it?

Right, this is a really long and complicated question. There’s a lot of information in it. So let’s start by underlining all the important bits. So we don’t miss anything out. Okay, so first things first, we know that we’ve got a hot-air balloon which has a mass of 350 kilograms, we know that two ropes were thrown out of the hot-air balloon and people have grabbed onto them in order to help pull the balloon to the ground.

Now, we’re told that before the ropes are being pulled, the balloon accelerates vertically downward at 0.25 meters per second squared. And when the ropes are pulled, it accelerates downwards still, but this time at 1.5 meters per second squared. So when people are pulling on the ropes, the balloon accelerates downwards even faster.

Now, we’re also told that each rope has the same force applied to it and each person pulling on the rope has a mass of 75 kilograms. What we need to do is to find out how many newtons each person apparently weighs while holding onto a rope. So this concept of how much does each person apparently weigh can be quite confusing. So we’ll come back to this in a second.

First though, let’s draw a diagram of our hot-air balloon. Alright, so here’s our balloon and we’re going to describe what it does before the ropes are being pulled. Firstly, we’ve been told that the mass of the balloon is 350 kilograms. We’ll call this value capital 𝑀. And we will say on the side that capital 𝑀 is equal to 350 kilograms.

Next, we’re told that before people start pulling on the ropes to help pull the balloon to the ground. The balloon is accelerating downwards, which we’ll call acceleration 𝑎 one. And this acceleration 𝑎 one is equal to 0.25 meters per second squared. And of course, this acceleration is vertically downward as we’ve been told in the question. Now, this is before the ropes have been pulled. Afterwards, however, there are two people pulling on ropes to help the balloon come down to the ground faster. So afterwards, the balloon still has the same mass. The mass is not changing. However, this time it’s accelerating downwards with an acceleration 𝑎 two. And we’ve been told that 𝑎 two is equal to 1.5 meters per second squared.

Next, we know that each rope has the same force applied to it. So we’ll say that each person is pulling on the rope with a force 𝐹. And this force is in a downward direction. Now, obviously, the ropes haven’t been drawn perfectly pointing in a downward direction. But we can imagine that the force 𝐹 is acting downwards. This is what’s causing the balloon to accelerate downwards even more.

Now, we don’t actually know what force 𝐹 is. So we’ll just leave it as 𝐹 for now. However, one thing that we do know is the mass of the people pulling on the rope. We’ll say that they have a mass lowercase 𝑚. And we’ve been told they both have the same mass, which happens to be 75 kilograms.

Now, what we’ve been asked to find in this question is the apparent weight of each person that’s holding a rope. And we want to find this apparent weight in newtons because we’ve been asked how many newtons does each person who holds a rope apparently weigh. So we’ll say that the apparent weight of each person in newtons is equal to question mark because that’s what we’re trying to find out. Oh! And we forgot to label this part of the diagram “after” when the people are pulling on the ropes.

Okay, so what we can do now is to look at the “before” situation. So what we’re considering is this situation here. Now, in this situation, all we know is the acceleration of the balloon and the mass of the balloon. This, however, is quite useful information because we can use what’s known as Newton’s second law of motion. Newton’s second law of motion tells us the 𝐹 sub res — the resultant force on an object — is equal to the mass of the object multiplied by the acceleration of that object.

And since we have the mass and acceleration of the balloon, we can work out the resultant force on the balloon before the ropes have been pulled. So we can say that the resultant force on the balloon 𝐹 sub res is equal to the mass of the balloon which is capital 𝑀 multiplied by the acceleration of the balloon which is 𝑎 sub one. And of course, we know the values of capital 𝑀 and 𝑎 sub one, which happen to be 350 kilograms and 0.25 meters per second squared, respectively. So we can sub those in and work out what 𝐹 sub res is. Now, this force happens to be 87.5 newtons.

And as we said already, this is the resultant force on the balloon. Now, this is the resultant force because there are two forces initially acting on the balloon: one of course is the weight of the balloon which acts downwards which we’ll call capital 𝑊 and the other is the upthrust from the balloon. Because of course, remember this contraption here is heating up the air in the balloon, which results in an upwards force, which we’ll call- let’s say the lift capital 𝐿.

So we’re balancing two forces here: the lift and the weight and the resultant force is 87.5 newtons. And we know that the weight must be bigger because the resultant force must be in a downward direction because the acceleration is in a downward direction. So the resultant force 𝐹 sub res is 87.5 newtons in the downward direction. So that’s what we say 𝐹 sub res is in a downward direction.

Okay, so why is this useful? Well, we can take this resultant force and apply it to the “after” situation because the weight of the balloon is not going to change and we haven’t been told that the lift of the balloon is changing either. So even though we haven’t explicitly worked out the weight and the lift, which we could do if we wanted to cause we know the mass and we can work out the weight that way but we don’t need to here, what we do know is the difference between the weight and the lift or the resultant force earlier.

In other words, 𝐹 sub res was calculated in the following way. We say 𝐹 sub res is equal to the weight which we’ll say is in a positive direction because positive we’ve taken to be downwards. And the lift is in the opposite direction. So we say minus 𝐿, where 𝐿 is the size of the lift force. So now, as we said in the “after” situation, 𝑊 and 𝐿 are not changing and hence 𝑊 minus 𝐿 will not change either.

So what we can do now is to work out the resultant force in the “after” situation. This time we have to take into account not only the lift and the weight, but also the forces being exerted on the balloon by the people pulling on the ropes — that’s the two forces 𝐹. So to help us differentiate between the resultant force in the “before” situation and the “after” situation, we’ll say that what we’ve already calculated as the resultant force is 𝐹 sub res comma one and in the “after” situation we’re calculating 𝐹 sub res comma two.

Now, we just need to balance the forces. 𝐹 sub res comma two is equal to first of all, there’s the weight in the downward direction. Then, there are the two forces 𝐹 also in the downward direction so plus two times 𝑓. And finally, there’s the lift force which is in the upward direction. So once again, we take it as negative. So now, this is the new resultant force on the hot-air balloon.

But now, we also know from Newton’s second law of motion that the resultant force on an object is equal to the mass of that object multiplied by its acceleration, which in this case is 𝑎 sub two. In other words, the new resultant force is equal to the mass of the balloon, which isn’t changing, multiplied by the new acceleration. So at this point, what we can do is to equate the right-hand sides of this equation and this equation because we’ve calculated 𝐹 sub res comma two in two different ways.

So we say that 𝑊 plus two 𝐹 minus 𝐿 is equal to 𝑀𝑎 sub two. Then, we can rearrange to find out what 𝑓 is. To do this, what we can do is to subtract 𝑊 and add 𝐿 to both sides of the equation, which will end up looking something like this. And then at this point, we can divide both sides of the equation by two and the twos on the left-hand side cancel. So on the left, we’re simply left with the force 𝐹. And remember 𝐹 and 𝐹 sub res are completely different. 𝐹 sub res is referring to the resultant force on the balloon, whereas the force 𝐹 is the force with which the people on the ground pull on the ropes.

Okay, anyway, so we’ve got an expression for the force 𝐹. Now, on the right-hand side of the equation, we know everything we need to. It might look like we don’t because we don’t know what 𝑊 and 𝐿 are. But remember we know what 𝑊 minus 𝐿 is. So we also know what minus 𝑊 minus 𝐿 is when we put brackets around it. Because when we multiply the negative sign by both the terms in the parenthesis, we find that that gives us minus 𝑊 plus 𝐿. And that’s exactly what we’ve got here.

Therefore, minus 𝑊 plus 𝐿 is equal to minus and then in brackets 𝑊 minus 𝐿. And this whole thing we know to be 𝐹 sub res comma one because 𝐹 sub res comma one is equal to 𝑊 minus 𝐿. So at this point, what we can do is to say that 𝐹 now is equal to 𝑀𝑎 sub two minus 𝐹 sub res comma one divided by two. And we can plug in all these values. So that gives us 𝑀 multiplied by 𝑎 sub two minus 𝐹 sub res comma one divided by two. So when we evaluate this fraction, we find that the force 𝐹 is equal to 218.75 newtons.

So we’ve gone to great lengths to find out the force 𝐹. But why is this useful in helping us find the apparent weight of each person in newtons? Well, let’s first write down that the force 𝐹 is equal to 218.75 newtons. And then, let’s look out one of the robes being pulled by a person in more detail. Okay, so now, we’re looking at the person and the rope in more detail.

Now, this person is pulling down on the rope with a force 𝐹. But what this means is that we can use Newton’s third law of motion. This law states that if an object exerts a force on a second object, then the second object exerts an equal and opposite force on the first object.

Now, in this scenario, the person is the first object and the rope is the second object. The person is exerting a downwards force 𝑓 on the rope. So by Newton’s third law, the rope will exert an upwards, an equal force 𝐹 on the person. Now, it’s very important that this force is equal. So it’s still 𝐹 and is in the opposite direction. So it’s upwards this time. And this force is being exerted by the rope on the person.

So we know that there’s now an upwards force on the person and the other thing we know is that the person’s weight must also be acting. We’ll call this weight 𝑊 sub 𝑝 for the weight of the person because we’ve used 𝑊 already to label the weight of the balloon. So we’ll just say 𝑊 sub 𝑝 instead. So now, we know that there are two forces acting on the person: the force upwards 𝐹 exerted by the rope on the person and the weight of the person 𝑊 sub 𝑝.

However, in this question, we’ve been asked to find the apparent weight of each person. Now, the apparent weight refers to what the person would feel like they weigh — in other words, the resultant force on the person — because their weight doesn’t actually change. It’s still 𝑊 sub 𝑝. However, exerting an extra force upwards 𝐹 will make it feel like they weigh less because they’ve been pulled to the ground with a slightly lesser force because once again the resultant force is no longer going to be the person’s weight. It’s now going to be the combination of the person’s weight and the force 𝐹.

Therefore, the apparent weight of the person, which we’ll call 𝑎 for apparent weight, is equal to the resultant force on the person which happens to be 𝑊 sub 𝑝 the weight of the person minus the upwards force 𝐹. And so that’s what apparent weight means. It’s the resultant force caused by the weight of the person plus any other forces acting on the person to make them feel either heavier or lighter. In this case, the person will feel lighter because the force 𝐹 is acting in the opposite direction to the weight.

So now, we might realize that we know what the force 𝐹 is, but we don’t know what the weight of the person is. Luckily, however, we can recall that the weight of an object 𝑊 is given by the mass of the object 𝑚 multiplied by the gravitational field strength of the Earth 𝑔. So in our case, 𝑊 sub 𝑝 — the weight of the person — is equal to the mass of the person, which is lowercase 𝑚 as we’ve called it over here. And we multiply the mass of the person by the gravitational field strength of the Earth 𝑔. And we can also recall that 𝑔 is equal to 9.8 meters per second squared. That’s the acceleration due to gravity on Earth.

So now, we can substitute in the values of 𝑚 and 𝑔 over here. And these values happen to be 75 kilograms, which we’ve been told in the question is the mass of each person and 9.8 metres per second squared as we’ve already seen over here. And that’s the expression for 𝑊 sub 𝑝. Evaluating it, we find that the weight of the person is 735 newtons.

And at this point, we know the values of 𝑊 sub 𝑝 and 𝐹. So we can calculate the apparent weight of the person 𝑎. So we say that the apparent weight of the person is equal to 𝑊 sub 𝑝, which is 735 newtons, minus force 𝐹, which is 218.75 newtons. And we find that 𝑎 is equal to 516.25 newtons and this is our final answer.

The apparent weight of each person pulling on the rope is 516.25 newtons.