Video: Distinguishing Weight from Other Forces

In this video, we will learn how to distinguish between the effects of the weight of an object and other forces acting on it.

14:27

Video Transcript

In this video, we will be looking at how to distinguish between the weight of an object and other forces acting on the object. The reason that we would want to do this is because it is often very easy to confuse the weight of an object with an overall or net downward force acting on the object. So let’s clear up some misconceptions.

Let us start by recalling what we mean when we talk about the weight of an object. Whenever we place an object, such as this tennis ball, into a gravitational field, such as this gravitational field of the Earth, that object experiences a force. And that force acts to pull the object towards the centre of mass of whatever is generating the gravitational field. In this case, the tennis ball placed in the gravitational field of the Earth experiences this gravitational force that pulls the tennis ball towards the centre of the Earth. This gravitational force is known as the weight of the ball.

Now in a constant gravitational field, this weight force is a constant as well. We can recall how to calculate the weight of an object by remembering that the weight of the object is defined as the mass of the object multiplied by the strength of the gravitational field. Which by the way is also sometimes known as the acceleration due to gravity.

Now this weight force, 𝑊, as we’ve called it, will continue to act on the object that’s placed in the gravitational field for as long as that object is in the gravitational field. Closer to the surface of the Earth, the gravitational field becomes almost a constant. And so we essentially treat the Earth’s gravitational field close to the surface as a constant field.

And we can also recall that the value of this gravitational field strength or acceleration due to gravity near the surface of the Earth is 9.8 metres per second squared. So in Earth’s gravitational field, near the surface of the Earth, the weight of any object is given by multiplying the mass of that object by the gravitational field strength, which is 9.8 metres per second squared.

Now there are a few misconceptions about the notion of an object’s weight that we can clear up in this video. The first of these misconceptions is the assumption that any downward net force on an object is the same thing as its weight. This is definitely incorrect. The weight of an object is very specifically the gravitational force exerted on that object. And that force is given by this equation. 𝑊 is equal to 𝑚𝑔, where 𝑊 is the weight, 𝑚 is the mass of the object, and 𝑔 is the gravitational field strength, as we saw earlier.

To illustrate this point, let’s think about a rocket. But let’s imagine that this rocket is flying in a downward direction, that is, pointing towards the surface of the Earth. And not only is it just falling downwards, but its thrusters are actually firing and providing an extra downward force onto the rocket. So what are the forces acting on the rocket?

Well, first of all, we’ve got the weight of the rocket, which is the gravitational force exerted on the rocket simply because it’s in the gravitational field of the Earth. And we can call this force 𝑊. Secondly, there’s an additional downward force on the rocket that is provided by the firing thrusters. And that force is also pushing the rocket in a downward direction. So we’ll call this second force 𝑇 for the thrust from the thrusters.

And then, of course, if we wanted to, we could also consider the air resistance on the rocket because this rocket is flying through the atmosphere. And so it’s going to experience a resistive force that acts in the opposite direction to its motion. Let’s call this resistive force 𝑅. And let’s assume that this resistive force is relatively small.

Now in this situation, it is very easy to assume that the weight of the rocket is the same thing as the overall or net or resultant force acting on the rocket. And to find out the resultant force acting on the rocket, we can first of all define arbitrarily that the rocket moving in a downward direction means that it’s moving in the positive direction. And therefore, anything acting in the upward direction is acting in the negative direction.

Then we can say that the net force on the rocket, which we’ll call 𝐹 subscript net, is equal to firstly the weight of the rocket. And that’s a positive force because it’s acting in the downward direction. Plus the thrust on the rocket from the boosters, once again a positive force. Minus the resistive force acting on the rocket, because it’s acting in the negative direction. So this is an expression for the overall force acting on the rocket.

And what this is equivalent to is just one force acting on the rocket that’s equal to the net force acting on the rocket. So we could replace these labels for 𝑅, 𝑇, and 𝑊 and say that this entire situation is equivalent to the rocket having just one force with a magnitude of 𝑊 plus 𝑇 minus 𝑅 acting on it. Because, to any of observer, that’s exactly what this looks like.

The rocket’s acceleration downwards will be at a rate given by Newton’s second law of motion, which tells us that the net force on an object, 𝐹 subscript net, is equal to the mass of that object multiplied by its acceleration. And so this rocket will not be accelerating downwards at the same rate as the acceleration due to gravity.

Remember, we assumed that gravity’s acceleration is 𝑔 is equal to 9.8 metres per second squared. This rocket will not be experiencing that specific acceleration. And this is because the rocket has other forces acting on it, that is, the thrust from the boosters and the air resistance. But that does not mean that the net force acting on the rocket is the same thing as its weight.

It’s very easy to imagine that, just because the rock is accelerating downwards at a certain rate 𝑎, that the force that causes this, which is actually the net force on the rocket, is equivalent to the weight of the rocket. But as we’ve seen already, this is not true. The weight of an object has a very specific magnitude given by multiplying the mass of the object by the strength of the gravitational field that the object is placed in. And therefore, any downward net force on an object is not the same thing as its weight. So that’s one misconception about weight dealt with. Let’s now look at another one.

This second misconception is that objects that are stationary on Earth’s surface do not have weight. Now this statement once again is most definitely not true. And this statement kind of links to the previous one as well. It’s related to a misunderstanding of the net force acting on an object as compared to the weight of an object.

Let’s think about an object such as this block of wood, for example, sitting stationary on the Earth’s surface. Well, if this block of wood is stationary, then we know that it has a velocity of zero. In other words, the block is not moving. And if the block is not moving, then there’s no way that it can be accelerating. So we can also say that the acceleration of the block is zero.

But then if we recall Newton’s second law of motion, which tells us that the net force on an object, 𝐹 subscript net, is equal to the mass of that object multiplied by the acceleration it experiences. And then we substitute in a value of zero for the acceleration. Then we see that the entire right-hand side becomes zero because we’ve got the mass of the block multiplied by zero.

And then if the right-hand side is zero, then the left-hand side must be zero as well. Therefore, the net force on this object, our wooden block, is zero. Now this might lead us to believe that the weight of the wooden block is zero because the block is just sitting there. We don’t see the effects of the weight of the block.

But remember, Newton’s second law of motion refers to the net force on the object. And actually, for however long this block is in the gravitational field of the Earth, there will be a gravitational force acting on the block. The only difference here is that because the block is sitting on the floor, the floor exerts an upward force onto the block, which is normal to the surface of the Earth. And this force, which we’ll call 𝐹, is also known as a contact force. This force only occurs because of the contact between the block and the surface of the Earth.

Now because the block is stationary and therefore not accelerating, we know that this means that the net force on the block is zero. Therefore, this force, 𝐹, which is the normal force on the block, must exactly cancel out the weight of the block. And the fact that these two forces cancel each other out means that the net force on the block is zero. And therefore, the block is not moving.

So the idea is that the gravitational force on the block is still acting. But in the situation where the block is placed on the floor, this force is counteracted by the normal force. So the misconception is that because the block is sitting on the floor, it’s stationary. We don’t see it falling through the Earth’s surface or anything like that. And so we don’t see the effects of the weight of the block. Hence, we assume that the weight of the block doesn’t exist.

But the fact of the matter is that, for however long any object with mass is placed into a gravitational field, that object will experience a gravitational force, otherwise known as the weight of the object. If we want our object which has mass to have zero weight, then we need to take it away from any gravitational fields. In other words, we need to take it deep into outer space away from any stars or planets or galaxies that could produce gravitational fields. And only then will our block of wood have a weight of zero. So now that we’ve cleared out these misconceptions, let’s have a look at an example question.

A rock with a mass of 2.5 kilograms is thrown vertically upward by an applied force of 60 newtons. What is the rock’s weight? What is the net vertically upward force applied to the rock? What is the rock’s rate of upward vertical acceleration just before it is released? What is the rock’s rate of upward vertical acceleration just after it is released?

Okay, so in this question, we’ve been told that we have a rock that is thrown vertically upward. We’ve been told that the rock is thrown vertically upward by an applied force of 60 newtons. So we can say that the hand, for example, doing the throwing of the rock is applying an upward force of 60 newtons. But as well as this, we need to recall that the rock is going to have a gravitational force acting on it because the rock is found in the gravitational field of the Earth.

Therefore, the rock is going to have a weight force, which we’ll call 𝑊, acting on it. And the first part of the question actually asks us to find the rock’s weight. So to do this, we can recall that the weight of an object is found by multiplying the mass of that object by the gravitational field strength of the Earth in this case. And from the question, we already know the mass of the rock. And we can recall that the gravitational field strength of the Earth is 9.8 metres per second squared.

Therefore, to find the weight of the rock 𝑊, we’ll say that this is equal to 2.5 kilograms — that’s the mass of the rock — multiplied by 9.8 metres per second squared, the gravitational field strength. Now before we evaluate this, we can see that we’re working in base units, kilograms for mass and metres per second square for the gravitational field strength, otherwise known as the acceleration due to gravity. Therefore, the weight of the rock is going to be in its own base unit, which is the newton. And this is because weight is a force.

And so the right-hand side of the equation becomes 24.5 newtons. Hence, we can say that the weight of the rock, the answer to the first part of the question, is 24.5 newtons.

Moving on then, we need to find the net vertically upward force applied to the rock. In other words, we need to find the overall force acting on the rock when the two individual forces acting on the rock are the 60-newton upward force applied by the hand and the downward gravitational force, otherwise known as the weight. And because in the question we’ve been asked to find the net vertically upward force, let’s arbitrarily choose the upward direction to be positive and the downward direction therefore will be negative.

Now it’s of course worth noting that we’re trying to find the net vertically upward force applied to the rock just before the rock has been let go from the hand. Because once the hand lets go of the rock, this 60-newton upward force will no longer be acting on the rock. However, because we’ve been asked to find the net force, we therefore need to consider the two forces acting on the rock. And this is just before the rock has been released.

Hence, we can say that the net force on the rock before it’s been released, which we’ll call 𝐹 subscript net, is equal to the upward 60-newton force, which is positive because it’s pointing in the upward direction, minus its weight, which we’ve seen is 24.5 newtons, because the weight is acting in the downward direction. And so the overall or resultant or net upward force is 35.5 newtons. And this is our answer to the second part of the question. So let’s move on to the third part.

What is the rock’s rate of upward vertical acceleration just before it is released?

Well, like we said already, just before the rock is released from the hand, it has two forces acting on it: the upward 60-newton force and the downward weight force. And as we’ve already seen, the net force on the rock because of this is 35.5 newtons upward. And that’s because this net force is positive and we said the positive direction was the upward direction.

Then we can recall Newton’s second law of motion, which tells us that the net force on an object is equal to the object’s mass multiplied by the acceleration it experiences. Now in this scenario, just before the rock is released from the hand, we already know the net force and we also know the mass of the rock. Therefore, we can solve for the acceleration of the rock. And coincidentally, because the net force on the rock is in the vertically upward direction, this means that using our equation will give us the rock’s rate of upward vertical acceleration, just as we need.

So to solve for the acceleration, let’s divide both sides of this equation by the mass of the rock 𝑚. Doing this means that the 𝑚 on the right-hand side cancels. And what we’re left with is that the net force on the rock divided by the mass of the rock is equal to the acceleration experienced by the rock.

Then when we plug in our values, we see that the acceleration experienced by the rock is equal to the net force, 35.5 newtons, divided by the mass, 2.5 kilograms. This evaluates to an acceleration of 14.2 metres per second squared. And therefore, we found the answer to the third part of our question. So we can move on to the final part.

What is the rock’s rate of upward vertical acceleration just after it is released?

Well, just after the rock is released, the hand is no longer exerting that upward 60-newton force on the rock. Therefore, the only force now acting on the rock is the weight of the rock. And so the new net force acting on the rock is simply the weight of the rock.

Writing this mathematically, we can say that the net force on the rock, 𝐹 subscript net, comma after the rock has been released from the hand is now simply given by the weight of the rock. But remember, the weight of the rock is acting in the downward or negative direction. Therefore, we can say that the net force on the rock is equal to negative 𝑊.

But then we know that the weight of an object is defined as the mass of that object multiplied by the gravitational field strength. Therefore, using that equation, we can say that negative 𝑊 is equal to negative the mass of the rock multiplied by the gravitational field strength of Earth. And we also know that the net force on an object at any time is given by multiplying the mass of that object by the acceleration experienced by the object. And the effect of all of this is to realise the mass of the rock multiplied by its acceleration is the same thing as negative the mass of the rock multiplied by the gravitational field strength of Earth.

Therefore, we can get rid of anything else in the middle of this equation. And we can say that 𝑚𝑎 is equal to negative 𝑚𝑔. Then we can divide both sides of the equation by the mass of the rock, 𝑚. And this results in 𝑚 cancelling on both sides. And so what we’re left with is that the acceleration experienced by the rock just after the rock is released from the hand is equal to negative 𝑔 or negative 9.8 metres per second squared. And that is our answer to the very final part of this question.

Okay, so now that we’ve had a look at an example question, let’s summarise what we’ve talked about in this lesson. Firstly, we’ve seen that an object’s weight is given by finding the product between its mass and the acceleration due to the gravitational field it is placed in. In other words, the weight of the object is equal to the mass of the object multiplied by the gravitational field strength. 𝑊 is equal to 𝑚𝑔.

Secondly, we’ve seen that the gravitational force acts on an object for however long it is in the gravitational field. In other words, the weight of the object doesn’t stop existing just because the object is not moving, for example, when the object is stationary on the surface of the Earth.

And finally, we’ve seen that the weight of an object is not necessarily the same as the net vertically downward force on the object. Now the weight of the object can be the same thing as the net vertically downward force on the object. But that’s only true if the weight is the only force acting in the upward or downward direction. However, in general, the net force downwards on the object is not the same thing as its weight.

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