Lesson video

Hello there, I'm Mr. Forbes, and welcome to this lesson from the "Forces make things change" unit.

This lesson's all about forces acting in two dimensions at right angles to each other and what that does to the movement of an object.

In this lesson, you're going to learn how to draw and interpret free body diagrams when there are forces acting at right angles to each other on an object.

You're also going to learn how to use those diagrams to find resultant forces and how to split one force into its vertical and horizontal components.

These are the keywords that'll help you through the lesson.

The first of them is resultant force.

And a resultant force is a single force that represents the effect of all the forces acting on an object.

The next keyword's equilibrium, and that's the state of an object when there's no resultant force acting on it.

So an object in equilibrium has no resultant force.

Vector diagrams, and vector diagrams are used to draw forces acting end-to-end and to a scale.

And we can use those to find the resultant force, its size and direction.

We can also split forces into components.

So the component of a force is one of the two perpendicular forces that when added together will give that single force.

And finally, resolving forces.

When we split a force up into two perpendicular forces that add up to that force.

So we take a force and split it down into two forces that act at right angles.

This lesson's in three parts.

In the first part, we're gonna concentrate on how to draw and interpret free body diagrams in two dimensions.

In the second part of the lesson, we're going to use those scale free body diagrams to find the resultant force in two dimensions.

And finally, we're going to look at how to split up a single force into its two perpendicular components, and that's the resolution of a force.

So let's start by looking at free body diagrams in two dimensions.

In many situations there are forces acting in parallel with each other, but that's not always the case.

There are sometimes horizontal forces and vertical forces acting at the same time.

So let's see an example of that.

So here's someone trying to push a box along the floor.

As they push, there's going to be a horizontal force acting on the box.

There we have 80 newtons.

But at the same time, there's going to be a downwards force, the force of gravity, pulling a box downwards there.

That's 300 newtons.

And as you can see, those forces are acting at right angles to each other.

So we've got a situation where the forces are not in opposite directions.

So here's a second example.

We're gonna hit a tennis ball horizontally by a tennis racket.

So you might think there's only a horizontal force, but there are vertical forces acting on the ball as well.

When the racket hits the ball, we've got a large horizontal force acting on it, so that's gonna push it sideways, but at the same time, there's gonna be a gravitational force acting on the ball because of the Earth pulling it down, the weight of the ball.

So we've still got two perpendicular forces, two forces acting at right angles.

We have perpendicular forces even when the ball's in flight or moving over the net.

So I've got a tennis ball here.

It's moving over the net, you can see the direction it's moving towards right there, but the ball is still attracted downwards towards the Earth because of the weight of the ball.

And as it's pushing its way through the air, there's also going to be a drag force acting on the ball.

So again, we've got perpendicular forces here, one acting downwards and one acting towards the left.

The situation gets a little bit more complicated when the ball hits the ground.

So we've got the ball hitting the ground, it was moving towards the right and downwards, and we've still got the gravitational force acting on the ball.

So it's still got its weight there.

It's still moving through the air, so we still have a drag force acting on the ball, but now it's touching the ground.

So there's going to be a reaction force acting on it due to the impact, and that'll cause the ball to bounce.

And as it's in contact with the ground and moving towards the right, it's gonna slip across the ground a little bit, so we're going to have a frictional force from the grass as well.

So we've got a situation here where we've got four forces acting, two vertically and two horizontally.

Let's have a look at another situation where we've got forces that are perpendicular.

I've got an aeroplane flying here.

So we've got a thrust force pushing the plane forwards through the air.

And of course we're gonna have the weight force pulling the plane downwards due to gravity there.

And as it's moving through air, there's going to be a drag force that's opposing the movement of the plane, so we've got a backwards there.

And finally, well, for the plane to be able to fly, there's going to be an upwards force, and that upwards force, called a lift force, is produced by the winds of the plane.

So again, four forces here, two horizontal and two vertical.

If we want to work out what's going to happen to the motion of the plane, we have to treat those sets of forces separately.

So we look at the horizontal forces and we look at the vertical forces separately, and that'll help us decide what's going to happen to the motion of the plane.

So if we look at the horizontal forces first, if I've got a thrust force and a drag force that are equal in size to each other but in opposite directions, I've got no overall resultant horizontal force there.

So it's not going to accelerate horizontally.

It's not going to speed up or slow down horizontally.

We can also look at the vertical forces, the lift and the weight.

And if they're equal to each other, again, it's not going to accelerate vertically.

It's gonna have a steady vertical speed.

Okay, let's see if you understand the forces acting on an aeroplane at constant speed.

So I've got an aeroplane here flying horizontally at a constant speed and vertically at constant speed.

Which of those force diagrams correctly shows the forces acting on the aeroplane? And I've got six there.

We've also got a key showing you the weight, thrust, drag, and lift forces.

So pause the video, decide which is the correct diagram, and restart, please.

Welcome back.

Well, as the plane isn't speeding up or slowing down, the horizontal forces must be thus equal and the vertical forces must be equal, so the correct diagram you should have chosen was diagram b here.

Well done if you've got that.

We describe an object that has no acceleration as being in equilibrium.

It's not speeding up or slowing down in any way, so it's in equilibrium.

And I've got a model aeroplane in equilibrium here.

If it's in equilibrium, it's not changing speed horizontally, it's not changing speed vertically, and it's also not changing direction.

And the only time that can happen is if there's no resultant horizontal force.

So the drag and the thrust forces must be equal to each other and in opposite directions.

And there must also be no resultant vertical force, so the lift and the weight forces are also equal to each other and in opposite directions.

That doesn't mean the lift and the thrust are the same.

It's only the vertical and the horizontal forces treated separately we need to look at.

So using what I've just said and examining these diagrams carefully, I'd like you to decide which of the following model boats is in equilibrium.

Pause the video, make your selection, and restart, please.

Welcome back.

You should have selected diagram c there.

And as you can see, the drag and the thrust forces are equal and opposite, so there's no resultant horizontal force.

And the up thrust and the weight are equal and opposite, so there's no resultant vertical force.

So that's in equilibrium.

The other two boats are not in equilibrium.

Well done if you got that.

If there are differences in the horizontal forces, then there's gonna be acceleration horizontally.

So I've got two planes here.

Let's look at them both.

In the first one, I've made the thrust larger than the drag.

So the thrust and the drag are different, and as you can see, the thrust is larger and that makes the plane accelerate.

It's going to speed up.

If I make the thrust smaller than the drag, then the plane's gonna slow down horizontally.

But if you look carefully at the vertical forces there, they're equal to each other, so there's not gonna be any acceleration vertical at all.

If I've got differences in the vertical forces, that's going to cause the aeroplane to accelerate vertically.

So in this example here, I've got the same plane, the thrust and the drag are the same as each other, so there's no change in the horizontal movement, but the lift is much greater than the weight in that diagram, so it's going to accelerate vertically upwards.

In this diagram, I've got a smaller lift than I have weight, and so it's going to accelerate vertically downwards.

Let's see if we can work out what happens to this aeroplane.

So I can look carefully in the diagram, compare the forces, and I can see the lift and the weight are equal to each other in size, and so it's not going to accelerate vertically.

I can see the drag is greater than the thrust.

That means that this aeroplane's going to slow down.

And I should say it slows down horizontally because the drag is larger than the thrust.

I'd like you to work out what's going to happen to this aeroplane and why.

So pause the video, work it out, and then restart, please.

Welcome back.

Well in this one, the drag and the thrust are the same size, and the lift is greater than the weight, so this plane's going to speed up vertically upwards.

It's gonna accelerate upwards because the lift is larger than the weight.

Well done if you've got that.

Sometimes the forces aren't acting at right angles to each other, so the situation could become a little more complicated.

I've got a car on a slope here, it's parked on that steep slope, and we're going to look at some of the forces acting on it.

Well, first of all, as it's parked and stationary, there must be a frictional force holding it in place to stop it sliding down that slope.

And then we've got the weight of the car itself.

The weight of the car acts directly downwards to the centre of the Earth, so we've got that.

And as you can see, they're not at right angles to each other.

And third, we've got the normal reaction force acting at right angles to the road surface.

And that again, that force isn't at right angles to the weight, but it is at right angles to the frictional force there.

And now a question for you to answer.

A child has lost the sledge in the snow, and it's sliding away down the hill on its own.

So you can see that in the picture.

And that sledge is in motion down the hill.

Which of those free body diagrams correctly shows the forces acting on the sledge? And as usual, I've got a key to name those forces there.

So pause the video, decide which of the six pictures shows the free body diagram, and then restart, please.

Welcome back.

Well, you should have selected diagram c.

As you can see there, I've got a frictional force that's along the line of the slope.

That's the pale yellow arrow there that's going up the slope.

That's the friction.

I've got the normal reaction force at right angles to the slope, and that's shown by the blue arrow.

And I've got the weight acting directly downwards.

So well done If you selected c.

And now it's time for the first task of the lesson.

And what I'd like you to do is to draw separate free body diagrams for an aeroplane in each of these scenarios.

Travelling at constant vertical speed while accelerating horizontally.

Travelling at constant horizontal speed while accelerating upwards.

Slowing down horizontally and accelerating downwards as it comes into land.

I'd also let you to describe and explain what's happening to the car shown in that diagram in terms of the forces acting on it.

So pause the video, answer those questions, and restart, please.

Welcome back.

And here are the free body diagrams for the aeroplane in each of those scenarios.

As you can see, I've not drawn the plane, I've just used a simplified model of it, which is just a circle.

So well done if you drew diagrams that look like this.

And this is the diagram of the car that you were meant to explain.

And as you can see, the force up the slope is much larger than the frictional force down the slope, so that car is going to be accelerating upwards along the slope.

Well done if you've got that.

And now we've reached the second part of the lesson, and in it, we're going to find the resultant of forces in two dimensions by using scale diagrams. Let's get on with that.

Now as a reminder, a resultant force is a single force that represents all the forces acting on the object, the overall effect of all those forces.

So when I've got a simple diagram like this where it's just in one dimension, so we've got forces to the left and to the right, we can add those forces and subtract forces depending on which direction they're going in.

So if all the forces act in a straight line, we can find the resultant quite simply by adding all the forces in one direction and subtracting all the forces in the other direction.

So in this diagram, we've got 1,300 newtons there pulling towards the left, and 1,150 newtons pulling towards the right, so the team of three is winning in that competition.

A resultant force of 150 newtons.

The resultant of a set of vertical and horizontal forces need to be dealt with in a slightly different way.

So I've got a model aeroplane, and we need to combine the vertical forces and combine the horizontal forces separately.

So there's my model aeroplane, and I've got some values for the lift, thrust, weight, and drag here.

I can look at the forces acting upwards and get a resultant of two newtons upwards because there's 20 newtons down and 22 newtons upwards, and subtracting those vertically, I get two newtons upwards as a resultant.

So we've just dealt with the vertical forces, then I can deal with the horizontal forces and do the same sort of stuff.

I've got 25 newtons thrust minus 20 newtons drag, so I get five newtons forwards as a resultant.

So I've got two separate resultant forces there for vertical and horizontal.

And now what I'd like you to do is find the.

(clear throat) And now what I'd like you to do is to find the resultant horizontal and vertical forces acting on this model aeroplane.

So pause the video, find the total resultant horizontally and the total resultant vertically, and then restart, please.

Welcome back.

And the resultant vertical forces here is four newton downwards.

Well done if you've got that.

And the resultant horizontal force is three newtons backwards.

So well done if you've got that as well.

Now let's see what we get when we've got forces acting at right angles to each other and how we can find the resultant.

So we're gonna start by considering this spaceship, and it's moving towards the right in this diagram, but it fires rocket thrusters in two separate directions at once.

So we've got two rocket thrusters firing here.

And what we're going to find is that's going to give an overall resultant force that's acting to the left, where that one thrust is going, and also downwards.

So we get a resultant force that looks a little like this.

I drawn with the blue arrow there.

Let's see if you can identify the direction of the resultant force when this force is acting at rectangles.

So I've got another spaceship here.

It's firing two thrusters again in the direction shown.

I'd like you to decide which direction would the resultant force be in when those two forces are combined together.

So pause the video, decide your answer, and restart, please.

And welcome back.

Well, the answer to that one, well, we've got downwards and to the right, and the arrow that shows that is arrow c.

Well done if you've got that.

So we have a rough idea of the direction of a resultant force, but we need to find that out more formally, and we also need to be able to find the size of that resultant force, and to do that, we use scale vector diagrams. I've got an example here.

So I've got a spaceship, and I've got two forces acting on it.

And they're at right angles, as you can see.

We've got 600 newtons and 300 newtons.

And I'm gonna draw a scale diagram of those forces using the grid I've drawn as well.

My scale is one centimetre is 100 newtons.

That seems like a sensible scale.

And what I need to do to find the resultant force is to draw each of those two forces end-to-end.

And what I mean by that is I start by drawing my 600 newton's force.

So I'll draw it on my paper there.

As you can see, six squares, 600 newtons.

And that's directly to the left.

And then at the end of that line, I draw the next force, and that's my 300 newtons downwards.

So I draw a 300 newtons force downwards, and my resultant force would be the arrow that joins the starting point to the end point.

So I'm gonna draw that arrow on, then I can measure the length of that, and I measured that to be 6.

7 centimetres.

So using my scale, that gives me a force of 670 newtons, and I can also see the direction.

Okay, now what I want you to do is find the magnitude of the two resultant forces shown in the diagram.

I've drawn two forces that act on an object.

They're at right angles.

One's 30 newtons, and one's 40 newtons.

And I've put it on a scale where one centimetre equals five newtons.

What I'd like you to do is to find the size of the resultant.

So we just need the size, not the direction.

Pause the video and restart when you've work out the size, please.

Welcome back.

And the answer to that one was 50 newtons.

If I draw a line to show the direction of the force and its length, it's that line.

And if I measure it, it's actually 10 centimetres long, so that gives me 50 newtons force.

So well done if you got that.

Now, this technique allows you to find a resultant force even if the forces aren't at right angles, as long as you carefully draw the diagram.

So we're gonna go through that process.

I've got here a book being pulled along a desk.

As you can see, I've got a frictional force of three newtons and I've got a pull through a string acting on the book of five newtons.

And that's acting at an angle of 30 degrees to the horizontal, and I've marked that on.

So what I've got to do is carefully draw those forces on my scale diagram, and that'll allow me to find the resultant of them.

So the first thing I'm gonna do is draw the pull force on my diagram.

I've chosen a scale of one centimetre is 0.

5 newtons.

So I draw that on measuring the angle at 30 degrees to the horizontal lines on my grid.

And drawing it, well, it's five newtons, so it's 10 centimetres long on my diagram.

So I've drawn that very carefully.

Then I draw the frictional force of three newtons, and that's horizontal, so I draw it like that.

And as usual, I can find the resultant force by drawing a line from the start point to the end point.

So I've drawn that there in blue.

And I can then measure the length of that line to find the size of the force and measure the angle of that line to find out what angle it acts at.

So I did that, and I found that my resultant force, I'm gonna name the FR, it's 2.

5 newtons.

So I've measured its length.

And I also use a protractor to measure the angle.

And I found that it's at 45 degrees to the horizontal, so I've mapped that on as well.

So I've used the process to find the resultant force.

Let's have a look at a second example.

So in this one, I've got a ship, and it's being pulled along by two tugboats, and those tugboats are producing 50 kilonewtons of force.

So I'm going to draw a scale diagram, and I've decided to use 10 kilonewtons per centimetre so that everything fits on.

And as usual, what I've gotta do is draw those two forces on.

So the first force that I've drawn in red there, the 50 kilonewtons, that's sort of going upwards slightly, I can draw it like that, measuring the angle to the horizontal and making sure it's 30 degrees.

And then at the end of that line, I can draw the other force, draw that on another 50 kilonewtons, and it's 30 degrees below that horizontal line.

So I'll draw that there.

And finally I can find a resultant of that by measuring the distance between the start and the endpoint.

So I'll draw that line on.

My resultant forces 87 kilonewtons.

And it's horizontal.

The direction works out to be horizontal there.

So I found the resultant force with that scale diagram.

Okay, we've reached another task in the lesson, and what I'd like you to do is to find some resultant forces using some vector scale diagrams, please.

So I've got a sledge being pulled across the flat ground using a rope, as shown in that diagram.

And you can see there's a pull force and a frictional force.

And I'd like you to work out the resultant force using a vector diagram.

And then I've got a more complicated example.

I've got two cranes lifting an object, and I've got three forces in that.

So I'd like you to use the same technique to find the resultant force there, please.

So pause the video, find the resultant forces, and restart.

Welcome back.

And this is what your vector scale diagram should look like for the sledge.

I've drawn both of those forces.

I've chosen a scale of one centimetre is 2.

5 newtons.

Your scale might be different, but your result should be the same.

You should have found the resultant of 15 newtons at 45 degrees to the horizontal there.

Well done if you've got that one.

And the second diagram.

And the second question is shown here.

Again, I've drawn all three of the forces from the diagram on my grid there, so the two red lines and the green line.

So I started with the green line, drew the two red lines, and then I find the gap between those.

So join the start to the end, and it's nine kilonewtons straight forwards.

Well done if you got that.

Now it's time for the final part of the lesson.

And we're going to look at resolving a force, the resolution of a force, which means splitting it into horizontal and vertical components.

So let's start on that.

In this diagram, I've drawn a box resting on a slope, and it's in equilibrium.

There's no resultant force acting on that box in any direction.

So let's identify the forces that are acting on it.

We've got a normal reaction force.

The normal reaction force is always at right angles to a surface and at the point of contact, so I've drawn that in.

We've also got the weight of the box acting vertically downwards.

Weight always acts directly downwards.

And I've drawn that from the centre of the box where the centre of its gravity will be.

And finally I've got a frictional force acting on the box that's preventing its sliding down 'cause as I said, the box is at rest, it's not slipping or sliding down the slope.

So the frictional force must act from the bottom of the box, and it must act in the direction of the slope.

Now, for that object to be in equilibrium as I've described it, there must be forces acting in the direction opposite to friction and in the opposite direction to the normal reaction force, but it's not clear what those forces are.

The force that is opposing friction is actually part of the weight of the object, and that's also the force that's opposing that normal reaction force.

We can split the weight force into two components.

One component acts along the plane, along the slope, in the direction of the friction, and one that acts at right angles to the slope similar to the normal force.

So we're going to cut that weight force into two parts.

We've got a perpendicular component of the weight that acts opposite to the normal force drawn like this, and we've got a parallel component of the weight that acts along the slope that acts like that.

So we're going to divide one force up into two, which is kind of the opposite of what we did earlier where we got two forces and combined it into a single force.

So we are going to split them up by considering this sort of shape, a rectangular shape, where we find the two parts of the force.

So let's see if you understand what I meant by parallel component to the weight of the box.

I've got a diagram here, and I've got all of the forces marked on it and labelled.

And I'd like to know what the size of the parallel component of the weight is.

That's the only one I've not labelled with its size.

So look carefully at the diagram and decide this box is in equilibrium, what size must that force be for it to be an equilibrium? Pause the video, make your decision, and then restart, please.

Welcome back.

Well, your answer should have been 30 newtons.

There must be a 30 newtons force acting down the slope to counteract the 30 newtons acting up the slope, the friction force there.

So well done if you selected that.

Okay, here's another example.

I'd like you to work out the perpendicular component of the weight.

Now, I've marked that on the diagram there.

What size must that force be for this box to be in equilibrium? So pause the video, make your selection, and restart.

Welcome back.

Well, you should have selected 60 newtons.

There's a 60 newtons normal reaction force.

There must be a 60 newton force that opposes that for this to be in equilibrium.

So that force is there 60 newtons.

Well done if you've got that.

If you've got forces that are not horizontal and vertical, you can break them down into two components, a separate horizontal and vertical component, and that process is known as resolving a force.

And we're going to look at how to resolve a force now.

So I've got a book here, and it's attached to a piece of string, and I'm sliding on the desk, and I've marked the angles on there.

I'm gonna put a pull force on it.

And that pull force is along the string there, and it's acting at 25 degrees to the horizontal.

And what I want to do is to split that pull force into horizontal and vertical components.

So I'm gonna find the part of that force that acts horizontally.

I'm gonna call that FH for force horizontal.

And I'm also gonna try and find the part of that force that acts vertically, or upwards in this case.

I'm gonna call that FV for force vertical.

So let's have a look at how to do that.

My diagram's here, and what I'm going to do is to draw that force on one of my scale of vector diagrams. So I'm gonna take that pull force and draw it on this scale diagram here.

I've chosen a scale of one centimetre is 0.

5 newtons.

So this line I'm drawing must be eight centimetres long, and I've got to draw it at 25 degrees to the horizontal.

So to find the two components, I'm gonna draw the force on my square grid at that length, eight centimetres, 25 degrees to the horizontal.

So there's my line, I've drawn it on, and then all I have to do is use that grid to read off its length horizontally and it's length vertically, or how many squares it takes in each direction.

So for the horizontal one, I can see it goes this many squares across.

I've got my horizontal component.

It's 3.

5 newtons because I've got one, two, three, four, five, six, seven squares, and seven times 0.

5 newtons is 3.

5 newtons.

And I can find the vertical part by doing similar vertically.

I draw my line into three boxes, so it's 1.

5 newtons there.

So I can always measure that with a ruler as well, and I can find the angle.

Okay, I'd like you to use that technique to find the horizontal and vertical components of the force I've drawn on the diagram.

So I've drawn the force in a big red diagonal arrow there, and I'd like you to use that scale to find its horizontal and vertical component.

And I've drawn those on for you to make it a little easier.

So pause the video, work out what those values must be according to the scale, and then restart, please.

Welcome back.

Well, the horizontal component was 160 newtons to the right.

It's eight boxes, and each box is 20 newtons, so it's 160 newtons.

And the vertical component, that's 7 times 20, so that's 140 newtons downwards.

Well done if you got those two.

Okay, it's time for the final task of the lesson, and I'd like you to answer these two questions.

I've got a car parked on a slope, and I'd like you to explain how the forces on it keep it in equilibrium.

So you can see the forces there, friction, normal reaction, and weight.

And then I'd like you to use a scale diagram to resolve the following forces.

So draw those three forces on a scale diagram and find the horizontal and vertical components for me, please.

So pause the video, try those questions, and restart.

Welcome back.

Your explanation about how the car stays in equilibrium should be something like this.

The weight of the car can be resolved into two components.

One component parallel to the slope in the opposite direction to the friction.

so counteracting the friction, and one at right angles to the slope and in opposite direction to the normal reaction force.

Well done if you've got something like that.

And here's my scale diagrams for the forces I asked about.

I've drawn each of them there and found the horizontal and vertical components, as you can see.

So well done if you got these answers.

And we've reached the end of the lesson.

And here's a quick summary of everything we've learned.

The forces of an object can connect at different angles.

When the object's in equilibrium, there's no resultant force in any direction.

We can get a resultant force, and we can find that using a scale vector diagram where we draw the forces end-to-end and then join the start and the end points, and that gives us the resultant force, which is FR drawn in this diagram here.

And forces can be resolved into a pair of perpendicular components, usually horizontal and vertical.

We split the force into two different components, as shown in that bottom diagram there.

Well done for reaching the end of the lesson.

I'll see you in the next one.