Movement on a slope

Introduction

Start the activity

In this activity you will:

• explore the behaviour of an object on a slope;
• look at a simple model and how to analyse the forces present.

Watch it go

Not Quite!

You need to press Start.

What did you see?

The object begins to accelerate.

The greater the angle, the further the object moves in 1.5 seconds.

In this walkthrough you'll see how to work out the acceleration given the mass of the object in this simple model.

Here you see an object on a slope.

Assume here that there is no friction between the slope and the object which is held in place to begin with.

Instructions

Move the green point to adjust the angle of the slope.

Press Start to release the object. It will move for 1.5 seconds. After this the animation stops.

Try this a few times, using Reset to repeat.

Observe the distance travelled and speed reached for various angles.

Finish when you are ready to move on.

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Weight and normal

Not Quite!

Look at the diagram again. Imagine the angle created between the line of direction of the normal and the slope.

Well Done!

You'll go on to see how it is useful to think about components of forces acting parallel and perpendicular to the slope.

Here you will think about the forces acting on the object in this simple model.

Instructions

There are two forces, the weight of the object and the normal reaction to the slope it is on.

Move the green point to adjust the angle of the slope.

Look at the forces on the object as you do this.

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Parallel and perpendicular

Not Quite!

Look carefully. Watch the size of the normal and the component of the weight perpendicular to the slope change as you change the angle.

Well Done!

The normal and the component of the weight perpendicular to the slope 'cancel' each other out.

This must be the case because there is no motion in that direction.

This means that, in this case, the component of the weight parallel to the slope is the resultant force on the object.

It's useful to resolve forces parallel and perpendicular to the slope in situations like this.

Instructions

On the right you can see the components of the weight parallel and perpendicular to the slope.

In this diagram the length of the vector represents the size of the force.

Move the green point to change the angle.

Watch the component of the weight perpendicular to the slope and the normal carefully.

Does the statement at the bottom appear to be True or False?

Choose and Submit to check.

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Resolving

Not Quite!

Make sure you have applied the trigonometric ratios correctly.

It might help to draw the right angled triangle in the diagram separately.

Well Done!

This helps you to work out the resultant force on the object from its weight.

On this page it is assumed that you have met the idea of resolving forces before.

Here you'll look at how to use trigonometry to resolve the weight force parallel and perpendicular to the slope.

Instructions

You can use the green point to adjust the slope.

Notice the the angle between the weight force and the direction perpendicular to the slope is the same as the angle of inclination of the slope. You might need to draw a diagram to convince yourself of this.

Answer the question below and Submit.

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Resultant force

Not Quite!

Adjust the slope and watch the size of the resultant force vector carefully.

Well Done!

This is probably as you would expect. A steeper slope will produce a greater resultant force and therefore a greater acceleration.

Here is an example where the resultant force is calculated.

The object shown has mass $10\text{kg}$ and $\text{g}=9.8\text{ms}^{-2}$ so its weight is $98\text{N}$.

Instructions

The normal and the component of the weight perpendicular to the slope have been faded in the diagram on the right since they sum to zero.

As you have seen, this means that the component of the weight parallel to the slope is the resultant force.

You can change the slope using the green point and the resultant force is calculated.

Is the statement true or false?

Choose a response and Submit.

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Acceleration

Not Quite!

Using $F=ma$ gives $9.8=2a$.

Well Done!

The example you have seen is very simple but the principles apply for analysing any movement on a slope.

The object on the right here has mass $2\text{kg}$ and $\text{g}=9.8\text{ms}^{-2}$ so its weight is $19.6\text{N}$.

The slope is inclined at an angle of $30^{\circ}$ and the resultant force is $9.8\text{N}$ as shown.

Instructions

Use $F=ma$ to calculate the acceleration of the object when it is released.