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Inclined planes

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Summary

Inclined planes

In a nutshell

Objects on inclined planes can be modelled using force diagrams. 



Solving problems with inclined planes

To solve problems which involve inclined planes, you resolve parallel and perpendicular to the plane using trigonometry. The angle that the object makes with the plane is equal to the angle of inclination of the plane. When resolving the weight of an object parallel to the plane you use sine, and when resolving perpendicular to the plane you use cosine. 


Example 1

A ball of mass 10kg10kg10kg​ rolls down a smooth slope angled at 30°30\degree30° to the horizontal. 

a: Find the magnitude of the normal reaction of the slope on the ball.

b: Find the acceleration of the ball.


Maths; Forces and friction; KS5 Year 13; Inclined planes

a: To find the normal reaction, find the component of the weight of the ball which is perpendicular to the plane:


mgcos⁡(θ)=10gcos⁡(30)=84.9N\begin{aligned}mg\cos(\theta)&=10g \cos(30)\\ &=84.9N\end{aligned}mgcos(θ)​=10gcos(30)=84.9N​​​


The magnitude of the normal reaction of the slope on the ball is 84.9N‾\underline{84.9N}84.9N​.


b: The weight of the ball rolling down the hill, resolving parallel to the plane:

mgsin⁡(θ)=10gsin⁡(30)mg\sin(\theta)=10g\sin(30)mgsin(θ)=10gsin(30)​​


The acceleration of the ball:

F=maa=Fm\begin{aligned}F&=ma\\ a&=\dfrac{F}{m}\end{aligned}Fa​=ma=mF​​​​


Solve to find acceleration:

a=10gsin⁡(30)10=gsin⁡(30)=4.9ms−2\begin{aligned}a&=\dfrac{10g\sin(30)}{10}\\ &=g\sin(30)\\ &=4.9ms^{-2}\end{aligned}a​=1010gsin(30)​=gsin(30)=4.9ms−2​​​


The acceleration of the ball is 4.9ms−2‾\underline{4.9ms^{-2}}4.9ms−2​.


Example 2

A block of mass 5kg5kg5kg is moving on a smooth slope and is being acted on by a force of 20N20N20N parallel to the slope in the direction shown in the diagram. The slope is inclined at an angle of 20°20\degree20°. Work out the acceleration of the block.


Maths; Forces and friction; KS5 Year 13; Inclined planes


The parallel component of the weight of the block:

mgsin⁡(θ)=5gsin⁡(20)mg\sin(\theta)=5g\sin(20)mgsin(θ)=5gsin(20)​​


Resolve forces parallel to the slope using F=maF=maF=ma​:

20−5gsin⁡(20)=5a3.41013...=5aa=0.648ms−2\begin{aligned}20-5g\sin(20)&=5a\\ 3.41013...&=5a\\ a&=0.648ms^{-2}\end{aligned}20−5gsin(20)3.41013...a​=5a=5a=0.648ms−2​​​


The block has an acceleration of 0.648ms−2‾\underline{0.648ms^{-2}}0.648ms−2​.



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FAQs - Frequently Asked Questions

How do you find the angle an object makes with a plane?

The angle that the object makes with the plane is equal to the angle of inclination of the plane.

How do you work out the parallel component of the weight of an object?

When resolving the weight of an object parallel to the plane you use sine.

How do you work out the perpendicular component of the weight of an object?

When resolving the weight of an object perpendicular to the plane you use cosine.

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