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Dynamics and inclined planes

Dynamics and inclined planes

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Summary

Dynamics and inclined planes

In a nutshell

To solve problems where there is motion on an inclined plane, you can use the same principles already seen in static equilibrium problems. However any motion in a particular plane can be considered by using F=maF=maF=ma in a particular direction. You can still use ∑F=0\sum F=0∑F=0​ for a plane where there is no motion as well as F=μRF=\mu RF=μR.


Equations

DESCRIPTION
EQUATION
When an object accelerates in a particular plane,
use Newton's second law
​F=m×a F = m \times aF=m×a​​
When an object does not accelerate in a particular plane,
the resultant force is zero.
​∑F=0\sum F = 0∑F=0​​
Friction
​F=μRF=\mu RF=μR​​


Variable definitions

QUANTITY NAME

SYMBOL

UNIT NAME

UNIT

​Force of frictionForce \ of \ frictionForce of friction​​
​FFF​​
​NewtonsNewtonsNewtons​​
​NNN​​
​Reaction forceReaction \ forceReaction force​​
​RRR​​
​NewtonsNewtonsNewtons​​
​NNN​​
​WeightWeightWeight​​
​WWW​​
​NewtonsNewtonsNewtons​​
​NNN​​
​AccelerationAccelerationAcceleration​​
aaa​​​
​Meters per second squaredMeters \ per \ second\ squaredMeters per second squared​​
​m.s−2m.s^{-2}m.s−2​​
​Coefficient of frictionCoefficient \ of \ frictionCoefficient of friction​​
​μ\muμ​​
​no unitsno \ unitsno units​​
​−-−​​



Dynamics and inclined planes

Procedure

1.
Draw a force diagram.
2.
Resolve the forces into different components; refer to the inclined plane.
3.
For the plane where there is no acceleration, use ∑F=0\sum F =0∑F=0 to form equations.
4.
For the plane where there is acceleration, use F=maF=maF=ma to form equations.​
5.
Solve the equations.


Example 1

A fox of mass mmm slides from rest on a slide inclined at θ=52∘\theta = 52^{\circ}θ=52∘ to the horizontal. The fox accelerates down the slide at a=2.4 m.s−2a=2.4 \ m.s^{-2}a=2.4 m.s−2. Calculate the coefficient of friction μ.\mu .μ.​


First draw a force diagram:

Maths; Application of forces; KS5 Year 13; Dynamics and inclined planes


Resolve the forces into components, parallel and perpendicular to the plane. You should add labels to your diagram as follows:

Maths; Application of forces; KS5 Year 13; Dynamics and inclined planes


Resolving perpendicular to the plane gives:

R=Wcos⁡(52)R=mgcos⁡(52)R=W \cos(52) \\ R=mg \cos(52)R=Wcos(52)R=mgcos(52)​​


Resolving parallel to the plane gives:

F=maP=Wsin⁡(52)−F=mamgsin⁡(52)−μ×mgcos⁡(52)=m×2.4\begin{aligned}F &=ma \\P = W \sin(52) - F &= ma \\mg \sin(52) - \mu \times mg \cos(52) &= m \times 2.4 \\\end{aligned}FP=Wsin(52)−Fmgsin(52)−μ×mgcos(52)​=ma=ma=m×2.4​​​


Solve for μ\muμ to give:

μ=mgsin⁡(52)−2.4mmgcos⁡(52)μ=0.882 (3 s.f.)‾\begin{aligned}\mu &= \dfrac{mg \sin(52)-2.4m}{mg \cos(52)} \\\\\mu&= \underline{0.882 \ (3 \ s.f.)}\end{aligned}μμ​=mgcos(52)mgsin(52)−2.4m​=0.882 (3 s.f.)​​​​



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Inclined planes

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FAQs - Frequently Asked Questions

What equations do you use when solving problems on inclined planes?

You can use the fact that the resultant force =0 for the plane where there is no motion and F=ma for plane where there is motion or acceleration.

How do you solve problems involving motion on an inclined plane?

To solve problems where there is motion on an inclined plane, you can use the same principles already seen in static equilibrium problems. However any motion in a particular plane can be considered by using F=ma in a particular direction.

What is an inclined plane?

An inclined plane is simply a sloping surface.

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