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Dynamics and inclined planes

Dynamics and inclined planes

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Tutor: Meera

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=ma in a particular direction. You can still use F=0\sum F=0​ for a plane where there is no motion as well as F=μRF=\mu R.


Equations

DESCRIPTION

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


Variable definitions

QUANTITY NAME

SYMBOL

UNIT NAME

UNIT

Force of frictionForce \ of \ friction​​
FF​​
NewtonsNewtons​​
NN​​
Reaction forceReaction \ force​​
RR​​
NewtonsNewtons​​
NN​​
WeightWeight​​
WW​​
NewtonsNewtons​​
NN​​
AccelerationAcceleration​​
aa​​​
Meters per second squaredMeters \ per \ second\ squared​​
m.s2m.s^{-2}​​
Coefficient of frictionCoefficient \ of \ friction​​
μ\mu​​
no 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 to form equations.
4.
For the plane where there is acceleration, use F=maF=ma to form equations.​
5.
Solve the equations.


Example 1

A fox of mass mm slides from rest on a slide inclined at θ=52\theta = 52^{\circ} to the horizontal. The fox accelerates down the slide at a=2.4 m.s2a=2.4 \ m.s^{-2}. Calculate the coefficient of friction μ.\mu .


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)​​


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}​​


Solve for μ\mu 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}​​



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FAQs - Frequently Asked Questions

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