PlanetPhysics/Frictionless Inclined Plane

The inclined plane is a common example of [[../Newtons3rdLaw/|Newton's laws of motion]]. It was used in Galileo's experiment to calculate the [[../Acceleration/|acceleration]] due to gravity and has been used by students for centuries to explore the laws of [[../CosmologicalConstant/|motion]]. Here we will examine a block sliding down a frictionless inclined plane as shown below. The y axis is perpendicular to the incline and the x axis is parallel to incline.

\vspace{20 pt} \includegraphics[scale=.85]{InclinePlane.eps} \vspace{20 pt}

(1)

The first thing to do is draw a free body [[../Commutativity/|diagram]] to describe the external [[../Thrust/|forces]] acting on our [[../TrivialGroupoid/|object]] and the coordinate [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] for the problem.

\vspace{20 pt} \includegraphics[scale=.85]{FreeBodyDiagram.eps} \vspace{20 pt}

(2)

Applying Newton's 2nd law to both x and y

$\sum \overrightarrow{F}=m\overrightarrow{a}$

{\mathbf y-dir} \\

First note that the block is stationary in the y direction, so we know that the acceleration is zero.

$\sum F_y=0$

The two forces in the y direction are due to the normal force applied by the incline and the force due to gravity.

$\sum F_y=N-mg\cos \theta =0$

{\mathbf x-dir} \\

With the block sliding down the frictionless incline, we get

$\sum F_x=mg\sin \theta =m{a}_x$

There are several quantities of interest, but let us start with the basics of acceleration, [[../Velocity/|velocity]] and displacement. For acceleration we see that it is constant in the x direction

$a_x=g\sin \theta$

Next, let us calculate the velocity at a given displacement. As usual, we integrate acceleration

$\frac{d{v}_x}{dt}=g\sin \theta$ $\underset{0}{\overset{v_x}{\int }}d{v}_x=\underset{0}{\overset{t}{\int }}g\sin \theta dt$

and carrying out the integration gives us the velocity of the block as a [[../Bijective/|function]] of time as it slides down the incline.

$v_x(t)=gt\sin \theta$

Next, integrate again to get displacement down the incline.

$\frac{dx}{dt}=gt\sin \theta$ $\underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}gt\sin \theta dt$

Carrying out the integration gives us the displacement of the block as a function of time as it slides down the incline.

$x(t)=\frac{1}{2}g{t}^2\sin \theta$

Now we can focus on other interesting quantities such as the velocity as a function of displacement. To get this solve (3) for t and plug it into (2)

$t=\sqrt{\frac{2x}{g\sin \theta }}$

$v_x(x)=g\sqrt{\frac{2x}{g\sin \theta }}\sin \theta$

putting all the terms under the [[../PiecewiseLinear/|square]] root gives

$v_x(x)=\sqrt{2gx\sin \theta }$

A fun and easy experiment is to measure the acceleration of gravity by rolling objects down an incline. All you have to do is solve equation (4) for g

$g=\frac{2x}{t^2\sin \theta }$

and then just measure $\theta$, $x$ and $t$. There is really nothing like carrying out experiments that helped shape physics, so go forth and measure.

{\mathbf [[../InertialSystemOfCoordinates/|equilibrium]].}

We already saw that the forces perpindicular to the plane in the y direction are in equilibrium

$N=mgcos\theta .$

For the x direction parallel to the plane, one must apply a force $F_a$ in the opposite direction (up the plane) to counter the force due to gravity. This equilibrium condition would mean our forces in the x direction would cancel so

$F_a+mgsin\theta =0$ $F_a=-mgsin\theta .$