Motion Along Inclined Planes
The inclined plane is a problem setting in which a massive object is on a slope, and only subject to motion in the direction down the incline.
Although gravity pulls an object straight down, the presence of the slope prevents this. Because objects can't move through the solid surface of the incline, the object is limited to movement along the surface of the incline.
Change of Coordinates
Solve the problem by changing the coordinate system. In the original coordinates, the \(y\)-axis is oriented vertically, the \(x\)-axis is oriented horizontally, and gravity acts purely in the \(y\)-direction. If instead the coordinates are oriented so that the \(x^\prime\)-axis is along the inclined plane and the \(y^\prime\)-axis is perpendicular to it, the problem becomes very simple.
Notice the decomposition of the gravitational force. Also present are a normal force and friction. [1]
The force of gravity along the plane is simply \(mg\sin\theta\), while the force of gravity perpendicular to the plane is \(mg\cos\theta:\)
\[\begin{align} F_{g\parallel} &= mg\sin\theta\\ F_{g\perp}&= mg\cos\theta. \end{align}\]
A \(\SI{5}{\kilo\gram}\) block is allowed to slide down a frictionless \(30^\circ\) incline. What is the block's acceleration?
As the block is on an incline, it may only accelerate parallel to the incline, so we will sum the forces in that direction. Since there is no force but gravity and the normal, the net force is simply
\[\begin{align} \Sigma F_{\parallel} = F_{g\parallel} = mg\sin\theta &= ma\\\\ \Rightarrow a &=g\sin\theta\\ &=(9.8)\sin 30^\circ\\ &=\SI[per-mode=symbol]{4.9}{\meter\per\second\squared}.\ _\square \end{align}\]
Two blocks \(A\) and \(B\) of equal masses are sliding down along parallel lines on a plane inclined at an angle of \(\theta =45^\circ\) with the horizontal, as shown in the figure. The coefficients of kinetic friction are \(0.2\) and \(0.3\) for \(A\) and \(B,\) respectively. At \(t=0,\) both the blocks are at rest, and block \(A\) is \(2\) meters behind block \(B\).
Find the time (in seconds) and distance (in meters) from the initial position of \(A\) to where the front faces of the blocks come in line on the inclined plane, as illustrated in the figure. \(\big(\)Use \(g=\SI[per-mode=symbol]{10}{\meter\per\second\squared}.\big)\)
Input your answer as the sum of the answers for the time and distance, to 3 decimals.
Source : JEE 2004
Skier on a Slope
A Taste of Energy:
A more powerful way of thinking about mechanics is energy. If a skier moves a distance of \(d\) along the surface, they will have moved through a vertical displacement of \(d\sin\theta\), which can be seen with simple geometry. For the moment, assume there is a quantity of interest called \(K\) which has the form \(\frac12 mv^2\). Now, if the object started from rest, this quantity will have the value
\[\Delta K = \frac12 mv^2.\]
Plugging in \(v,\)
\[\Delta E = \frac12 mg^2t^2\sin^2\theta. \]
The time, \(t\), can be found by solving the simple kinematics relation \(d = \frac12 at^2\) or
\[t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2d}{g\sin\theta}}.\]
Therefore,
\[\Delta E = mgd\sin\theta.\]
Evidently, by falling a vertical distance \(h = d\sin\theta\) under gravity, the object has picked up energy
\[\Delta E = mgh.\]
References
- Mets501, . Free body. Retrieved May 4, 2016, from https://commons.wikimedia.org/wiki/File:Free_body.svg
