![](\"resources/question-resources/inclined_solution.png\")
What is the acceleration in $\\mathrm{ms^{-2}}$ of the particle as it slides down the plane if the plane is smooth?
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", "unitTests": [], "mustBeReduced": false, "maxValue": "9.8*sin(radians(theta))-mu*9.8*cos(radians(theta))"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Finding the acceleration of a particle on an inclined plane which is at an angle to the horizontal. Smooth then with value for coefficient of friction.
"}, "name": "Simon's copy of inclined plane 1", "advice": "You can draw a diagram of the forces acting upon the particle.
\n
(a)
\nWhen the plane is smooth we have that the coefficient of friction $\\mu = 0$.
\nWe then resolve $F=ma$ in the direction down the slope.
\n\\begin{align} F &= ma, \\\\
mg \\cos ((90 - \\theta)^{\\circ}) & = ma, \\\\
\\var{mass} \\times 9.8 \\times \\cos (\\var{90 - theta}^{\\circ}) & = \\var{mass} \\times a, \\\\
9.8 \\cos (\\var{90 - theta}^{\\circ}) & = a, \\\\
\\var{precround(9.8*cos(radians(90-theta)),3)} \\mathrm{ms^{-2}} & = a. \\end{align}
So the acceleration of the particle when the slope is smooth is $ \\var{precround(9.8*cos(radians(90-theta)),3)} \\mathrm{ms^{-2}}$.
\n\n
(b)
\nIf the plane is rough and the coefficient of friction is $\\mu = \\var{mu}$ then we first resolve the forces perpendicular to the plane to find $R$.
\n\\begin{align} F & = ma, \\\\
R - mg \\cos \\theta & = 0, \\\\
R &= mg \\cos \\theta, \\\\
& = \\var{mass} \\times 9.8 \\cos( \\var{theta}^{\\circ})=\\var{precround(mass*9.8*cos((pi/180)*theta),3)}.
\\end{align}
We then resolve $F=ma$ in the direction down the slope, using the value we have for $R$.
\n\\begin{align} F & = ma, \\\\
mg \\sin \\theta - \\mu R & = ma, \\\\
mg \\sin \\theta - \\mu mg \\cos \\theta & = ma, \\\\
g \\sin \\theta - \\mu g \\cos \\theta & = a, \\\\
\\left(9.8\\sin(\\var{theta}^{\\circ})\\right) - \\left(\\var{mu}\\times 9.8 \\cos(\\var{theta}^{\\circ})\\right)& = a, \\\\
\\var{precround(9.8*sin(radians(theta))-mu*9.8*cos(radians(theta)),3)} \\mathrm{ms^{-2}} & = a. \\end{align}
So the acceleration of the acceleration of the particle is $\\var{precround(9.8*sin(radians(theta))-mu*9.8*cos(radians(theta)),3)} \\mathrm{ms^{-2}}$ when the coefficient of friction is $mu=\\var{mu}$.
\n", "preamble": {"css": "", "js": ""}, "statement": "Suppose that there is a plane which is inclined to the horizontal at an angle of $\\var{theta}^{\\circ}$. A particle of mass $\\var{mass}\\mathrm{kg}$ is placed upon the plane.
\n![](\"resources/question-resources/inclined.png\")
All parts should be answered to 3d.p and the acceleration due to gravity should be taken as $g = 9.8 \\mathrm{ms^{-2}}$.
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