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We can draw a diagram showing the weight, normal reaction $R$ and the force of friction, $F$.

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Here $\\theta = \\var{theta}^{\\circ}$ and $m = \\var{mass}$. 

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a)

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To find the normal reaction, $R$, we resolve perpendicular to the plane.

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\\begin{align}
R - mg \\cos \\theta & = 0, \\\\
R & = mg \\cos \\theta, \\\\
& = \\var{mass} \\times 9.8 \\cos (\\var{theta}^{\\circ}), \\\\
& = \\var{precround(mass*9.8*cos(radians(theta)),3)} \\ \\mathrm{N}.
\\end{align}

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b)

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To find the coefficient of friction we can resolve parallel to the plane and use our 3d.p. value of $R$ from part a) and the fact that friction is limiting ($F = \\mu R$) as we are in equilibrium.

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\\begin{align}
F - mg \\sin \\theta & = 0, \\\\
F & = mg \\sin \\theta, \\\\
\\mu R & = mg \\sin \\theta, \\\\[0.5em]
\\mu & = \\frac{mg \\sin \\theta}{R}, \\\\[0.5em]
& = \\frac{ \\var{mass} \\times 9.8 \\sin (\\var{theta}^{\\circ})}{\\var{R}}, \\\\[0.5em]
& = \\var{precround( (mass*9.8*sin(radians(theta)))/R,3)}.
\\end{align}

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Find the normal reaction, $R$, between the ball and the plane, in $\\mathrm{N}$.

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$R = $ [[0]]

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Using the value of $R$ from part a) find the coeffiecient of friction, $\\mu$, between the ball and the plane.

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$\\mu = $ [[0]]

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A ball of mass $\\var{mass} \\ \\mathrm{kg}$ rests in limiting equilibrium on a rough plane inclined at $\\var{theta}^{\\circ}$ above the horizontal.

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Suppose that the acceleration due to gravity is $g= 9.8 \\mathrm{ms^{-2}}$.

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the normal reaction R, 3d.p

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A mass is resting on a rough plane. Given the angle of the plane and the mass of the object, find the normal reaction force and hence the coefficient of friction.

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