![](\"resources/question-resources/force_component_image_6.png\"/)
If $P$ is applied horizontally to the mass which lies at rest, we can resolve $F = ma$ with acceleration $a=0$ to find the normal reaction $R$, and from this find the maximum frictional force $F_{\\text{MASS}}=\\mu R$.
\nIn the upwards direction we have
\n\\begin{align}
F & = ma \\\\
R - mg & = 0 \\\\
R & = mg \\\\
& = \\var{mass} \\times 9.8 \\\\
& = \\var{mass*9.8}
\\end{align}
The maximum possible frictional force is $F_{\\text{MASS}} = \\mu R = \\var{mu} \\times \\var{R} = \\var{precround(mu*R,3)} \\ \\mathrm{N}$. Therefore to achieve movement $P$ must exceed $\\var{precround(mu*R,3)} \\ \\mathrm{N}$.
\nNote that if $ \\tan \\alpha = \\frac{\\var{O}}{\\var{A}}$ then using SOHCAHTOA where $O = \\var{O}$, $A = \\var{A}$ and $H = \\sqrt{\\var{O}^2 + \\var{A}^2}$ we have that $\\sin \\alpha = \\frac{\\var{O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}$ and $\\cos \\alpha = \\frac{\\var{A}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}$.
\n$P$ is applied at an angle above the horizontal as shown in the diagram. We can resolve $F=ma$ in the upward direction where acceleration $a=0$.
\n\\begin{align} R + P \\sin \\alpha - mg & = ma \\\\
R + \\frac{\\var{O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}P - (\\var{mass} \\times 9.8)& = 0 \\\\
R & = \\var{mass*9.8} - \\frac{\\var{O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}P
\\end{align}
We can now resolve $F=ma$ in the horizontal direction, where acceleration is also zero and $F_{\\text{MAX}} = \\mu R = \\var{mu} \\times \\left( \\var{mass*9.8} - \\frac{\\var{O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}P \\right)$.
\n\\begin{align} P \\cos \\alpha - F_{\\text{MAX}} &= ma \\\\
P \\cos \\alpha & = F_{\\text{MAX}} \\\\
\\frac{\\var{A}}{\\sqrt{\\var{O}^2 + \\var{A}^2}} P & = \\var{mu} \\times \\left( \\var{mass*9.8} - \\frac{\\var{O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}P \\right) \\\\
& = \\var{mu*mass*9.8} - \\frac{\\var{mu*O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}} P \\\\
\\left( \\frac{\\var{A}}{\\var{precround(H,3)}} + \\var{precround(mu*O/H,3)} \\right) P & = \\var{mu*mass*9.8} \\\\
P & = \\frac{ \\var{mu*mass*9.8}}{ \\left( \\frac{\\var{A}}{\\var{precround(H,3)}} + \\var{precround(mu*O/H,3)} \\right)} \\\\
& = \\var{precround((mu*mass*9.8)/(A/H + mu*O/H),3)} \\end{align}
Therefore $P$ must exceed $\\var{precround((mu*mass*9.8)/(A/H + mu*O/H),3)} \\ \\mathrm{N}$ in order for the mass to start moving.
\nNote that if $\\tan \\alpha = \\frac{\\var{O2}}{\\var{A2}}$ then using SOHCAHTOA where $O = \\var{O2}$, $A = \\var{A2}$ and $H = \\sqrt{\\var{O2}^2 + \\var{A2}^2}$ we have that $\\sin \\alpha = \\frac{\\var{O2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}$ and $\\cos \\alpha = \\frac{\\var{A2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}$.
\n![](\"resources/question-resources/force_component_image_7.png\"/)
We can resolve $F = ma$ in the upward direction where acceleration is zero.
\n\\begin{align}
R - mg - P \\sin \\alpha & = m \\times 0 \\\\
R & = mg + P \\sin\\alpha \\\\
& = \\var{mass*9.8} + \\frac{\\var{O2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}P
\\end{align}
From this we can calculate $F_{\\text{MAX}} = \\mu \\times R = \\var{mu} \\times \\left(\\var{mass*9.8} + \\frac{\\var{O2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}P\\right)$.
\nWe then resolve $F = ma$ in the horizontal direction, where once again acceleration is zero.
\n\\begin{align} P \\cos \\alpha - F_{\\text{MAX}} & = m \\times 0 \\\\
\\frac{\\var{A2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}P & = \\var{mu}\\left(\\var{mass*9.8} + \\frac{\\var{O2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}P\\right) \\\\
\\left(\\frac{\\var{A2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}} - \\frac{\\var{mu} \\times \\var{O2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}\\right)P & = \\var{mu} \\times \\var{mass*9.8} \\\\
\\var{precround((A2/H2) - (mu*O2/H),3)} P & = \\var{mu*9.8*mass} \\\\
P & = \\var{precround((mu*9.8*mass)/(A2/H2 - mu*O2/H),3)} \\end{align}
Therefore $P$ must exceed $\\var{precround((mu*9.8*mass)/(A2/H2 - mu*O2/H),3)}N$ in order for the mass to start moving.
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", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(mu*mass*g)/(cos1+mu*sine1)", "minValue": "(mu*mass*g)/(cos1+mu*sine1)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "Find the magnitude in Newtons (to 3 d.p.) that $P$ must exceed to achieve movement if $P$ is applied at an angle $\\alpha$ below the horizontal, where $\\tan \\alpha = \\frac{\\var{O2}}{\\var{A2}}$.
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\nThe acceleration due to gravity is $\\var{g} \\, \\mathrm{ms^{-2}}$.
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