// Numbas version: finer_feedback_settings {"name": "Dynamics - maximum frictional force", "duration": 0, "metadata": {"description": "

To be used on the Mechanics wiki page under Dynamics - maximum frictional force page

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a)

The maximum frictional force is $F_{\\text{MAX}} = \\mu R$, where $R$ is found by resolving the forces in the upwards direction. Here, acceleration is zero.

\\begin{align}
F &= ma \\\\
R - mg & = ma \\\\
R - \\left(\\var{mass} \\times 9.8 \\right) & = \\var{mass} \\times 0 \\\\
R & = \\var{mass} \\times 9.8 \\\\
&= \\var{mass*9.8}
\\end{align}

So $R = \\var{mass*9.8}$.

Therefore when the coefficient of friction $\\mu = \\var{friction1}$ we have

\\begin{align}
F_{\\text{MAX}} & = \\mu R \\\\
& = \\var{friction1} \\times \\var{mass*9.8} \\\\
& = \\var{precround(friction1*mass*9.8,3)}
\\end{align}

So the maximum frictional force is $F_{\\text{MAX}} = \\var{precround(friction1*mass*9.8,3)}N.$

b)

We have from part a) that $R = \\var{mass*9.8}$.

Therefore when the coefficient of friction $\\mu = \\var{friction2}$ we have

\\begin{align}
F_{\\text{MAX}} & = \\mu R \\\\
& = \\var{friction2} \\times \\var{mass*9.8} \\\\
& = \\var{precround(friction2*mass*9.8,3)}
\\end{align}

So the maximum frictional force is $F_{\\text{MAX}} = \\var{precround(friction2*mass*9.8,3)}N.$

c)

Now we have that $F_{\\text{MAX}} = \\var{precround(friction2*mass*9.8,3)}N$ so any horizontal force less than or equal to $F_{\\text{MAX}}$ will not move the mass.

The frictional force acting on the mass will be equal to the horizontal force in order to prevent the mass from moving.

d)

The force applied to the mass and the frictional force cancel each other out, so acceleration is zero.

e)

Now we have that $F_{\\text{MAX}} = \\var{precround(friction2*mass*9.8,3)}N$ so any horizontal force less than or equal to $F_{\\text{MAX}}$ will not move the mass.

When the horizontal force is equal to the maximum frictional force the friction will need to be at its maximum value to stop the mass from moving.

f)

The mass is in limiting equilibrium but still isn't moving so acceleration is zero.

g)

Now that a horizontal force greater than $F_{\\text{MAX}}$ is applied the mass will move.

The friction will be unable to stop the mass moving but it will remain at its maximum value of $F_{\\text{MAX}}$.

h)

The acceleration of the mass can be found by resolving $F = ma$.

\\begin{align}
F & = ma \\\\
\\var{2*a + Fmax} - \\var{Fmax} & = \\var{mass} \\times a \\\\
a & = \\frac{\\var{2*a + Fmax} - \\var{Fmax}}{\\var{mass}} \\\\
& = \\var{precround(2*a/mass,3)}
\\end{align}

So the acceleration of the mass is $\\var{precround(2*a/mass,3)} \\, \\mathrm{ms^{-2}}$.

", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "

Find, in $\\mathrm{N}$ to 3 decimal places, the maximum frictional force which can act on the mass when the coefficient of friction is $\\var{friction1}$.

", "precisionMessage": "

You have not given your answer to the correct precision.

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Find, in $\\mathrm{N}$ to 3 decimal places, the maximum frictional force which can act on the mass when the coefficient of friction is $\\var{friction2}$.

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Suppose a horizontal force of $\\var{P1}N$ is applied to the mass. The coefficient of friction is still $\\var{friction2}$.

What is the magnitude of the frictional force acting on the mass, in $\\mathrm{N}$ to 3 decimal places?

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What is the acceleration of the mass, in $\\mathrm{ms^{-2}}$ to 3 decimal places?

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Suppose a horizontal force of $\\var{Fmax}N$ is applied to the mass. The coefficient of friction is still $\\var{friction2}$.

What is the magnitude of the frictional force acting on the mass, in $\\mathrm{N}$ to 3 decimal places?

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What is the acceleration of the mass, in $\\mathrm{ms^{-2}}$ to 3 decimal places?

Suppose a horizontal force of $\\var{Fmax+2*a}N$ is applied to the mass. The coefficient of friction is still $\\var{friction2}$.

What is the magnitude of the frictional force acting on the mass, in $\\mathrm{N}$ to 3 decimal places?

What is the acceleration of the mass, in $\\mathrm{ms^{-2}}$ to 3 decimal places?

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "2*a/mass", "minValue": "2*a/mass", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "

A $\\var{mass} \\, \\mathrm{kg}$ mass is resting on a rough horizontal plane. The acceleration due to gravity is $\\var{g} \\mathrm{ms^{-2}}$.

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Maximum frictional force when coefficient of friction is friction2

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A block lies on rough horizontal ground. The coefficient of friction is given and students find the maximum value of friction $F_{\\text{MAX}} = \\mu R$ to determine if the block will move when three different magnitudes of force act on the block horizontally.

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a)

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 $F_{\\text{MASS}}=\\mu R$.

In the upwards direction we have

\\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 $\\var{precround(mu*R,3)} \\ \\mathrm{N}$.

b)

Note 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 $\\cos \\alpha = \\frac{\\var{A}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}$.

$P$ is applied at an angle above the horizontal as shown in the diagram. We can resolve $F=ma$ in the upward direction where acceleration $a=0$.

\\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 $F_{\\text{MAX}} = \\mu R = \\var{mu} \\times \\left( \\var{mass*9.8} - \\frac{\\var{O}}{\\sqrt{\\var{O}^2 + \\var{A}^2}}P \\right)$.

\\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.

c)

Note 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 $\\cos \\alpha = \\frac{\\var{A2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}$.

We can resolve $F = ma$ in the upward direction where acceleration is zero.

\\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 $F_{\\text{MAX}} = \\mu \\times R = \\var{mu} \\times \\left(\\var{mass*9.8} + \\frac{\\var{O2}}{\\sqrt{\\var{O2}^2 + \\var{A2}^2}}P\\right)$.

We then resolve $F = ma$ in the horizontal direction, where once again acceleration is zero.

\\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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Find the magnitude in Newtons (to 3 d.p.) that $P$ must exceed to achieve movement if $P$ is applied horizontally.

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You have not given your answer to the correct precision.

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Find the magnitude in Newtons (to 3 d.p.) that $P$ must exceed to achieve movement if $P$ is applied at an angle $\\alpha$ above the horizontal, where $ \\tan \\alpha = \\frac{\\var{O}}{\\var{A}}$.

", "precisionMessage": "

You have not given your answer to the correct precision.

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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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A block of mass $\\var{mass} \\ \\mathrm{kg}$ lies at rest on a rough horizontal floor. The coefficient of friction between the mass and the floor is $\\var{mu}$. A force $P \\ \\mathrm{N}$ is applied to the mass which can push or pull the mass horizontally along.

The acceleration due to gravity is $\\var{g} \\, \\mathrm{ms^{-2}}$.

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Adjacent side in the first angle.

"}, "O2_raw": {"definition": "random(1..5#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "O2_raw", "description": "

Opposite side in the second angle.

A random number is generated, then we divide by the gcd with A2_raw to make sure they're coprime.

"}, "g": {"definition": "9.8", "templateType": "number", "group": "Ungrouped variables", "name": "g", "description": "

acceleration due to gravity

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Hypotenuse in the second angle.

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Hypotenuse in the first angle.

"}, "sine2": {"definition": "O2/H2", "templateType": "anything", "group": "Ungrouped variables", "name": "sine2", "description": "

Sine of the second angle.

"}, "A_raw": {"definition": "random(7..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "A_raw", "description": "

adjacent side in the first angle.

A random number is generated, then we divide by the gcd with O_raw to make sure they're coprime.

"}, "O": {"definition": "O_raw/gcd(O_raw,A_raw)", "templateType": "anything", "group": "Ungrouped variables", "name": "O", "description": "

Opposite side in the first angle.

"}, "O2": {"definition": "O2_raw/gcd(O2_raw,A2_raw)", "templateType": "anything", "group": "Ungrouped variables", "name": "O2", "description": "

Opposite side in the second angle.

"}, "mu": {"definition": "random(0.01..0.99#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "mu", "description": "

coefficient of friction

"}, "R": {"definition": "mass*g", "templateType": "anything", "group": "Ungrouped variables", "name": "R", "description": "

reaction force against the block

"}, "mass": {"definition": "random(3..10#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "mass", "description": "

mass of the block

"}, "O_raw": {"definition": "random(1..6#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "O_raw", "description": "

Opposite side in the first angle.

A random number is generated, then we divide by the gcd with A_raw to make sure they're coprime.

"}, "cos1": {"definition": "A/H", "templateType": "anything", "group": "Ungrouped variables", "name": "cos1", "description": "

Cosine of the first angle

"}, "sine1": {"definition": "O/H", "templateType": "anything", "group": "Ungrouped variables", "name": "sine1", "description": "

Sine of the first angle

"}, "A2_raw": {"definition": "random(6..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "A2_raw", "description": "

Adjacent side in the second angle.

A random number is generated, then we divide by the gcd with O2_raw to make sure they're coprime.

"}, "max_friction": {"definition": "mu*R", "templateType": "anything", "group": "Ungrouped variables", "name": "max_friction", "description": "

Maximum frictional force on the object

"}, "A2": {"definition": "A2_raw/gcd(O2_raw,A2_raw)", "templateType": "anything", "group": "Ungrouped variables", "name": "A2", "description": "

Adjacent side in the second angle.

"}, "cos2": {"definition": "A2/H2", "templateType": "anything", "group": "Ungrouped variables", "name": "cos2", "description": "

Cosine of the second angle.

"}}, "metadata": {"description": "

A block lies on rough horizontal ground. The coefficient of friction is given and students find the maximum value of friction $F_{\\text{MAX}} = \\mu R$. This is then used to determine the magnitude of force that must be exceeded to achieve movement when the force is applied horizontally and at two different angles to the horizontal.

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