// Numbas version: finer_feedback_settings {"name": "Maximum frictional force acting on a mass", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["mass", "friction1", "R", "g", "friction2", "Fmax", "P1", "a"], "name": "Maximum frictional force acting on a mass", "tags": [], "advice": "

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?

", "allowFractions": false, "variableReplacements": [], "maxValue": "P1", "minValue": "P1", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "

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

", "allowFractions": false, "variableReplacements": [], "maxValue": "0", "minValue": "0", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"prompt": "

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}], "extensions": [], "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.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}]}]}], "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}]}