![](\"resources/question-resources/pulley2.png\")
An intermediary value to simplify the advice.
"}, "m2": {"definition": "random(0.5..5.5#0.25)", "name": "m2", "templateType": "randrange", "group": "Ungrouped variables", "description": ""}, "theta": {"definition": "random(10..70#1)", "name": "theta", "templateType": "randrange", "group": "Ungrouped variables", "description": ""}, "acceleration": {"definition": "(m1*9.8 - m2*9.8*cos(radians(90-theta))-mu*R)/(m1+m2)", "name": "acceleration", "templateType": "anything", "group": "Ungrouped variables", "description": ""}}, "ungrouped_variables": ["m1", "m2", "theta", "mu", "R", "acceleration", "tension", "t"], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Two particles connected by a string which passes over a pulley at the top of an inclined plane. Find the acceleration of the masses and the tension in the string. Can not model the whole system as a single particle.
"}, "variable_groups": [], "extensions": [], "tags": [], "name": "Ed's copy of Masses connected through a pulley", "preamble": {"css": "", "js": ""}, "functions": {}, "showQuestionGroupNames": false, "rulesets": {}, "type": "question", "statement": "A light inextensible string connects two boxes $A$ and $B$, of masses $m_1 = \\var{m1} \\mathrm{kg}$ and $m_2 = \\var{m2} \\mathrm{kg}$ respectively.
\nAt the top of a rough inclined plane there is fixed a small smooth pulley over which the string passes. The plane is inclined to the horizontal at an angle $\\theta = \\var{theta}^{\\circ}$.
\nBox $A$ hangs on the edge of the plane with the string vertical and taut, whereas box $B$ rests on the inclined plane.
\n
You are told that the coefficient of friction between $B$ and the plane is $\\mu = \\var{mu}$ and the acceleration due to gravity is $g = 9.8\\mathrm{ms^{-2}}$.
", "advice": "We can draw a diagram to show all the forces acting on each box and their accelerations. The surface is rough and friction will be limiting so equal to $\\mu R$. ![](\"resources/question-resources/pulley3.png\")
Here, the mass of box $A$ is $m_1 = \\var{m1}\\mathrm{kg}$ and the mass of box $B$ is $m_2 = \\var{m2}\\mathrm{kg}$. The acceleration acts in the direction shown as box $A$ is heavier.
\nAs the boxes are moving in different directions we can not model the whole system as a single particle.
\nTo find the normal reaction $R$ between box $B$ and the plane we resolve the forces perpendicular to the plane, where there is no acceleration.
\n\\begin{align}
R - m_2g \\cos \\theta & = 0, \\\\
R &= m_2 g \\cos \\theta, \\\\
& = \\var{m2} \\times 9.8 \\cos ( \\var{theta}{^\\circ}), \\\\
& = \\var{precround(R,3)} \\mathrm{N}.
\\end{align}
The normal reaction between the box $B$ and the plane is $\\var{precround(R,3)} \\mathrm{N}$.
\nTo find the acceleration we treat the boxes separately to get two equations involving the unknowns $T$ and $a$, then add them in order to cancel $T$.
\nLooking at box $B$ and resolving parallel to the plane in the direction of acceleration we have equation (1):
\n\\begin{align}
T - m_2g \\cos(90^{\\circ} - \\theta) - \\mu R & = m_2 a, \\\\
T - (\\var{m2} \\times g \\cos(\\var{90-theta}^{\\circ})) - (\\var{mu} \\times \\var{precround(R,3)}) & = \\var{m2}a, \\\\
\\simplify{T-{precround(t,3)}} &= \\var{m2}a.
\\end{align}
Looking at box $A$ and resolving in the direction of acceleration we have equation (2):
\n\\begin{align}
m_1g - T & = m_1a, \\\\
\\var{m1}g - T & = \\var{m1}a.
\\end{align}
Adding equations (1) and (2) gives
\n\\begin{align}
\\simplify[basic]{T - {precround(t,3)} +{m1}g -T} & = \\var{m2}a + \\var{m1}a, \\\\
\\var{m1}g - \\var{precround(m2*9.8*cos(radians(90-theta)) - mu*R,3)} & =(\\var{m2} + \\var{m1} )a, \\\\
\\var{precround(m1*9.8-m2*9.8*cos(radians(90-theta)) - mu*R,3)} & = \\var{m2+m1}a, \\\\
a & = \\var{precround((m1*9.8-m2*9.8*cos(radians(90-theta)) - mu*R)/(m2+m1),3)} \\mathrm{ms^{-2}}.
\\end{align}
The acceleration of the system is $\\var{precround(acceleration,3)} \\mathrm{ms^{-2}}$.
\nWe can find the tension in the string by substituting our value for acceleration into either equation (1) or (2).
\nThis gives
\n\\begin{align}
\\var{m1}g - T & = \\var{m1}a, \\\\
T & = \\var{m1}g - \\var{m1}a, \\\\
& = \\var{m1} (9.8 - \\var{precround(acceleration,3)}), \\\\
& = \\var{precround(m1*(9.8 - acceleration),3)} \\mathrm{N}.
\\end{align}
The tension in the string is $\\var{precround(tension,3)} \\mathrm{N}.$
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\n$R = $ [[0]]
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", "maxValue": "R", "scripts": {}, "type": "numberentry", "showPrecisionHint": false, "strictPrecision": true, "marks": 1}], "variableReplacements": []}, {"allowFractions": false, "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "variableReplacements": [{"variable": "R", "must_go_first": false, "part": "p0g0"}], "minValue": "(m1*9.8 - m2*9.8*cos(radians(90-theta))-mu*R)/(m1+m2)", "precision": "3", "precisionType": "dp", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "maxValue": "(m1*9.8 - m2*9.8*cos(radians(90-theta))-mu*R)/(m1+m2)", "prompt": "The system is released from rest. Using your value of $R$ from part a) find the acceleration of the system, in $\\mathrm{ms^{-2}}$ to 3 decimal places.
", "type": "numberentry", "scripts": {}, "showPrecisionHint": false, "strictPrecision": false, "marks": 1}, {"allowFractions": false, "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "variableReplacements": [{"variable": "acceleration", "must_go_first": false, "part": "p1"}], "minValue": "tension", "precision": "3", "precisionType": "dp", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.
", "maxValue": "tension", "prompt": "Using the answers to the previous parts, find the tension in the string, in Newtons ($\\mathrm{N}$) to 3 decimal places.
", "type": "numberentry", "scripts": {}, "showPrecisionHint": false, "strictPrecision": false, "marks": 1}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}, {"name": "Ed Southwood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2415/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}, {"name": "Ed Southwood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2415/"}]}