![](\"resources/question-resources/improvement1.png\"/)
You have not given your answer to the correct precision.
", "precisionPartialCredit": 0, "correctAnswerStyle": "plain", "minValue": "K*a+FA", "mustBeReducedPC": 0, "allowFractions": false, "scripts": {}, "extendBaseMarkingAlgorithm": true, "correctAnswerFraction": false, "customMarkingAlgorithm": "", "marks": 1, "notationStyles": ["plain", "en", "si-en"], "precision": "3", "showCorrectAnswer": true, "type": "numberentry", "showPrecisionHint": false, "unitTests": [], "mustBeReduced": false, "maxValue": "K*a+FA"}], "variableReplacements": [], "sortAnswers": false, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "prompt": "Suppose that the acceleration $a = \\var{a}$ and the mass of the particle $K = \\var{K} \\ \\mathrm{kg}$. Find the force $C \\ \\mathrm{N}$ if $A = \\var{FA}\\ \\mathrm{N}$.
\n$C = $ [[0]] $\\mathrm{N}$
", "unitTests": [], "scripts": {}, "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true}, {"showFeedbackIcon": true, "showPrecisionHint": false, "variableReplacements": [], "strictPrecision": false, "variableReplacementStrategy": "originalfirst", "precisionType": "dp", "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "correctAnswerStyle": "plain", "minValue": "FD*9.8", "mustBeReducedPC": 0, "allowFractions": false, "scripts": {}, "extendBaseMarkingAlgorithm": true, "correctAnswerFraction": false, "customMarkingAlgorithm": "", "marks": 1, "notationStyles": ["plain", "en", "si-en"], "precision": "3", "showCorrectAnswer": true, "type": "numberentry", "prompt": "Find the magnitude of the force $B$ if $D =\\simplify{{FD}g} \\ \\mathrm{N}$.
", "unitTests": [], "mustBeReduced": false, "maxValue": "FD*9.8"}, {"showFeedbackIcon": true, "showPrecisionHint": false, "variableReplacements": [], "strictPrecision": false, "variableReplacementStrategy": "originalfirst", "precisionType": "dp", "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "correctAnswerStyle": "plain", "minValue": "FA2-K2*a2", "mustBeReducedPC": 0, "allowFractions": false, "scripts": {}, "extendBaseMarkingAlgorithm": true, "correctAnswerFraction": false, "customMarkingAlgorithm": "", "marks": 1, "notationStyles": ["plain", "en", "si-en"], "precision": "3", "showCorrectAnswer": true, "type": "numberentry", "prompt": "Consider another particle which has mass $\\var{K2} \\ \\mathrm{kg}$ and is now decelerating horizontally at $\\var{a2}\\ \\mathrm{ms^{-2}}$. If $A = \\var{FA2}\\ \\mathrm{N}$ what is $C$?
", "unitTests": [], "mustBeReduced": false, "maxValue": "FA2-K2*a2"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Using $F=ma$ to find magnitudes of unknown forces acting on an accelerating body.
"}, "name": "Simon's copy of Particle in vertical equilibrium, accelerating horizontally", "advice": "To find the resultant force in a certain direction when there is more than one force acting on an object you can resolve the forces. You resolve in the direction of acceleration. Using the letter $R$, with an arrow will indicate the direction the forces are being resolved in, for example $R(\\rightarrow)$.
\nIn this case $R(\\rightarrow)$ means apply $F=ma$ in the horizontal direction, where we treat forces and accelerations to the right as positive, and to the left as negative. So we need to resolve the equation
\n\\begin{align}F & = ma \\\\
C - \\var{FA} & = \\var{K} \\times \\var{a} \\\\
C & = \\var{K*a} + \\var{FA} \\\\
& = \\var{K*a + FA}. \\end{align}
Here the forces, $F=C - \\var{FA}$ because $\\var{FA}$ is acting in the opposite direction to $C$. The mass is $m=\\var{K}$ and the acceleration is $a=\\var{a}$. Therefore $C = \\var{K*a+FA}\\ \\mathrm{N}$.
\n\nIn this case we resolve in the direction perpendicular to acceleration so we use $R(\\uparrow)$ which means apply $F=ma$ in the vertical direction, where acceleration, $a=0$ because there is no vertical acceration.
\n\\begin{align} F & = ma \\\\
B - \\var{FD}g & = \\var{K} \\times 0 \\\\
B & = 0 + \\left(\\var{FD} \\times 9.8\\right) \\\\
& = \\var{FD*9.8} \\end{align}
Here the forces, $F=B - \\var{FD}g$ because $\\var{FD}g$ is acting downwards which is the opposite direction to $B$. Therefore $B = \\var{FD*9.8} \\ \\mathrm{N}$.
It is easier to take the positive direction as the direction of the acceleration. Therefore we use $R(\\leftarrow)$ because we are decelerating.
\n\\begin{align} F & = ma \\\\
\\var{FA2} - C & = \\var{K2} \\times \\var{a2} \\\\
C & = \\var{FA2} - \\left( \\var{K2} \\times \\var{a2} \\right) \\\\
& = \\var{FA2 - K2*a2}. \\end{align}
Here the forces, $F = \\var{FA2} - C$ because $\\var{FA2}$ is in the direction of acceleration and $C$ is acting in the opposite direction. The mass is $m = \\var{K2}$ and $a=\\var{a2}$. Therefore $C = \\var{FA2 - K2*a2} \\ \\mathrm{N}$.
\n", "preamble": {"css": "", "js": ""}, "statement": "Consider the following diagram of a particle of mass $K \\ \\mathrm{kg}$, moving left-to-right. The forces $A, B, C$ and $D$ are acting upon it, producing a horizontal acceleration of $a \\, \\mathrm{ms^{-2}}$
\n
The acceleration due to gravity is $9.8 \\, \\mathrm{ms}^{-2}$. Answer all the following questions to 3 decimal places.
", "extensions": [], "ungrouped_variables": ["a", "K", "FA", "FD", "a2", "K2", "FA2", "downforce", "FD2"], "variables": {"K": {"group": "Ungrouped variables", "templateType": "randrange", "name": "K", "definition": "random(2..9#1)", "description": "mass
"}, "downforce": {"group": "Ungrouped variables", "templateType": "randrange", "name": "downforce", "definition": "random(3..23#1)", "description": "downforce for part d)
"}, "K2": {"group": "Ungrouped variables", "templateType": "randrange", "name": "K2", "definition": "random(1..5#1)", "description": "mass 2
"}, "a": {"group": "Ungrouped variables", "templateType": "randrange", "name": "a", "definition": "random(0.5..7#0.5)", "description": "acceleration
"}, "FD": {"group": "Ungrouped variables", "templateType": "anything", "name": "FD", "definition": "K", "description": "Force D
"}, "FD2": {"group": "Ungrouped variables", "templateType": "anything", "name": "FD2", "definition": "K2", "description": "downforce D 2
"}, "FA": {"group": "Ungrouped variables", "templateType": "randrange", "name": "FA", "definition": "random(0.5..4.5#0.5)", "description": "Force A
"}, "FA2": {"group": "Ungrouped variables", "templateType": "randrange", "name": "FA2", "definition": "random(40..80#0.5)", "description": ""}, "a2": {"group": "Ungrouped variables", "templateType": "randrange", "name": "a2", "definition": "random(0.25..5#0.25)", "description": "acceleration
"}}, "functions": {}, "variable_groups": [], "rulesets": {}, "type": "question", "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": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "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": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}