![](\"resources/question-resources/statics1.png\")
You have not given your answer to the correct precision.
", "extendBaseMarkingAlgorithm": true, "showPrecisionHint": false}], "marks": 0, "prompt": "$F_2 = $ [[0]]
\n"}, {"sortAnswers": false, "variableReplacements": [], "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "scripts": {}, "showFeedbackIcon": true, "type": "gapfill", "gaps": [{"precisionType": "dp", "mustBeReduced": false, "strictPrecision": false, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "customMarkingAlgorithm": "", "scripts": {}, "mustBeReducedPC": 0, "precision": "3", "showFeedbackIcon": true, "minValue": "F3", "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "precisionPartialCredit": 0, "showCorrectAnswer": true, "unitTests": [], "maxValue": "F3", "allowFractions": false, "marks": 1, "correctAnswerStyle": "plain", "correctAnswerFraction": false, "precisionMessage": "You have not given your answer to the correct precision.
", "extendBaseMarkingAlgorithm": true, "showPrecisionHint": false}], "marks": 0, "prompt": "$F_3 = $ [[0]]
\n"}], "preamble": {"css": "", "js": ""}, "variables": {"F1": {"group": "Ungrouped variables", "description": "", "templateType": "randrange", "definition": "random(2..4#0.25)", "name": "F1"}, "theta3": {"group": "Ungrouped variables", "description": "", "templateType": "randrange", "definition": "random(2..30#1)", "name": "theta3"}, "F2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "definition": "(F3*cos(radians(theta3))-F1*cos(radians(theta1)))/(cos(radians(theta2)))", "name": "F2"}, "check": {"group": "Ungrouped variables", "description": "Total horizontal force. Should be zero for equilibrium
", "templateType": "anything", "definition": "F1*cos(radians(theta1))+F2*cos(radians(theta2))-F3*cos(radians(theta3))", "name": "check"}, "theta2": {"group": "Ungrouped variables", "description": "", "templateType": "randrange", "definition": "random(10..70#1)", "name": "theta2"}, "check2": {"group": "Ungrouped variables", "description": "Total vertical force. Should be zero for equilbrium.
", "templateType": "anything", "definition": "F1*sin(radians(theta1))-F2*sin(radians(theta2))-F3*sin(radians(theta3))", "name": "check2"}, "F3": {"group": "Ungrouped variables", "description": "F3 calculated by the student
", "templateType": "anything", "definition": "(F1*sin(radians(theta1))+F1*(cos(radians(theta1))/cos(radians(theta2)))*sin(radians(theta2)))/((sin(radians(theta3))+(cos(radians(theta3))*sin(radians(theta2)))/cos(radians(theta2))))", "name": "F3"}, "theta1": {"group": "Ungrouped variables", "description": "", "templateType": "randrange", "definition": "random(10..60#1)", "name": "theta1"}}, "rulesets": {}, "functions": {}, "ungrouped_variables": ["F1", "theta1", "F2", "theta2", "F3", "theta3", "check", "check2"], "statement": "The diagram shows a particle in equilibrium under the forces shown.
\n
You are told that $F_1 = \\var{F1} \\ \\mathrm{N}$ and that the angles are $\\theta_1 = \\var{theta1}^{\\circ}, \\ \\theta_2 = \\var{theta2}$ and $\\theta_3 = \\var{theta3}^{\\circ}$.
\nBy resolving forces, find the magnitude of the forces $F_2 $ and $F_3 $, in Newtons, and input them to 3 d.p. below.
", "variable_groups": [], "tags": [], "variablesTest": {"condition": "F2>0.1 and F3>0.1", "maxRuns": 100}, "advice": "As the particle is in equilibrium we equate the sum of the forces to zero.
\nMethod 1
\nThe quickest method to find $F_2$ is to resolve in the direction of $F_3$. Then:
\n
$F_1 \\sin(\\theta_1-\\theta_3) = F_2 \\sin(\\theta_2+\\theta_3)$
$F_2 = F_1 \\frac{\\sin(\\theta_1-\\theta_3)}{\\sin(\\theta_2+\\theta_3)}$
\n$F_2 = \\var{F1} \\frac{\\sin(\\var{theta1}-\\var{theta3})}{\\sin(\\var{theta2}+\\var{theta3})}=\\var{F1} \\frac{\\sin(\\var{theta1-theta3})}{\\sin(\\var{theta2+theta3})}=\\var{precround(F1*sin((pi/180)*(theta1-theta3))/sin((pi/180)*(theta2+theta3)),3)}$
\n
Similarly, the quickest method to find $F_3$ is to resolve in the direction of $F_2$. Then:
$F_1 \\sin(\\theta_1+\\theta_2) = F_3 \\sin(\\theta_2+\\theta_3)$
$F_3 = F_1 \\frac{\\sin(\\theta_1+\\theta_2)}{\\sin(\\theta_2+\\theta_3)}$
\n$F_3 = \\var{F1} \\frac{\\sin(\\var{theta1}+\\var{theta2})}{\\sin(\\var{theta2}+\\var{theta3})}=\\var{F1} \\frac{\\sin(\\var{theta1+theta2})}{\\sin(\\var{theta2+theta3})}=\\var{precround(F1*sin((pi/180)*(theta1+theta2))/sin((pi/180)*(theta2+theta3)),3)}$
\n\n\n\nMethod 2
\nAlternatively, we can find $F_2$ and $F_3$ by resolving horizontally and vertically as follows:
\n\nResolving horizontally we have
\n\\begin{align}
F_1 \\cos \\theta_1 + F_2 \\cos \\theta_2 - F_3 \\cos \\theta_3 & = 0, \\\\
\\var{F1} \\cos \\var{theta1}^{\\circ} + F_2 \\cos \\var{theta2}^{\\circ} - F_3 \\cos \\var{theta3}^{\\circ} & = 0, \\\\
F_2 \\cos \\var{theta2}^{\\circ} & = F_3 \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}, \\\\
F_2 & = \\frac{F_3 \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}}.
\\end{align}
Resolving vertically, we have
\n\\begin{align}
F_1 \\sin \\theta_1 - F_2 \\sin \\theta_2 - F_3 \\sin \\theta_3 & = 0, \\\\
\\var{F1} \\sin \\var{theta1}^{\\circ} - F_2 \\sin \\var{theta2}^{\\circ} - F_3 \\sin \\var{theta3}^{\\circ} & = 0, \\\\
F_3 \\sin \\var{theta3}^{\\circ}& = \\var{F1} \\sin \\var{theta1}^{\\circ} - F_2 \\sin \\var{theta2}^{\\circ}.
\\end{align}
Substituting our value for $F_2$ into the formula $F_3 \\sin \\var{theta3}^{\\circ}$ and rearranging gives
\n\\begin{align}
F_3 \\sin \\var{theta3}^{\\circ} & = \\var{F1} \\sin \\var{theta1}^{\\circ} - F_2 \\sin \\var{theta2}^{\\circ} \\\\
F_3 \\sin \\var{theta3}^{\\circ} & = \\var{F1} \\sin \\var{theta1}^{\\circ} - \\left( \\frac{F_3 \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}}{\\cos \\var{theta2}^{\\circ}} \\right) \\sin \\var{theta2}^{\\circ}, \\\\
F_3 \\sin \\var{theta3}^{\\circ} & = \\var{F1} \\sin \\var{theta1}^{\\circ} - F_3 \\frac{ \\cos \\var{theta3}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}} \\sin \\var{theta2}^{\\circ} + \\var{F1} \\frac{ \\cos \\var{theta1}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}} \\sin \\var{theta2}^{\\circ}, \\\\
F_3 \\left( \\sin \\var{theta3}^{\\circ} + \\frac{ \\cos \\var{theta3}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}} \\sin \\var{theta2}^{\\circ} \\right) & = \\var{F1} \\sin \\var{theta1}^{\\circ} + \\var{F1} \\frac{ \\cos \\var{theta1}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}} \\sin \\var{theta2}^{\\circ} \\\\
F_3 & = \\frac{\\var{F1} \\sin \\var{theta1}^{\\circ} + \\var{F1} \\frac{ \\cos \\var{theta1}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}} \\sin \\var{theta2}^{\\circ}}{ \\sin \\var{theta3}^{\\circ} + \\frac{ \\cos \\var{theta3}^{\\circ}}{ \\cos \\var{theta2}^{\\circ}} \\sin \\var{theta2}^{\\circ}}.
\\end{align}
Therefore, to 3 s.f. we have $F_3 = \\var{precround(F3,3)} \\mathrm{N}. $
\nWe can then use our $F_3$ value in our equation for $F_2$
\n\\begin{align}
F_2 & = \\frac{F_3 \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}}{\\cos \\var{theta2}^{\\circ}}, \\\\
& = \\frac{\\var{precround(F3,3)} \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}}{\\cos \\var{theta2}^{\\circ}}, \\\\
& = \\var{precround(F2,3)} \\mathrm{N}.
\\end{align}
Three forces act on a particle. You're given the magnitude of one, and the directions of all three. Find the magnitudes of the other two forces.
", "licence": "Creative Commons Attribution 4.0 International"}, "name": "Simon's copy of Resolve forces on particle in equilibrium", "type": "question", "contributors": [{"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": "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/"}]}