// Numbas version: finer_feedback_settings {"name": "Resolve forces on particle in equilibrium", "extensions": [], "custom_part_types": [], "resources": [["question-resources/statics1.png", "/srv/numbas/media/question-resources/statics1.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["F1", "theta1", "F2", "theta2", "F3", "theta3", "check", "check2"], "name": "Resolve forces on particle in equilibrium", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

As the particle is in equilibrium we equate the sum of the forces to zero. 

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Resolving horizontally we have 

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\\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}

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Resolving vertically, we have

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\\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}

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Substituting our value for $F_2$ into the formula $F_3 \\sin \\var{theta3}^{\\circ}$ and rearranging gives

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\\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} 

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Therefore, to 3 s.f. we have $F_3 = \\var{siground(F3,3)} \\mathrm{N}. $

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We can then use our $F_3$ value in our equation for $F_2$

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\\begin{align} 
F_2 & = \\frac{F_3 \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}}{\\cos \\var{theta2}^{\\circ}}, \\\\
& = \\frac{\\var{siground(F3,3)} \\cos \\var{theta3}^{\\circ} - \\var{F1} \\cos \\var{theta1}^{\\circ}}{\\cos \\var{theta2}^{\\circ}}, \\\\
& = \\var{siground(F2,3)} \\mathrm{N}.
\\end{align}
  
 
              

                    

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

$F_2 = $ [[0]]

\n

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You have not given your answer to the correct precision.

", "allowFractions": false, "variableReplacements": [], "maxValue": "F2", "strictPrecision": false, "minValue": "F2", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"prompt": "

$F_3 = $ [[0]]

\n

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You have not given your answer to the correct precision.

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The diagram shows a particle in equilibrium under the forces shown.

\n

\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 $\\theta_3 = \\var{theta3}^{\\circ}$.

\n

By resolving horizontally and vertically, find the magnitude of the forces $F_2 $ and $F_3 $, in Newtons, and input them to 3 significant figures below.

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"theta2": {"definition": "random(10..70#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "theta2", "description": ""}, "F1": {"definition": "random(2..4#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "F1", "description": ""}, "F2": {"definition": "(F3*cos(radians(theta3))-F1*cos(radians(theta1)))/(cos(radians(theta2)))", "templateType": "anything", "group": "Ungrouped variables", "name": "F2", "description": ""}, "theta1": {"definition": "random(10..60#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "theta1", "description": ""}, "theta3": {"definition": "random(2..30#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "theta3", "description": ""}, "F3": {"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))))", "templateType": "anything", "group": "Ungrouped variables", "name": "F3", "description": "

F3 calculated by the student

"}, "check2": {"definition": "F1*sin(radians(theta1))-F2*sin(radians(theta2))-F3*sin(radians(theta3))", "templateType": "anything", "group": "Ungrouped variables", "name": "check2", "description": "

Total vertical force. Should be zero for equilbrium.

"}, "check": {"definition": "F1*cos(radians(theta1))+F2*cos(radians(theta2))-F3*cos(radians(theta3))", "templateType": "anything", "group": "Ungrouped variables", "name": "check", "description": "

Total horizontal force. Should be zero for equilibrium

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

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.

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