Adjusts all angles to 0 < $\\theta$ < 360.
\nAccepts '°' and 'deg' as units.
\nPenalizes if not close enough or no units.
\n90° = -270° = 450°
", "help_url": "", "input_widget": "string", "input_options": {"correctAnswer": "plain_string(settings['expected_answer']) ", "hint": {"static": true, "value": ""}, "allowEmpty": {"static": true, "value": false}}, "can_be_gap": true, "can_be_step": true, "marking_script": "original_student_scalar:\nmatchnumber(studentAnswer,['plain','en'])[1]\n\nstudent_scalar:\nmod(original_student_scalar,360)\n\n\nstudent_unit:\nstudentAnswer[len(matchnumber(studentAnswer,['plain','en'])[0])..len(studentAnswer)]\n\ninterpreted_unit:\nif(trim(student_unit)='\u00b0','deg',student_unit)\n\ninterpreted_answer:\nqty(mod(student_scalar,360),'deg')\n\nclose:\nwithintolerance(student_scalar, correct_scalar,decimal(settings['close_tol']))\n\ncorrect_scalar:\nmod(scalar(settings['expected_answer']),360)\n\nright:\nwithintolerance(student_scalar, correct_scalar, decimal(settings['right_tol']))\n\ngood_unit:\nsame(qty(1,interpreted_unit),qty(1,'deg'))\n\nmark:\nassert(close,incorrect('Incorrect.');end());\nif(right,correct('Correct angle.'), set_credit(1 - settings['close_penalty'],'Angle is close.'));\nassert(good_unit,sub_credit(settings['unit_penalty'], 'Missing or incorrect units.'))", "marking_notes": [{"name": "original_student_scalar", "description": "Retuns the scalar part of students answer (which is a quantity) as a number.
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", "definition": "if(trim(student_unit)='\u00b0','deg',student_unit)"}, {"name": "interpreted_answer", "description": "A value representing the student's answer to this part.", "definition": "qty(mod(student_scalar,360),'deg')"}, {"name": "close", "description": "", "definition": "withintolerance(student_scalar, correct_scalar,decimal(settings['close_tol']))"}, {"name": "correct_scalar", "description": "Normalize expected_answer with mod 360
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", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "{applet}
\nThree forces, $A = \\var{A F_units}$, $C = \\var{C F_units}$ and $D = \\var{F F_units}$, act on the rigid, bent bar which can pivot freely on a pin at $B$.
\nThe segment lengths are: $AB = \\var{ab D_units}$, $BC = \\var{bc D_units}$, and $CD= \\var{cd D_units}$.
", "advice": "From the geometry of triangle $BCD$:
\n$\\phi = \\tan^{-1}\\left( \\dfrac{BC}{CD}\\right) =\\var{precround(phi,1)}°$
$d = \\sqrt{BC^2 +CD^2} = \\var{siground(abs(r_d),4) D_units}$
From a moment balance around point $B$, the clockwise moment equals the counterclockwise moments:
\n$\\begin{align}
M_D &= M_A + M_C \\\\
&= (\\var{AB})\\ A_y + (\\var{BC})\\ C_x\\\\
&= \\simplify[!collectNumbers]{({siground(M_A[2],4)} +{siground(M_C[2],4)})}\\\\
&= \\var{siground(-M D_units F_units,4)} \\text{ clockwise}
\\end{align}$
With the required moment $M_D$ and the magnitude of force $D$ known, the perpendicular distance and the angle between segment $BD$ and force $D$ are:
\n$d_\\perp = M_D / D = \\dfrac{ \\var{siground(-M D_units F_units,4)}}{\\var{F F_units}} = \\var{siground(dperp,4) D_units}$
$\\alpha =\\sin^{-1}\\left( \\dfrac{d_\\perp}{d}\\right)=\\var{precround(alpha1,1)}°$
Since $\\sin \\alpha = \\sin (180°-\\alpha)$ another possible angle is:
\n$\\alpha' = (180° - \\var{precround(alpha1,1)}) = \\var{precround(alpha2,1)}°$
\nFinally the angle from the positive $x$ axis to force $D$ can be determined. There are two possible solutions:
\n$\\theta =180° - \\alpha - \\phi = \\var{precround(-theta1,1)}° $ or
$\\theta = 180 - \\alpha' - \\phi = \\var{precround(-theta2,1)}$
Note: You can use the red slider on the diagram to switch between the two possible solutions.
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\n$\\theta = $ [[0]] (Measure $\\theta$ CW from the positive $x$ axis.)
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