// Numbas version: exam_results_page_options {"name": "Equilibrium of a three-force body: triangle", "extensions": ["geogebra", "quantities", "weh"], "custom_part_types": [{"source": {"pk": 19, "author": {"name": "William Haynes", "pk": 2530}, "edit_page": "/part_type/19/edit"}, "name": "Engineering Accuracy with units", "short_name": "engineering-answer", "description": "

A value with units marked right if within an adjustable % error of the correct value.  Marked close if within a wider margin of error.

Does clumsy substitution to

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1. replace '-' with ' '

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2. replace '°' with ' deg'

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to allow answers like 10 ft-lb and 30°

", "name": "student_units"}, {"definition": "try(\ncompatible(quantity(1, student_units),correct_units),\nmsg,\nfeedback(msg);false)\n", "description": "", "name": "good_units"}, {"definition": "switch(not good_units, \n student_scalar * correct_units, \n not right_sign,\n -quantity(student_scalar, student_units),\n quantity(student_scalar,student_units)\n)\n \n", "description": "

This fixes the student answer for two common errors.

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If student_units are wrong  - replace with correct units

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If student_scalar has the wrong sign - replace with right sign

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If student makes both errors, only one gets fixed.

", "name": "student_quantity"}, {"definition": "try(\nscalar(abs((correct_quantity - student_quantity)/correct_quantity))*100 \n,msg,\nif(student_quantity=correct_quantity,0,100))\n ", "description": "", "name": "percent_error"}, {"definition": "percent_error <= settings['right']\n", "description": "", "name": "right"}, {"definition": "right_sign and percent_error <= settings['close']", "description": "

Only marked close if the student actually has the right sign.

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99  N is accepted.", "default_value": "75", "label": "Close with units.", "help_url": "", "name": "C1"}, {"input_type": "percent", "hint": "Partial credit for forgetting units or using wrong sign.
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This value would be close if the expected units were provided.  If the correct answer is 100 N, and close is ±1%,
99 is accepted.", "default_value": "25", "label": "Close, no units.", "help_url": "", "name": "C3"}], "public_availability": "restricted", "published": false, "extensions": ["quantities"]}, {"source": {"pk": 24, "author": {"name": "William Haynes", "pk": 2530}, "edit_page": "/part_type/24/edit"}, "name": "Angle quantity", "short_name": "angle-quantity-from-reference", "description": "

Angle as a quantity in degrees.

Find the reactions of a rigid body (a triangular plate) at a pin and roller, using the three-force body principle.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "extensions": ["geogebra", "quantities", "weh"], "preamble": {"css": "", "js": "Numbas.extensions.weh.scope.ggbApplet.then(function(applet){\n applet.setValue('show',false);\n applet.setValue('debug',false);\n});"}, "variablesTest": {"condition": "not (theta in [0,180,180+alpha1,360+alpha1,360-alpha1])", "maxRuns": 100}, "rulesets": {}, "variables": {"theta": {"definition": "random(300..380#5)", "templateType": "anything", "description": "

reference direction.  Set by student choice.

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this is the y-coordinate of the intersection point

", "group": "Ungrouped variables", "name": "h"}}, "name": "Equilibrium of a three-force body: triangle", "variable_groups": [{"variables": ["theta", "units", "alpha1", "FB", "L"], "name": "Inputs"}], "advice": "
\n
1. Draw a free body diagram.
2. \n
3. Apply $\\Sigma M_A$=0 to find the reaction at $C$.  There's no x-component at $C$, because the support there is a roller.  This should have been indicated on your free body diagram.
4. \n
5. Once $C$ and is known, apply $F_x = 0$ and $F_y=0$ to find components $A_x$ and $A_y$.
6. \n
7. With $A_x$ and $A_y$ known, use the pathagorean theorem to calculate the magnitude of $A$, and use trig to get the its direction.
8. \n
", "statement": "

The triangular plate is secured by a pin at A and a roller at C and is subjected to a {FB} load as shown.

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Determine the reactions at the pin and the roller using the three-force body principle.

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A three-force body is an object acted upon by exactly three forces.  When a three-force body is in equilibrium the lines of action of the three forces must either intersect at a common point or be parallel to each other.  We can use this idea to find the reaction forces for three-force bodies.

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In this problem the lines of action of force B and force C are known and their intersection point X may be determined.

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Use the given geometric information to determine distance h.

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$h$ = [[0]]

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h = {h}

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With h known, the direction of force A can be found since its line of action must pass through X.

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Use the given geometric information and determine $\\angle CAX$, which we will call $\\theta_A$

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$\\theta_A=$ [[0]]

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$\\theta_a$ = {theta_a}

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The object is in equilibrium so when the three forces are added tip-to-tail they form a closed triangle as shown below.

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Determine the three angles in the force triangle.

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$\\alpha$ = [[0]]  $\\beta$ = [[1]]   $\\gamma$ = [[2]]

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$\\alpha$ = {alpha} $\\beta$ = {beta}  $\\gamma$ = {gamma}

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A = {FA}  B = {FB}  C = {FC}

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Use the law of sines to determine the magnitudes of forces A and C.

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A = [[0]]

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C = [[1]]

\n

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