// Numbas version: finer_feedback_settings {"name": "Resolve forces into components", "extensions": [], "custom_part_types": [], "resources": [["question-resources/force_component_image_4.png", "/srv/numbas/media/question-resources/force_component_image_4.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "
Resolve each force from the positive $x$-direction (pointing to the right). For the force $P$ this is at $90^{ \\circ}$ to the $x$-axis therefore has a contribution of $P \\times \\cos 90^{\\circ} = 0$ to the sum of components.
\nThe force $F$ is at $(180-\\var{theta1})^{\\circ}=\\var{90 + 90 - theta1}^{\\circ}$ to the positive $x$-direction therefore has a contribution of $F \\times \\cos \\var{180-theta1}^{\\circ} = \\var{force1} \\times \\cos \\var{180-theta1}^{\\circ} = \\var{precround(force1*cos(radians(180-theta1)),3)}$ to the sum of components. This will be negative as you can imagine if you were moving in the positive $x$-direction this force is acting in the opposite direction and pulling you back!
\nThe force $Q$ is at $\\var{theta2}^{\\circ}$ to the positive $x$-direction therefore has a contribution of $Q \\times cos \\var{theta2}^{\\circ} = \\var{force3} \\times \\cos\\var{theta2}^{\\circ} = \\var{precround(force3*cos(radians(theta2)),3)}$. This is positive as it is acting in the same direction as the positive.
\nTherefore the sum of components in the $x$-direction is $0 - \\var{precround(-force1*cos(radians(180-theta1)),3)} + \\var{precround(force3*cos(radians(theta2)),3)} = \\var{precround(force1*cos(radians(180-theta1)) + force3*cos(radians(theta2)),3)}$.
\nResolve each force from the positive $y$-direction (upwards). For the force $P$ this is acting completely in the positive direction, at no angle. Therefore it's contribution is $\\var{force2}$. Note that this is the same as $\\var{force2} \\times \\cos 0^{\\circ}$.
\nThe force $F$ is at $(90 - \\var{theta1})^{\\circ} = \\var{90 - theta1}^{\\circ}$ to the positive $y$-direction therefore has a contribution of $F \\times \\cos \\var{90 - theta1}^{\\circ} = \\var{force1} \\times \\cos \\var{90 - theta1}^{\\circ} = \\var{precround(force1*cos(radians(90-theta1)),3)}$ to the sum of components. This is positive as it is acting in the same direction to the positive.
\nThe force $Q$ is at $(90+\\var{theta2})^{\\circ}=\\var{90 + theta2}^{\\circ}$ to the positive $y$-direction therefore has a contribution of $Q \\times \\cos \\var{90 + theta2}^{\\circ} = \\var{force3} \\times \\cos \\var{90 + theta2}^{\\circ} = \\var{precround(force3*cos(radians(90+theta2)),3)}$ to the sum of components. This is negative as it is acting downwards, in the opposite direction to the positive.
\nTherefore the sum of components in the $y$-direction is $\\var{force2} + \\var{precround(force1*cos(radians(90-theta1)),3)} - \\var{-precround(force3*cos(radians(90+theta2)),3)} = \\var{precround(force2 + force1*cos(radians(90-theta1)) + force3*cos(radians(90+theta2)),3)}$.
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Find the component of $F$ in the $x$-direction
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "\nIn the diagram above, $F = \\var{force1} \\, \\mathrm{N}$, $P = \\var{force2} \\, \\mathrm{N}$ and $Q = \\var{force3} \\, \\mathrm{N}$. The angles are $\\theta = \\var{theta1}^{\\circ}$ and $\\theta^{\\ast} = \\var{theta2}^{\\circ}$.
\nGive your answers to the following questions in Newtons to 3 decimal places.
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