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a) - c)

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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.

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The 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!

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The 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.

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Therefore the sum of components in the $x$-direction is $0 - \\var{precround(-force1*cos(radians(180-theta1)),3)} +  \\var{precround(force3*cos(radians(theta2)),3)} = \\var{precround(force1*cos(radians(180-theta1)) + force3*cos(radians(theta2)),3)}$.

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d) - g)

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Resolve 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 $\\var{force2} \\times \\cos 0^{\\circ}$. 

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The 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.

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The 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.

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Therefore 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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You have not given your answer to the correct precision.

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Find the component of $Q$ in the $x$-direction.

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Find the resultant force in the $x$-direction.

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Find the component of $P$ in the $y$-direction.

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Find the component of $F$ in the $y$-direction.

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Find the component of $Q$ in the $y$-direction.

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Find the resultant force in the $y$-direction.

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In 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 $\\theta^{\\ast} = \\var{theta2}^{\\circ}$. 

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Give your answers to the following questions in Newtons to 3 decimal places.

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Find the $x$ and $y$ components of the resultant force on an object, when multiple forces are applied at different angles.

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