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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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\n

$F = \\var{force1} \\, \\mathrm{N}$ at an angle of $\\theta = \\var{theta1}^{\\circ}$ clockwise from the left horizonatal.
$P = \\var{force2} \\, \\mathrm{N}$ vertically upwards.
$Q = \\var{force3} \\, \\mathrm{N}$ at an able of $\\theta^{\\ast} = \\var{theta2}^{\\circ}$ clockwise from the right horizontal. 

\n

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

", "advice": "

a) - c)

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Resolve each force in the HORIZONTAL:

\n

$P$ doesn't contribute.

\n

The force $F$ is at $\\var{theta1}^{\\circ}$ to the horizontal, therefore has a contribution of $F \\times \\cos \\var{theta1}^{\\circ} = \\var{force1} \\times \\cos \\var{theta1}^{\\circ} = \\var{precround(force1*cos(radians(theta1)),3)}$.  This force is acting to the left.

\n

The force $Q$ is at $\\var{theta2}^{\\circ}$ to the horizontal, 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 acting to the right.

\n

Therefore the sum of components in the $x$-direction is

\n

forces acting to the right - forces acting to the left

\n

\\[\\var{precround(force3*cos(radians(theta2)),3)} - \\var{precround(-force1*cos(radians(180-theta1)),3)}\\]
\\[= \\var{precround(force1*cos(radians(180-theta1)) + force3*cos(radians(theta2)),3)}\\]

\n

(if this value is positive then it is acting to the right, if it is negative it is acting to the left)

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d) - g)

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Resolve each force from the VERTICAL:

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For the force $P$ this is acting completely in the positive direction, at no angle. Therefore it's contribution is $\\var{force2}$.

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The force $F$ is at $\\var{theta1}^{\\circ}$ to the horizontal, therefore has a contribution of $F \\times \\sin \\var{theta1}^{\\circ} = \\var{force1} \\times \\sin \\var{theta1}^{\\circ} = \\var{precround(force1*sin(radians(theta1)),3)}$.  This force is acting upwards.

\n

The force $Q$ is at $\\var{theta2}^{\\circ}$ to the horizontal, 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 acting to the right.

\n

Therefore the sum of components in the $y$-direction is

\n

forces acting up - forces acting down

\n

\\[\\var{precround(force3*sin(radians(theta2)),3)} - \\var{precround(-force1*sin(radians(180-theta1)),3)}\\]
\\[= \\var{precround(force1*sin(radians(180-theta1)) + force3*sin(radians(theta2)),3)}\\]

\n

(if this value is positive then it is acting to the right, if it is negative it is acting to the left)

\n

\n

 

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