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A rectangle $ABCD$ has $AB=3cm$ and $BC=4cm$. Forces, all measured in newtons and of magnitudes $2$, $4$, $6$, $8$ and $k$, act along $AB$, $BC$, $CD$, $DA$ and $AC$ respectively, the direction of each force being shown by the order of the letters. The resultant of the five forces is parallel to $BD$. Find $k$ and show that the resultant has magnitude $$\frac{5}{6}$$ newtons.

($k$ should equal $\frac{35}{6}$).

How do you find $k$ when there are 4 forces of different magnitudes in operation around the perimeter of the rectangle?

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  • $\begingroup$ Firstly: how do you find $k$ when there are 4 forces of different magnitudes in operation around the perimeter of the rectangle? $\endgroup$ – J132 Oct 19 '16 at 22:34
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Drawing a rectangle of forces with $AB$ (=3cm) drawn horizontally, then $BC$ drawn vertically from the tail of $AB$ etc,

Also, let $θ$ be the angle between $AC$ and $AB$ which gives $$sin(θ) = \frac{4}{5}$$ and $$cos(θ) = \frac{3}{5}$$

The horizontal component of the resultant is:

Fx = $Fcos(θ)$ = $\frac{5}{6}$ x $\frac{4}{5}$ = $\frac{2}{3}$

The vertical component of the resultant is:

-Fy = $Fsin(θ)$ = $\frac{5}{6}$ x $\frac{3}{5}$ = $\frac{1}{2}$

We have -Fy because we are told that the resultant is parallel to $BD$. It will point downwards in a negative y-direction.

So $Fy = -\frac{1}{2}$

Using this, I get the following:

Resolving horizontally: $$2 + kcos(θ) -6 =\frac{2}{3}$$ from which $$k= \frac{35}{6}N$$

Resolving vertically: $$4 + ksin(θ) -8 = -\frac{1}{2}$$ from which $$k= \frac{35}{6}N$$

From these 2 results, the value of $k$ is $\frac{35}{6}N$

The magnitude of the resultant is:

$$R = \sqrt((-0.5)^2 + (2/3)^2) = \frac{5}{6}N$$

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  • $\begingroup$ There is a mistake in your instructions for the polygon below. Where it says 'Now draw a line parallel to the AD segment through the head of the vector corresponding to the DA force', it should say 'Now draw a line parallel to the AC segment through the head of the vector corresponding to the DA force', which worked when I tried this method. If the line is parallel to the AD segment I do not get the lines to intersect. $\endgroup$ – J132 Oct 30 '16 at 16:39
  • $\begingroup$ Now it is solved $\endgroup$ – la flaca Oct 30 '16 at 17:59

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