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