2. a) determine the resultant r of the four - force coplanar force system of figure 3.7. compute its…

Explanation:

Step1: Resolve each force into x - components

For the 5 - kip force: $F_{x1}=5\cos(140^{\circ})\approx - 3.83$ kips (since the angle with the positive x - axis is $140^{\circ}$). For the 8 - kip force: $F_{x2}=8\cos(25^{\circ})\approx7.25$ kips. For the 12 - kip force: $F_{x3}=12\cos(330^{\circ}) = 12\cos(30^{\circ})\approx10.39$ kips. For the 10 - kip force: $F_{x4}=10\cos(240^{\circ})=- 5$ kips. $\sum F_{x}=F_{x1}+F_{x2}+F_{x3}+F_{x4}\approx - 3.83 + 7.25+10.39 - 5=8.81$ kips.

Step2: Resolve each force into y - components

For the 5 - kip force: $F_{y1}=5\sin(140^{\circ})\approx3.21$ kips. For the 8 - kip force: $F_{y2}=8\sin(25^{\circ})\approx3.38$ kips. For the 12 - kip force: $F_{y3}=12\sin(330^{\circ})=-6$ kips. For the 10 - kip force: $F_{y4}=10\sin(240^{\circ})=-8.66$ kips. $\sum F_{y}=F_{y1}+F_{y2}+F_{y3}+F_{y4}\approx3.21 + 3.38-6 - 8.66=-8.07$ kips.

Step3: Calculate the magnitude of the resultant force

$R=\sqrt{(\sum F_{x})^{2}+(\sum F_{y})^{2}}=\sqrt{(8.81)^{2}+(-8.07)^{2}}\approx\sqrt{77.62 + 65.12}=\sqrt{142.74}\approx11.94$ kips.

Step4: Calculate the angle of the resultant force with the x - axis

$\theta=\tan^{-1}\left(\frac{\sum F_{y}}{\sum F_{x}}\right)=\tan^{-1}\left(\frac{-8.07}{8.81}\right)\approx\tan^{-1}(-0.916)\approx - 42.5^{\circ}$ (the negative sign indicates that the angle is measured clock - wise from the positive x - axis).

Step5: Fill the table

Force (kips)	$F_{x}$ (kips)	$F_{y}$ (kips)
5	- 3.83	3.21
8	7.25	3.38
12	10.39	- 6
10	- 5	- 8.66
Total	8.81	- 8.07

Answer:

Magnitude of resultant $R\approx11.94$ kips. Sense: The resultant force has a positive x - component and a negative y - component. Angle of inclination $\theta\approx - 42.5^{\circ}$ with the horizontal X - axis. Table:

Force (kips)	$F_{x}$ (kips)	$F_{y}$ (kips)
5	- 3.83	3.21
8	7.25	3.38
12	10.39	- 6
10	- 5	- 8.66
Total	8.81	- 8.07