Question

Find the resultant force And angle

Answer 1

The first force has magnitude 60 N and makes an angle of -30° relative to the positive horizontal axis (i.e. to the immediate right), while the second force has magnitude 150 N and direction +30° from the horizontal.

In component form, we have

$\vec F_1 = F_{1,x}\,\vec\imath + F_{1,y}\,\vec\jmath \\\\ \vec F_1 = (60\,\mathrm N)\cos(-30^\circ)\,\vec\imath + (60 \,\mathrm N)\sin(-30^\circ)\,\vec\jmath \\\\ \vec F_1 \approx (52.0\,\mathrm N)\,\vec\imath - (30\,\mathrm N)\,\vec\jmath$

$\vec F_2 = F_{2,x}\,\vec\imath + F_{2,y}\,\vec\jmath \\\\ \vec F_2 = (150\,\mathrm N)\cos(30^\circ)\,\vec\imath + (150\,\mathrm N)\sin(30^\circ)\,\vec\jmath \\\\ \vec F_2 \approx (130\,\mathrm N)\,\vec\imath + (75\,\mathrm N)\,\vec\jmath$

The resultant force is the vector

$\vec F = \vec F_1 + \vec F_2 \\\\ \vec F \approx (182\,\mathrm N)\,\vec\imath + (45\,\mathrm N)\,\vec\jmath$

Its magnitude is

$R = \|\vec F\| \\\\ R \approx \sqrt{(182\,\mathrm N)^2 + (45\,\mathrm N)^2} \\\\ R = 187.35\,\mathrm N \approx \boxed{190\,\mathrm N}$

Its direction $\theta$ is such that

$\tan(\theta) \approx \dfrac{45\,\mathrm N}{182\,\mathrm N}$

Because the components of the resultant force are both positive, that means the angle $\vec F$ makes with the horizontal is between 0° and 90°. So

$\theta \approx \tan^{-1}\left(\dfrac{45\,\mathrm N}{182\,\mathrm N}\right) \approx \boxed{14^\circ}$