Question aecf3

Jan 27, 2018

$20 N$ to ${45}^{\circ}$ with the x-axis

Explanation:

Well, they are in opposite directions, and the ${45}^{\circ}$ one has more force, so it wins.

Thus, the resultant force is $30 N - 10 N = 20 N$.

Since the ${45}^{\circ}$ is stronger, the resultant force with be $20 N$ with a direction of ${45}^{\circ}$ with the x-axis.

Jan 27, 2018

$39.74 N \text{at } {41.3}^{\circ}$ to the x axis.

Explanation:

I will assume that the 30^@ "and 45^@# angles are both counterclockwise from the +x axis. (Or both could be clockwise -- my result would not be different if so.)

The x axis component of the 10 N force would be
${F}_{\text{1x}} = 10 N \cdot \cos {30}^{\circ} = 8.66 N$
The y axis component of the 10 N force would be
${F}_{\text{1y}} = 10 N \cdot \sin {30}^{\circ} = 5.0 N$

The x axis component of the 30 N force would be
${F}_{\text{2x}} = 30 N \cdot \cos {45}^{\circ} = 21.21 N$
The y axis component of the 30 N force would be
${F}_{\text{2y}} = 30 N \cdot \sin {45}^{\circ} = 21.21 N$

The x component of the resultant is
$8.66 N + 21.21 N = 29.87 N$
The y component of the resultant is
$5.0 N + 21.21 N = 26.21 N$

The magnitude of the resultant is
$\sqrt{{\left(29.87 N\right)}^{2} + {\left(26.21 N\right)}^{2}} = 39.74 N$

The direction of the resultant is ${\text{tan}}^{-} 1 \left(\frac{26.21}{29.87}\right) = {41.3}^{\circ}$

I hope this helps,
Steve