To solve this problem, we'll use the principles of vector addition and trigonometry.

Given:

• Vertical component of the force ($F_y$) = 130 N
• Horizontal component of the force ($F_x$) = 150 N

Part (a): Find the magnitude of the force $F$

The magnitude of the force $F$ can be found using the Pythagorean theorem because the force components form a right triangle.

$F = \sqrt{F_x^2 + F_y^2}$

Substitute the values:

$F = \sqrt{(150)^2 + (130)^2}$

$F = \sqrt{22500 + 16900} = \sqrt{39400} \approx 198.49 \, \text{N}$

Part (b): Find the angle $\theta$

The angle $\theta$ that the force makes with the horizontal can be found using the tangent function, which relates the angle to the opposite and adjacent sides of the triangle.

$\tan(\theta) = \frac{F_y}{F_x}$

$\theta = \tan^{-1}\left(\frac{130}{150}\right)$

$\theta = \tan^{-1}(0.867) \approx 40.93^\circ$

Summary:

• (a) The magnitude of the force $F$ is approximately 198.49 N.
• (b) The angle ( \theta