Question

a block is pulled by two forces. the first force is \\(45.0\text{ n}\\) at an angle of \\(49.0^{\circ}\\) and the second is \\(55.0\text{ n}\\) at an angle of \\(56.3^{\circ}\\).

what is the y-component of the total force acting on the block?

\\\vec{f}_y = ?\text{ n}\\

Explanation:

⚡ Using what you learned: Vectors in Plane and Space (components, magnitude, direction)

Step 1: Identify individual y-components

The vertical component of a force vector \( F \) acting at an angle \( \theta \) relative to the horizontal is given by:

$$ F_y = F \sin(\theta) $$

For the first force:

$$ F_{1y} = 45.0 \sin(49.0^\circ) $$

$$ F_{1y} \approx 45.0 \times 0.7547 = 33.96\text{ N} $$

For the second force:

$$ F_{2y} = 55.0 \sin(56.3^\circ) $$

$$ F_{2y} \approx 55.0 \times 0.8320 = 45.76\text{ N} $$

Step 2: Sum the components

The total y-component of the force is the sum of the individual y-components:

$$ F_y = F_{1y} + F_{2y} $$

$$ F_y = 33.96\text{ N} + 45.76\text{ N} = 79.72\text{ N} $$

Rounding to three significant figures:

$$ F_y \approx 79.7\text{ N} $$

Answer:

79.7