Question

a block is pulled by two horizontal forces. the first force is (500\text{ n}) at an angle of (65.0^{circ}) and the second is (415\text{ n}) at an angle of (270^{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: Find the y-component of the first force

$$ F_{1y} = 500 \sin(65.0^\circ) $$

$$ F_{1y} \approx 500 \times 0.9063 = 453.15\text{ N} $$

Step 2: Find the y-component of the second force

$$ F_{2y} = 415 \sin(270^\circ) $$

$$ F_{2y} = 415 \times (-1) = -415\text{ N} $$

Step 3: Calculate the total y-component

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

$$ F_y = 453.15 - 415 = 38.15\text{ N} $$

Rounding to three significant figures:

$$ F_y \approx 38.2\text{ N} $$

Answer:

38.2