A block is pulled by two horizontal forces. The first force is 124 N at an angle of \( 29.6^{\circ} \) and the second is 187 N at an angle of \( 194^{\circ} \). What is the \( y \)-component of the total force acting on the block? \[ \overrightarrow{\mathrm{F}_{y}}=[?] \mathrm{N} \]

To find the \( y \)-component of the total force acting on the block, we can break down each force into its \( y \)-components using trigonometric functions. 1. For the first force (124 N at \( 29.6^{\circ} \)): \[ F_{1y} = 124 \cdot \sin(29.6^{\circ}) \approx 124 \cdot 0.491 = 61.0 \, \text{N} \] 2. For the second force (187 N at \( 194^{\circ} \)): \[ F_{2y} = 187 \cdot \sin(194^{\circ}) \approx 187 \cdot (-0.515) = -96.4 \, \text{N} \] 3. Now, add the \( y \)-components together to find the total: \[ \overrightarrow{F_{y}} = F_{1y} + F_{2y} = 61.0 - 96.4 = -35.4 \, \text{N} \] Thus, the \( y \)-component of the total force acting on the block is: \[ \overrightarrow{\mathrm{F}_{y}} = -35.4 \, \text{N} \]