# Step 1: **Express Force Components in Terms of Angles** For each case, de the force \( F \) into its \( x \) and \( y \) components using trigonometry. Assume \( F \) is the magnitude of the force, and \( \beta \) (or \( \alpha \)) is the angle as shown. --- ### (a) - \( F \) is directed *upward and to the left*. - The angle \( \beta \) is measured **from the positive \( x \)-axis toward the force**. **Components:** - \( F_x = -F \cos \beta \) - \( F_y = F \sin \beta \) **Explanation:** Since the force is to the left, the \( x \)-component is negative. The \( y \)-component is positive (upward). --- ### (b) - \( F \) is directed *downward and to the right*. - The angle \( \beta \) is measured **from the positive \( x \)-axis toward the force**. **Components:** - \( F_x = F \cos \beta \) - \( F_y = -F \sin \beta \) **Explanation:** The force is to the right (positive \( x \)), but downward (negative \( y \)). --- ### (c) - \( F \) is directed *downward and to the left*. - The angle \( \beta \) is measured **from the positive \( x \)-axis toward the force**. **Components:** - \( F_x = -F \cos \beta \) - \( F_y = -F \sin \beta \) **Explanation:** Both components are negative: left and down. --- ### (d) - \( F \) is directed *upward and to the right*. - The angle \( \alpha \) is measured **from the positive \( x \)-axis toward the force**. **Components:** - \( F_x = F \cos \alpha \) - \( F_y = F \sin \alpha \) **Explanation:** Both \( x \) and \( y \) components are positive: right and up. --- # Step 2: **Summary Table** | Part | \( F_x \) | \( F_y \) | |------|----------------------|----------------------| | (a) | \( -F \cos \beta \) | \( F \sin \beta \) | | (b) | \( F \cos \beta \) | \( -F \sin \beta \) | | (c) | \( -F \cos \beta \) | \( -F \sin \beta \) | | (d) | \( F \cos \alpha \) | \( F \sin \alpha \) | --- ## **Final Answer** \[ \boxed{ \begin{aligned} &\text{(a)} \quad F_x = -F \cos \beta,\quad F_y = F \sin \beta \\ &\text{(b)} \quad F_x = F \cos \beta, \quad F_y = -F \sin \beta \\ &\text{(c)} \quad F_x = -F \cos \beta, \quad F_y = -F \sin \beta \\ &\text{(d)} \quad F_x = F \cos \alpha,\quad F_y = F \sin \alpha \\ \end{aligned} } \]