To determine the $x$-component of the total force acting on the block, we need to calculate the $x$-components of the two forces separately and then sum them.

Given:

1. Force 1: $F_1 = 207 \, \mathrm{N}$, angle $\theta_1 = 60.5^\circ$
2. Force 2: $F_2 = 175 \, \mathrm{N}$, angle $\theta_2 = 282^\circ$

The $x$-component of a force is calculated as: $F_x = F \cos(\theta)$

Step 1: Calculate the $x$-components of each force

Force 1:

$F_{1x} = 207 \cos(60.5^\circ)$ Using $\cos(60.5^\circ) \approx 0.4961$: $F_{1x} = 207 \times 0.4961 = 102.7 \, \mathrm{N}$

Force 2:

$F_{2x} = 175 \cos(282^\circ)$ Since $\cos(282^\circ) = \cos(282^\circ - 360^\circ) = \cos(-78^\circ) \approx 0.2079$: $F_{2x} = 175 \times 0.2079 = 36.4 \, \mathrm{N}$

Step 2: Add the $x$-components

Because the angle $282^\circ$ points slightly to the left of the $y$-axis, its $x$-component is negative. Thus: $F_{\text{total},x} = F_{1x} + F_{2x} = 102.7 - 36.4 = 66.3 \, \mathrm{N}$

Final Answer:

The $x$-component of the total force is: $\boxed{66.3 \, \mathrm{N}}$

Let me know if you want further clarification!

Relative Questions:

1. How do angles greater than $180^\circ$ affect $x$- and $y$-components of a force?
2. What is the $y$-component of the total force in this problem?
3. How would the problem change if the block were on an inclined plane?
4. How can vector components be visualized graphically?
5. Why is it useful to resolve forces into components?

Tip:

Always double-check the quadrant of your angle to correctly assign positive or negative signs to components!