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A block is pulled by a force of [tex][tex]$177 N$[/tex][/tex] directed at a [tex]85.0^{\circ}[/tex] angle from the horizontal.

What is the [tex]x[/tex]-component of the force acting on the block?

[tex]\overrightarrow{F_x} = [?] N[/tex]


Sagot :

To find the [tex]\( x \)[/tex]-component of the force acting on the block, we need to resolve the force into its horizontal and vertical components. Here's the step-by-step process:

1. Identify the given data:
- The magnitude of the force, [tex]\( F = 177 \)[/tex] Newtons.
- The angle from the horizontal, [tex]\( \theta = 85.0^\circ \)[/tex].

2. Convert the angle from degrees to radians:
We know that [tex]\( 1^\circ = \frac{\pi}{180} \)[/tex] radians. Therefore,
[tex]\[ \theta \text{ in radians} = 85.0 \times \frac{\pi}{180} = 1.4835 \text{ radians} \][/tex]

3. Use the cosine function to find the [tex]\( x \)[/tex]-component:
The [tex]\( x \)[/tex]-component of the force can be found by:
[tex]\[ F_x = F \cos(\theta) \][/tex]

4. Substitute the known values into the formula:
[tex]\[ F_x = 177 \cos(1.4835) \][/tex]

5. Calculate [tex]\( \cos(1.4835) \)[/tex]:
From trigonometric tables or a calculator, [tex]\( \cos(1.4835) \)[/tex] is approximately [tex]\( 0.0872 \)[/tex].

6. Multiply the force magnitude by the cosine of the angle:
[tex]\[ F_x = 177 \times 0.0872 \approx 15.42656646633549 \text{ Newtons} \][/tex]

Hence, the [tex]\( x \)[/tex]-component of the force acting on the block is approximately [tex]\( 15.43 \)[/tex] Newtons.