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A block is pulled by a force of [tex][tex]$124 N$[/tex][/tex] directed at a [tex]29.6^{\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 determine the [tex]\( x \)[/tex]-component of the force acting on the block, we can start by understanding that the force is directed at an angle and we need to break it down into its horizontal and vertical components. In this case, we are interested in the horizontal, or [tex]\( x \)[/tex]-component.

The force is given as [tex]\( 124 \, \text{N} \)[/tex] and the angle is [tex]\( 29.6^\circ \)[/tex]. The [tex]\( x \)[/tex]-component of the force can be calculated using the cosine of the given angle, as cosine relates the adjacent side (the horizontal component in this context) to the hypotenuse (the force).

1. Given value of the force:
[tex]\[ F = 124 \, \text{N} \][/tex]

2. Given angle with the horizontal:
[tex]\[ \theta = 29.6^\circ \][/tex]

3. Formula to calculate [tex]\( x \)[/tex]-component:
[tex]\[ F_x = F \cos(\theta) \][/tex]

4. Convert the angle from degrees to radians because trigonometric functions in most mathematical and programming environments use radians.

5. Calculate [tex]\( \cos(29.6^\circ) \)[/tex]:

6. Multiply the force by the cosine of the angle:
[tex]\[ F_x = 124 \times \cos(29.6^\circ) \][/tex]

Using the known cosine value for [tex]\( 29.6^\circ \)[/tex]:

[tex]\[ F_x \approx 124 \times 0.869 \][/tex]

Calculations show that the [tex]\( x \)[/tex]-component force [tex]\( F_x \)[/tex] comes out to be:

[tex]\[ \overrightarrow{F_x} \approx 107.82 \, \text{N} \][/tex]

So, the [tex]\( x \)[/tex]-component of the force acting on the block is [tex]\( 107.82 \, \text{N} \)[/tex].