Answered

What are the horizontal and vertical components of a

300 N force that is applied along a rope at 60° to the

horizontal and used to drag an object across a yard?


help! can u explain the solve? idk why my answer is always reverse, like the answer for horizontal, not for vertical. the answer for vertical, not for horizontal. so, i am confusing. pls help!

if u got the answer, thanks​

Answer :

cryssatemp

Answer:

In the figure attached we can see in a better way this situation. If we are told the rope is at 60 degress to the horizontal, we can place this rope in the origin of the X-Y coordinate system.

If we observe in detail, we have a right triangle where the opposite side [tex]OS[/tex] to the angle [tex]60\º[/tex] is the vertical component [tex]F_{y}[/tex] and the adjacent side [tex]AS[/tex] is the horizontal component [tex]F_{x}[/tex] . So, we can use trigonometric functions to find the force [tex]F[/tex] (the hypotenuse [tex]h[/tex]) applied to the rope.

For [tex]F_{x}[/tex]:

We can use the trigonometric function cosine, which is defined as:

[tex]cos(60\º)=\frac{AS}{h}[/tex]

This means [tex]F_{x}=Fcos(60\º)[/tex]

[tex]F_{x}=300Ncos(60\º)[/tex]

[tex]F_{x}=150N[/tex]

For [tex]F_{y}[/tex]:

We can use the trigonometric function sine, which is defined as:

[tex]sin(60\º)=\frac{OS}{h}[/tex]

This means [tex]F_{y}=Fsin(60\º)[/tex]

[tex]F_{y}=300Nsin(60\º)[/tex]

[tex]F_{y}=259.8N[/tex]

Finally the horizontal and vertical components are 150 N and 259.8 N

${teks-lihat-gambar} cryssatemp

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