Answer:
The force parallel to the horizontal is 26.24 N
Explanation:
She pulls on the leash with a force F = 30 N, this force, since its at an angle of 29° (i will cal this angle [tex]\theta[/tex]), it has a force component on x (the horizontal, i will call this force [tex]F_{x}[/tex]) and a force component on y (the vertical, i will call this [tex]F_{y}[/tex] ).
This can be seen in the attached picture.
Since we are asked about the force parallel to the horizontal, we need to find the component of the force [tex]F_{x}[/tex], since [tex]F_{x}[/tex] is the adjacent angle, we need to use cosine:
[tex]F_{x}=Fcos \theta[/tex]
since [tex]F=30N[/tex] and [tex]\theta=29[/tex]
[tex]F_{x}=(30N)cos(29)[/tex]
[tex]F_{x}=(30N)(0.8746)[/tex]
[tex]F_{x}=26.24N[/tex]
The force parallel to the horizontal is 26.24 N
