Answer:
F'=12.1243557 N
Step-by-step explanation:
Let the horizontal component of force he is pulling with be F' Newtons and the force applied be F Newtons.
[tex]\therefore \cos \theta = \frac{Horizontal \: Force}{Force \: applied}[/tex]
[tex] \therefore \cos \theta = \frac{F'}{F} [/tex]
[tex] \therefore F'=F\cos \theta[/tex]
[tex] \therefore F'=14.\cos 30\degree [/tex]
[tex] \therefore F'=14.\frac{\sqrt 3}{2} [/tex]
[tex] \therefore F'=7\sqrt 3 [/tex]
[tex] \therefore F'=12.1243557 \: N[/tex]
