(a) This can be found by calculating the horizontal component of the force applied and then dividing that by the mass of the box. As you may have noticed, horizontal components usually are found by using cos. it's no different here.
F_{hor}=30N*cos(30º)
F_{hor}=26N
Now that we have F_{hor} we divide it by the mass of the box to get acceleration:
a=F_{hor}/m
a=26N/28kg
a=0.93m/s^{2}
(b) The normal force, as we recall, is equal and opposite of the force of gravity. F_{g}=mg; m is the mass of the box and g is the acceleration due to gravity so we have:
F_{g}=28kg*9.81m/s^{2}
F_{g}=274.9N
So since F_{N}=F_{g} we have that F_{N}=274.9N.
But we also need to take into account the veritcal component of the pull on the box (this force counteracts F_{N).} To do that we find the vertical component through the Soa part of SoaCahToa:
sinΘ=o/h
sin(30º)=F_{ver}/30N
30N*sin(30º)=F_{ver}
15N=F_{ver}
Now we just add F_{ver} and F_{N} to get the real F_{N}.
Enjoy! =]
11/24/2012

Mykola V.
Comments
Thanks :)
You're welcome :)