Recent content by pxw

Oh, ok! I got it now! Thanks a lot! :D

Yes, I understand that I have to break up F into its x and y components, however, I'm not sure how to do that without a value... the equation I have is Fa(sin330)=Fay. Sticking that into the net force equation, I have 0=Fn - 50.97 + Fa(sin330). I can't seem to get any further than that...

1. A block with a mass of 6 kg is held in equilibrium on an inclined plane of angle θ=30 degrees by a horizontal force F. Find the magnitudes of normal force on the block and of F (ignore friction). So far, I've found Fg=6(-9.81)=-58.86 N I then found Fgy=-58.86(sin240)=50.97 and...