The site accepts the answer 49N for #6c, but I'm not sure why. 2. Relevant equations
[tex]F_g=mg[/tex]
[tex]F=\sqrt{F_x^2+F_y^2}[/tex]
[tex]F_x=F\times \cos \theta[/tex] 3. The attempt at a solution
The answer for #6c appears to be equal to
[tex]10\textrm{kg}\times 9.80\textrm{m/s}^2\times \textrm{cos }60^\circ=49\textrm{N,}[/tex]
but that doesn't make sense to me; shouldn't we combine the weight of the picture with the rightward pull? In that case, how do we find the rightward pull?

Or, is there an imaginary diagonal force--equal in magnitude to the gravitational force--of which we want the horizontal component? In that case, why?

The answer to #6b is "No," but I'm not sure why. (I'm guessing that once I understand #6c, then #6b will follow.)

Actually, I suppose that combining the force of gravity in this case (98N) with a horizontal vector couldn't reduce the magnitude to just 49N....
Is it the second option that I proposed, then? [tex]\sqrt{F_g^2\ \textrm{cos}^2\theta +F_g^2\ \textrm{cos}^2(90^\circ-\theta)}=F_g\textrm{,}[/tex] so it seems plausible. The forces on the ropes, not including from the wall or ceiling where they're above, should add up to the force exerted by the picture, right?

Still, even if I've found the way to do the problem, I don't feel like I have a sufficient understanding of it.