## Force Acting on Rope

1. The problem statement, all variables and given/known data
From http://library.thinkquest.org/10796/index.html (#6b, 6c)
 A 10 kg picture is hanging on a wall by two ropes. ... b. Are the forces acting on the ropes A and B equal in magnitude? c. How big is the force acting on the rope A?
The site accepts the answer 49N for #6c, but I'm not sure why.
2. Relevant equations
$$F_g=mg$$
$$F=\sqrt{F_x^2+F_y^2}$$
$$F_x=F\times \cos \theta$$
3. The attempt at a solution
The answer for #6c appears to be equal to
$$10\textrm{kg}\times 9.80\textrm{m/s}^2\times \textrm{cos }60^\circ=49\textrm{N,}$$
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.)

Thank you!
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 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? $$\sqrt{F_g^2\ \textrm{cos}^2\theta +F_g^2\ \textrm{cos}^2(90^\circ-\theta)}=F_g\textrm{,}$$ 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.
 You can resolve mg to 2 components of opposing direction to A and B since they are 90° apart. To me it is mgCos30°.

## Force Acting on Rope

 Quote by azizlwl You can resolve mg to 2 components of opposing direction to A and B since they are 90° apart. To me it is mgCos30°.
I'm sorry; I don't understand what you mean. Would you mind drawing it?

Would it be something like this?

$$mg\textrm{cos(30}^\circ \textrm{)}$$

EDIT - whoops, sorry, I edited without realizing you had posted again.

 Quote by intdx I'm sorry; I don't understand what you mean. Would you mind drawing it?
Just extend line A and B downward. These 2 directions can be components of mg.
 Hello intdx, Try Lami's theorem here.Or otherwise resolve the weight along OA and OB to check for equilibrium. regards Yukoel

 Tags force, gravity, picture, ropes

 Similar discussions for: Force Acting on Rope Thread Forum Replies Introductory Physics Homework 0 Classical Physics 4 Introductory Physics Homework 3 Introductory Physics Homework 7 Introductory Physics Homework 8