Coefficient of Friction at constant speed

[Tensaiga]

Hello, please help me out on this:
Two equall masses lie on each side of a rough wedge, connected by a string, the wedge makes anlges 53, and 47, to the horizontal, find the coefficient of friction for the masses to move at a constant speed.

MY THINKING:
since masses are equal:
mgsin(53) = mgcos(53)(u) + mgcos(47)(U) + mgsin(47) ---> these are all the opposing forces, and force of fricton of each mass, since the slop is different for them.
the mg(s) cancels out. Leaving:
sin(53)-sin(47) = (cos53+cos47)(u)
then i got u = 0.052
which is wrong...

 

[PPonte]

I do not know what a wedge* is, but I imagine the situation which was more probable with the given data.

Is $$\mu_k = 0.098$$?
If it is, then I will explain my thoughts.

*My english needs improvements.

 

[Doc Al]

MY THINKING:
since masses are equal:
mgsin(53) = mgcos(53)(u) + mgcos(47)(U) + mgsin(47) ---> these are all the opposing forces, and force of fricton of each mass, since the slop is different for them.
the mg(s) cancels out. Leaving:
sin(53)-sin(47) = (cos53+cos47)(u)
then i got u = 0.052
which is wrong...

Your thinking looks OK to me. Why do you think it's wrong?