Finding the Coefficient of Friction?

[Brodo17]

A boy pulls a sled of mass 5kg with a rope that makes an angle of 60 degrees w/ respect to the horizonal surface of the frozen pond. The boy pulls on the rope with a force of 10N and the sled moves with constant velocity. What is the coefficient of friction between the sled and the ice?


F=ma
f=mewFN


First I found the horizonatal force that the boy is exerting on the sled:
Cos60 X 10 = 5N
Then I thought that since the acceleration is constant that the sum of the forces must be zero. This would mean that the frictional force must be equal to the applied force of 5N

So, I said 5N = mew (5kg X 9.8)
And I found mew to be .102

However apparently the answer is .12

Thanks!

 

[Bluesroo]

What is the force of gravity acting on the sled?
What is the normal force before the boy pulls on the sled? What about after?

I hope these help.

 

[glueball8]

Both forces in the x and y direction must sum to 0, since its in constant velocity.