Net Force on a Sled: 17.7 m/s^2 at 68.5°

[MCATPhys]

A 10 kg sled is pulled at 35 degrees east of north with 100 N. There is a force of 150 N due east. And a friction force of 50 N. What is the acceleration of the sled, and its direction?

So I basically added up all the forces in the x direction, and the y direction, and used the pythagorean theorem to get the net force, and divided it by the mass to get the acceleration. But somehow, I don't get the right answer, which is 17.3 m/s^2 at 68.5 degrees east of north.

My work:
Net x force = cos(55)(100) + 150 - 50 = 157.36 N
Net y force = sin(55)(100) = 81.9 N
tan (x) = 81.9/157.36; x = 27.5 degrees
square root of (157.36^2 + 81.9 ^2) = 177.4 N; 17.7 m/s^2

 

[rl.bhat]

You have to subtract the frictional force from the resultant force, because the frictional force acts in the opposite direction of the motion.

 

[MCATPhys]

You have to subtract the frictional force from the resultant force, because the frictional force acts in the opposite direction of the motion.

"Net x force = cos(55)(100) + 150 - 50 = 157.36 N"

But I did subtract it. I subtracted it from the x force, because it acts in the opposite direction of the x axis.

 

[MCATPhys]

Nevermind. I understood what you meant. I subtracted from the angled force and got the right answer... thank you so much!