How Do I Solve for the Normal Force in this Cart Problem?

[|345+4|2D]

hi, i know its a really easy thing to figure out normal force or at least i might have been mislead.

Here's the problem

A cart loaded with bricks has a total mass of 24.9 kg and is pulled at constant speed by a rope. The rope is inclined at 21.8* degrees above the horizontal and the cart moves 13.9 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.549.
The acceleration of gravity is 9.8 m/s^2. What is the normal force exerted on the cart by the floor. Answer in units of N.

Ok this is what i did, i said n = mg. (which i could've swore is right -- since its in the book and the study guide book in like 5 different places). so i got n = 244.02 N. Tell me please -- am i finding the wrong thing? because n is normal force, correct?

 

[Päällikkö]

Draw a diagram. All forces should cancel out (F = ma, F = 0 <=> v = constant).
With N = mg forces in y-direction do not equal 0 (inclined rope). So, N is not mg.

 

[siddharth]

In this case, N will not be mg.
It really is very useful to draw the Free body force diagram for any problem of this type. Once you do that, draw the forces acting on the cart.
Notice that the force due to the rope has components in both the x-direction AND the y-direction.
So, when you say the net force (sum of all the forces) is zero in the y-direction, you will have to include the y-component of the force due to the rope!
So, you will not get N=mg.