Question

http://i.imgur.com/4o31s.jpg

Can someone check my answers for this?

I got
T1 = 2(49N)=98N
T2 = 49cos(9.46)=48.3N
and
Normal Force = 48.3N

is the normal force the and tensions should be the same?

Answer 1

I think the normal force would only be equal to the x component of the tension.

Answer 2

I attached a picture. The y-direction forces are everything going up, minus everything going down. From that, you get T2 is mg divided by cos(theta). You are multiplying by cos(theta).
The x-direction forces are everything pulling right minus everything pushing left. So the Normal force is T2 times sin(theta).

Answer 3

okay i got it now, thanks