# [Physics] Tension of string attached to ball on inclined plane

homework-and-exercisesstatics

A sphere of weight 50N and radius 4cm is held by a string of length 8 cm against a smooth wall inclined at an angle of $70^{\circ}$.
Find the tension in the string and the force between the ball and the wall.

I setup a coordinate system where The Y axis is parallel to the normal force. So that the X component of N would be zero. The 30 degrees is the angle between the string and plane of incline.

$\theta = 70^{\circ}$

$\sum{F_x} = Tcos(30) – mgsin(\theta) = 0$

$\sum{F_y} = N – Tsin(30) – mgcos(\theta) = 0$

$T = \frac{mgsin70^{\circ}}{cos(30)} = 54.25$N

$N = 44.226$N

The answer is given as $T = 49.8$N and $N = 33.7$N

What am I doing wrong?

First of all, where are you getting the term $\cos(30^{\circ})$? What two things here have an angle of $30^{\circ}$?
By having a term $mg\sin(\theta)$, I think you are assuming that the normal force from the wall is completely described by the weight of the ball. In many simple cases, this is true. But because the ball also has the force from the string, it's not true here.
You have 3 forces on the ball, not 2: The tension in the string $T$, the weight of the ball $W$, and the normal force $N$. In your first equation, you are assuming facts about N that you do not yet have.
In your second equation you have $N$, but without a $\cos$ or $\sin$ term. The normal force does not act vertically, so this is incorrect.