What angle will the string make with the vertical?

For this problem I suggest drawing a picture. You know that the component of gravity in the direction perpendicular to the slope must equal the component of tension in that direction. You also know the component of gravity down the slope plus the component of tension up the slope must equal 3m. (You know that the angle is less than 30 degrees because it must diminish the force of gravity down the slope, so the tension acts up-slope.) This gave me the following two equations:

$mg\cos(30\deg)=Tcos(\phi)$

$mg\sin(30\deg)-Tsin(\phi)=3m$

Which I combined like this:

$\frac{mg\cos(30\deg)}{\cos(\phi)}=T=\frac{mg\sin(30\deg)-3m}{\sin(\phi)}$

Which gives:

$tan(\phi)=\frac{g-6}{g\sqrt{3}}$

Which yields:

$\phi=12.6$ degrees

Now, for reasons which are become clear if you draw the picture; the angle, $\theta$ , which the string makes with the vertical is $\theta=30\deg-\phi$ which means $\theta$ , to the nearest integer, is 17 degrees.

Y accelration = $g - 3*sin30 = 8.3 m/s²$ and the horiz accel = $3*\cos30 = 2.6 m/s²$ where g=9.8 Angle= $\arctan(2.6/8.3) = 17º$