Step 1: Isolate the two forces that act on the blade face.
As the inclined blade moves through the soil at a lift angle of 30 degrees from the direction of travel, the soil pushes back on the blade with a normal reaction N perpendicular to the face and a friction force \( \mu N \) acting along the face, opposing the relative sliding of soil over the blade. Here \( N = 1000 \text{N} \) and \( \mu = 0.3 \).
Step 2: Resolve both forces along the direction of travel.
The horizontal (draft) component of the normal force is \( N \sin\alpha \) and the horizontal component of the friction force is \( \mu N \cos\alpha \), so the draft is \( D = N(\sin\alpha + \mu\cos\alpha) \).
Step 3: Substitute the numbers.
With \( \alpha = 30^\circ \), \( \sin 30^\circ = 0.5 \) and \( \cos 30^\circ = 0.866 \) as given, \( D = 1000 (0.5 + 0.3 \times 0.866) = 1000(0.5 + 0.2598) = 759.8 \text{N} \), which rounds to 760 N.
Step 4: Note the data that is not needed.
The soil density, angle of internal friction, blade dimensions and cutting resistance per unit length are all extra information for this particular calculation, since the draft here follows directly from the normal load, the friction coefficient and the lift angle alone.
\[ \boxed{760 \ N} \]