Question

The free body diagram shows a box being pulled to the left up a 25-degree incline.

A free body diagram with 4 forces vectors. The first pointing downward labeled 50 N, the second pointing northwest labeled 75 N, the third pointing northeast labeled 45 N, and the fourth pointing southeast labeled 20 N.

The magnitude of the normal force is
N.

Answer 1

Take the unit normal pointing perpendicular to the plane, at 65° above the +x (east) direction (the plane is at 155°). Project each force onto that normal:

Angles (math convention): 75 N at 135°, 45 N at 45°, 20 N at 315°, 50 N at 270°.
Projection = F cos(θ - 65°).

- 50 N: 50 cos(270°–65°) = 50 cos 205° = −50 cos25° = −45.315 N (into the plane)
- 75 N: 75 cos(135°–65°) = 75 cos70° = +25.651 N (outward)
- 45 N: 45 cos(45°–65°) = 45 cos(−20°) = +42.286 N (outward)
- 20 N: 20 cos(315°–65°) = 20 cos250° = −6.840 N (into the plane)

Sum of other forces' normal components = −45.315 + 25.651 + 42.286 − 6.840 = +15.78 N (net outward).

Because the net perpendicular tendency of the other forces is outward (they would pull the box off the plane), the contact normal cannot be negative; the block loses contact and the normal force is zero.

Answer: N = 0 N.