A block of mass 2 kg is placed on an incline with angle 37 degrees (sin = 0.6, cos = 0.8) and static friction coefficient mu_s = 0.5. A variable force F is applied parallel to the incline. Taking g = 10 m/s^2, what is the range of F for which the block remains in static equilibrium?

Answer and explanation

Correct answer: C. 4 N <= F <= 20 N (directed up the incline)

  1. Step 1: Normal force: N = mgcos(37) = 2 10 0.8 = 16 N.
  2. Step 2: Component of gravity down the incline: F_g = mgsin(37) = 2 10 0.6 = 12 N.
  3. Step 3: Maximum static friction: f_{s,max} = mu_s N = 0.5 16 = 8 N.
  4. Step 4: To prevent sliding down: F + f_{s,max} >= F_g, so F >= 12 - 8 = 4 N.
  5. Step 5: To prevent sliding up: F - f_{s,max} <= F_g, so F <= 12 + 8 = 20 N.
  6. Step 6: Therefore, 4 N <= F <= 20 N.
Answer
4 N <= F <= 20 N (directed up the incline).
Why not the others: 0 N <= F <= 12 N assumes friction is 12 N; 8 N <= F <= 16 N uses friction limits 4 N; 6 N <= F <= 24 N uses incorrect trigonometry values.

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