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ExamsJEE AdvancedPhysics

A block of mass m rests in equilibrium on a rough inclined plane inclined at angle theta to the horizontal. Three forces act on it: force F1 acts perpendicular to the inclined surface (pushing into the surface), force F2 acts along the inclined surface pointing upward along the slope, and force F3 acts vertically downward. Which set of relations is correct?

  1. F1 + F2 + F3 = 0 (vector), F1 = F3 cos(theta), F2 = F3 sin(theta), F1² + F2² = F3²
  2. F1 + F2 + F3 = 0 (vector), F1 = F3 cos(theta), F2 = F3 sin(theta), F1² - F2² = F3²
  3. F1 + F2 + F3 = 0 (vector), F1 = F3 sin(theta), F2 = F3 cos(theta), F1² + F2² = F3²
  4. F1 + F2 + F3 = 0 (vector), F1 = F3 sin(theta), F2 = F3 cos(theta), F1² - F2² = F3²

Correct answer: F1 + F2 + F3 = 0 (vector), F1 = F3 cos(theta), F2 = F3 sin(theta), F1² + F2² = F3²

Solution

A vertical force F3 resolved onto an inclined plane gives F3 cos(theta) perpendicular to the surface and F3 sin(theta) along the surface. For equilibrium these must be balanced by F1 and F2 respectively, and by Pythagoras F1² + F2² = F3².

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