Free Body Diagram Physics Example 3

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Example 3

medium
A 4 kg4 \text{ kg} block slides down a frictionless ramp inclined at 45°45°. Identify all forces in the free-body diagram and find the acceleration. Use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution

  1. 1
    Forces: weight W=mg=39.2 NW = mg = 39.2 \text{ N} (vertically down) and normal force NN (perpendicular to ramp). No friction.
  2. 2
    Component of weight along the ramp: W∥=mgsin⁡45°=39.2×0.707=27.7 NW_{\parallel} = mg\sin 45° = 39.2 \times 0.707 = 27.7 \text{ N}.
  3. 3
    Acceleration down the ramp: a=gsin⁡45°=9.8×0.707=6.93 m/s2a = g\sin 45° = 9.8 \times 0.707 = 6.93 \text{ m/s}^2

Answer

a≈6.93 m/s2a \approx 6.93 \text{ m/s}^2
On a frictionless incline, the FBD has only weight and normal force. The acceleration depends on the component of gravity parallel to the ramp surface, which equals gsin⁡θg\sin\theta.

About Free Body Diagram

A simplified diagram that isolates a single object and represents all external forces acting on it as labelled arrows originating from the object's centre of.

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