Free Body Diagram Physics Example 4

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Example 4

hard
A 6 kg6 \text{ kg} block hangs from two ropes: one horizontal (attached to a wall) and one at 60°60° above the horizontal (attached to the ceiling). Find the tension in each rope. Use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution

  1. 1
    FBD forces: weight W=6×9.8=58.8 NW = 6 \times 9.8 = 58.8 \text{ N} (down), tension T1T_1 (horizontal, toward wall), tension T2T_2 at 60°60° above horizontal.
  2. 2
    Vertical equilibrium: T2sin⁡60°=W  ⟹  T2=58.8sin⁡60°=58.80.866=67.9 NT_2 \sin 60° = W \implies T_2 = \frac{58.8}{\sin 60°} = \frac{58.8}{0.866} = 67.9 \text{ N}.
  3. 3
    Horizontal equilibrium: T1=T2cos⁡60°=67.9×0.5=33.9 NT_1 = T_2 \cos 60° = 67.9 \times 0.5 = 33.9 \text{ N}.

Answer

T1≈33.9 N,T2≈67.9 NT_1 \approx 33.9 \text{ N}, \quad T_2 \approx 67.9 \text{ N}
Drawing a careful free-body diagram and resolving each tension into components allows us to apply equilibrium conditions in both the horizontal and vertical directions independently.

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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