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Difficulty: EasyEquilibrium of Forces, Center of Gravity and Moments

A rigid rod ABAB of length 2.0 m2.0\text{ m} is hinged at end AA. A force of 50 N50\text{ N} is applied at end BB at an angle of 30∘30^\circ to the rod. What is the moment of this force about the hinge AA?

  1. A
    25 N m25\text{ N m}
  2. 50 N m50\text{ N m}Answer
  3. C
    86.6 N m86.6\text{ N m}
  4. D
    100 N m100\text{ N m}

Answer

The moment of the force about hinge AA is 50 N m50\text{ N m}.
The moment of a force about a pivot is calculated by multiplying the force magnitude by the perpendicular distance to the pivot line of action (F×Lsin⁡θF \times L \sin\theta). Substituting F=50 NF = 50\text{ N}, L=2.0 mL = 2.0\text{ m}, and sin⁡(30∘)=0.5\sin(30^\circ) = 0.5 gives 50×2.0×0.5=50 N m50 \times 2.0 \times 0.5 = 50\text{ N m}.

Step-by-Step Solution

1
Identify the formula for the moment of a force applied at an angle.
Moment=F×d⊥=F×Lsin⁡θ\text{Moment} = F \times d_\perp = F \times L \sin\theta
Moment is defined as the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the pivot point.
2
Substitute the given values into the formula.
Moment=50 N×2.0 m×sin⁡(30∘)\text{Moment} = 50\text{ N} \times 2.0\text{ m} \times \sin(30^\circ)
The force F=50 NF = 50\text{ N}, distance L=2.0 mL = 2.0\text{ m}, and angle θ=30∘\theta = 30^\circ.
3
Calculate the final moment value.
Moment=50×2.0×0.5=50 N m\text{Moment} = 50 \times 2.0 \times 0.5 = 50\text{ N m}
Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, the perpendicular distance is 1.0 m1.0\text{ m}, giving a moment of 50 N m50\text{ N m}.

Key Concept

Moment of a Force at an Angle
Estimated Time:45s
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Topics

Equilibrium of Forces, Center of Gravity and Moments

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