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NEET 2018 · Torque, Couple & Equilibrium of Rigid Bodies

Torque, Couple & Equilibrium of Rigid Bodies question from NEET 2018

Q36NEET 2018Easy

Torque, Couple & Equilibrium of Rigid Bodies

The moment of the force, F⃗=4i^+5j^−6k^\vec{F} = 4\hat{i}+5\hat{j}-6\hat{k} at

(2, 0, –3), about the point (2, –2, –2), is given by

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Answer: (d)

  1. Position relative to the point: r⃗=(2,0,−3)−(2,−2,−2)=(0,2,−1)\vec{r} = (2,0,-3)-(2,-2,-2) = (0,2,-1).
  2. τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F} with F⃗=(4,5,−6)\vec{F} = (4,5,-6).
  3. τx=2(−6)−(−1)(5)=−7\tau_x = 2(-6)-(-1)(5) = -7, τy=−[0(−6)−(−1)(4)]=−4\tau_y = -[0(-6)-(-1)(4)] = -4, τz=0(5)−2(4)=−8\tau_z = 0(5)-2(4) = -8
  4. So the answer is (d) −7i^−4j^−8k^-7\hat{i}-4\hat{j}-8\hat{k}.
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