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Q.Find the torque of a force 7î + 3ĵ − 5k̂ about the origin. The force acts on a particle whose position vector is î − ĵ + k̂.

Nagaland NbseNagaland Board of School Education (Class XI) 2021Subjective· 2mImportance★★★★★
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Computing the cross product r×Fr \times F for the given position vector and force gives τ=2i^+12j^+10k^ N⋅m\tau = 2\hat{i}+12\hat{j}+10\hat{k}\ N\cdot m.

Torque about the origin is τ=r×F\tau = r \times F, where r=i^−j^+k^r = \hat{i} - \hat{j} + \hat{k} and F=7i^+3j^−5k^F = 7\hat{i} + 3\hat{j} - 5\hat{k}. Using the determinant form,

τ=∣i^j^k^1−1173−5∣\tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix}

τx=(−1)(−5)−(1)(3)=5−3=2\tau_x = (-1)(-5) - (1)(3) = 5 - 3 = 2 …

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