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ExamsNEETPhysics

The moment of the force, \( \mathbf{F} = 4\mathbf{i} + 5\mathbf{j} - 6\mathbf{k} \) at \((2, 0, -3)\), about the point \((2, -2, -2)\), is given by

  1. \(-8\mathbf{i} - 4\mathbf{j} - 7\mathbf{k}\)
  2. \(-4\mathbf{i} - \mathbf{j} - 8\mathbf{k}\)
  3. \(-7\mathbf{i} - 4\mathbf{j} - 8\mathbf{k}\)
  4. \(-7\mathbf{i} - 8\mathbf{j} - 4\mathbf{k}\)

Correct answer: \(-7\mathbf{i} - 4\mathbf{j} - 8\mathbf{k}\)

Solution

The moment of a force is given by the cross product of the position vector (from the point about which the moment is calculated to the point of application of the force) and the force vector. The position vector is \\(\mathbf{r} = (2 - 2)\mathbf{i} + (0 - (-2))\mathbf{j} + (-3 - (-2))\mathbf{k} = 0\mathbf{i} + 2\mathbf{j} - 1\mathbf{k}\\. The moment is \\(\mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix} = -7\mathbf{i} - 4\mathbf{j} - 8\mathbf{k}\\.

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