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Moment of a force of magnitude 20 N acti...

Moment of a force of magnitude 20 N acting along positive x-direction at point `(3m,0,0)` about the point `(0,2,0)` (in N-m) is

A

20

B

60

C

40

D

30

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The correct Answer is:
To solve the problem of finding the moment of a force about a point, we can follow these steps: ### Step 1: Identify the Given Information - **Force (F)**: 20 N acting along the positive x-direction. This can be represented as a vector: \[ \vec{F} = 20 \hat{i} \] - **Point of application of force**: \( (3 \, \text{m}, 0, 0) \) - **Point about which we are calculating the moment**: \( (0, 2, 0) \) ### Step 2: Calculate the Position Vector (r) The position vector \( \vec{r} \) from the point about which we are calculating the moment to the point of application of the force is given by: \[ \vec{r} = (3 - 0) \hat{i} + (0 - 2) \hat{j} + (0 - 0) \hat{k} \] This simplifies to: \[ \vec{r} = 3 \hat{i} - 2 \hat{j} \] ### Step 3: Use the Moment Formula The moment \( \vec{M} \) of a force about a point is given by the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \): \[ \vec{M} = \vec{r} \times \vec{F} \] ### Step 4: Set Up the Cross Product We can express this in determinant form: \[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 0 \\ 20 & 0 & 0 \end{vmatrix} \] ### Step 5: Calculate the Determinant Calculating the determinant, we have: \[ \vec{M} = \hat{i} \begin{vmatrix} -2 & 0 \\ 0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 0 \\ 20 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -2 \\ 20 & 0 \end{vmatrix} \] Calculating each of these 2x2 determinants: - For \( \hat{i} \): \( (-2)(0) - (0)(0) = 0 \) - For \( \hat{j} \): \( (3)(0) - (20)(0) = 0 \) - For \( \hat{k} \): \( (3)(0) - (-2)(20) = 40 \) Putting it all together: \[ \vec{M} = 0 \hat{i} - 0 \hat{j} + 40 \hat{k} = 40 \hat{k} \] ### Step 6: Find the Magnitude of the Moment The magnitude of the moment \( |\vec{M}| \) is: \[ |\vec{M}| = 40 \, \text{N-m} \] ### Final Answer The moment of the force about the point \( (0, 2, 0) \) is: \[ \boxed{40 \, \text{N-m}} \]

To solve the problem of finding the moment of a force about a point, we can follow these steps: ### Step 1: Identify the Given Information - **Force (F)**: 20 N acting along the positive x-direction. This can be represented as a vector: \[ \vec{F} = 20 \hat{i} \] - **Point of application of force**: \( (3 \, \text{m}, 0, 0) \) ...
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DC PANDEY ENGLISH-ROTATION-Check point 9.2
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