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Notely Maths 12
Class 12 Maths
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7.5 Q-15

Question Statement

A force F‾=7i^+4j^−3k^\underline{F} = 7 \hat{i} + 4 \hat{j} - 3 \hat{k} is applied at point P(1,−2,3)P(1, -2, 3). Find its moment about the point Q(2,1,1)Q(2, 1, 1).

Background and Explanation

The moment of a force (or torque) about a point measures the rotational effect of that force relative to that point. The moment is calculated using the cross product of the position vector PQ→\overrightarrow{PQ} (from the point where the moment is to be calculated to the point where the force is applied) and the force vector F→\overrightarrow{F}.

The formula for the moment is:

M→=PQ→×F→\overrightarrow{M} = \overrightarrow{PQ} \times \overrightarrow{F}

Where:

  • PQ→\overrightarrow{PQ} is the displacement vector from point QQ to point PP,
  • F→\overrightarrow{F} is the force vector,
  • The cross product ×\times gives the moment vector.

Solution

Step 1: Define the given vectors

We are given the following information:

  • The force vector F→=7i^+4j^−3k^\overrightarrow{F} = 7 \hat{i} + 4 \hat{j} - 3 \hat{k},
  • Point P(1,−2,3)P(1, -2, 3) where the force is applied,
  • Point Q(2,1,1)Q(2, 1, 1) where we are calculating the moment.

Step 2: Calculate the displacement vector PQ→\overrightarrow{PQ}

The displacement vector PQ→\overrightarrow{PQ} is calculated by subtracting the coordinates of point Q(2,1,1)Q(2, 1, 1) from point P(1,−2,3)P(1, -2, 3):

PQ→=P−Q=(1,−2,3)−(2,1,1)\overrightarrow{PQ} = P - Q = (1, -2, 3) - (2, 1, 1) PQ→=(1−2,−2−1,3−1)=(−1,−3,2)\overrightarrow{PQ} = (1 - 2, -2 - 1, 3 - 1) = (-1, -3, 2)

Thus, the displacement vector is:

PQ→=−i^−3j^+2k^\overrightarrow{PQ} = -\hat{i} - 3 \hat{j} + 2 \hat{k}

Step 3: Compute the cross product PQ→×F→\overrightarrow{PQ} \times \overrightarrow{F}

The moment is calculated by the cross product of PQ→\overrightarrow{PQ} and F→\overrightarrow{F}:

PQ→×F→=∣i^j^k^−1−3274−3∣\overrightarrow{PQ} \times \overrightarrow{F} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} -1 & -3 & 2 7 & 4 & -3 \end{array} \right|

Now, expand the determinant:

PQ→×F→=i^((−3)(−3)−(2)(4))−j^((−1)(−3)−(2)(7))+k^((−1)(4)−(−3)(7))\overrightarrow{PQ} \times \overrightarrow{F} = \hat{i} \left( (-3)(-3) - (2)(4) \right) - \hat{j} \left( (-1)(-3) - (2)(7) \right) + \hat{k} \left( (-1)(4) - (-3)(7) \right)

Simplifying each term:

=i^(9−8)−j^(3−14)+k^(−4+21)= \hat{i} (9 - 8) - \hat{j} (3 - 14) + \hat{k} (-4 + 21) =i^(1)−j^(−11)+k^(17)= \hat{i} (1) - \hat{j} (-11) + \hat{k} (17) =i^+11j^+17k^= \hat{i} + 11 \hat{j} + 17 \hat{k}

Thus, the moment of the force about point QQ is:

M→=i^+11j^+17k^\overrightarrow{M} = \hat{i} + 11 \hat{j} + 17 \hat{k}

Key Formulas or Methods Used

  • Moment of Force (Torque):
M→=PQ→×F→ \overrightarrow{M} = \overrightarrow{PQ} \times \overrightarrow{F}

The moment is the cross product of the displacement vector PQ→\overrightarrow{PQ} and the force vector F→\overrightarrow{F}.

  • Cross Product: The cross product of two vectors A⃗=Axi^+Ayj^+Azk^\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} and B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} is calculated as:
A⃗×B⃗=(AyBz−AzBy)i^−(AxBz−AzBx)j^+(AxBy−AyBx)k^ \vec{A} \times \vec{B} = (A_y B_z - A_z B_y) \hat{i} - (A_x B_z - A_z B_x) \hat{j} + (A_x B_y - A_y B_x) \hat{k}

Summary of Steps

  1. Calculate the displacement vector PQ→\overrightarrow{PQ} by subtracting the coordinates of point QQ from point PP.
  2. Compute the cross product PQ→×F→\overrightarrow{PQ} \times \overrightarrow{F} to find the moment of the force.
  3. Simplify the cross product to obtain the moment vector M→=i^+11j^+17k^\overrightarrow{M} = \hat{i} + 11 \hat{j} + 17 \hat{k}.