7.5 Q-14

Question Statement

Find the moment about point $A(1,1,1)$ of each of the concurrent forces:

1. $\underline{F_1} = \hat{i} - 2 \hat{j}$
2. $\underline{F_2} = 3 \hat{i} + 2 \hat{j} - \hat{k}$
3. $\underline{F_3} = 5 \hat{j} + 2 \hat{k}$

Where the point of concurrency is $P(2,0,1)$.

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Background and Explanation

The moment of a force (or torque) about a point is a measure of how much a force will cause an object to rotate about that point. The moment is given by the cross product of the position vector $\overrightarrow{A \vec{P}}$ (from the point of rotation to the point of force application) and the force vector $\overrightarrow{F}$.

For a set of concurrent forces, we can find the total force $\overrightarrow{R}$ by adding up all the individual forces. Then, the moment about a point is the cross product of the position vector from the point to the point of concurrency $\overrightarrow{A \vec{P}}$ and the resultant force vector $\overrightarrow{R}$:

$$\overrightarrow{M} = \overrightarrow{A \vec{P}} \times \overrightarrow{R}$$

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Solution

Step 1: Define the given vectors

We are given the following forces:

• $\overrightarrow{F_1} = \hat{i} - 2 \hat{j} + 0 \hat{k}$,
• $\overrightarrow{F_2} = 3 \hat{i} + 2 \hat{j} - \hat{k}$,
• $\overrightarrow{F_3} = 0 \hat{i} + 5 \hat{j} + 2 \hat{k}$.

The point of concurrency is $P(2, 0, 1)$, and the point where we are calculating the moment is $A(1, 1, 1)$.

Step 2: Calculate the resultant force $\overrightarrow{R}$

The resultant force is the vector sum of $\overrightarrow{F_1}$, $\overrightarrow{F_2}$, and $\overrightarrow{F_3}$:

$$\overrightarrow{R} = \overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3}$$

Substituting the values of the forces:

$$\overrightarrow{R} = (\hat{i} - 2 \hat{j} + 0 \hat{k}) + (3 \hat{i} + 2 \hat{j} - \hat{k}) + (0 \hat{i} + 5 \hat{j} + 2 \hat{k})$$

Now, add the components:

$$\overrightarrow{R} = (1 + 3 + 0) \hat{i} + (-2 + 2 + 5) \hat{j} + (0 - 1 + 2) \hat{k}$$ $$\overrightarrow{R} = 4 \hat{i} + 5 \hat{j} + \hat{k}$$

Thus, the resultant force is:

$$\overrightarrow{R} = 4 \hat{i} + 5 \hat{j} + \hat{k}$$

Step 3: Calculate the position vector $\overrightarrow{A \vec{P}}$

The displacement vector $\overrightarrow{A \vec{P}}$ is found by subtracting the coordinates of point $A(1, 1, 1)$ from the coordinates of point $P(2, 0, 1)$:

$$\overrightarrow{A \vec{P}} = P - A = (2, 0, 1) - (1, 1, 1)$$ $$\overrightarrow{A \vec{P}} = (2 - 1, 0 - 1, 1 - 1) = (1, -1, 0)$$

Thus, the position vector is:

$$\overrightarrow{A \vec{P}} = \hat{i} - \hat{j}$$

Step 4: Calculate the moment of the resultant force

The moment of the resultant force about point $A$ is given by the cross product:

$$\overrightarrow{M} = \overrightarrow{A \vec{P}} \times \overrightarrow{R}$$

Substitute the values of the position vector and the resultant force:

$$\overrightarrow{M} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 1 & -1 & 0 4 & 5 & 1 \end{array} \right|$$

Now, expand the determinant:

$$\overrightarrow{M} = \hat{i} \left( (-1)(1) - (0)(5) \right) - \hat{j} \left( (1)(1) - (0)(4) \right) + \hat{k} \left( (1)(5) - (-1)(4) \right)$$

Simplifying each term:

$$= \hat{i} (-1 - 0) - \hat{j} (1 - 0) + \hat{k} (5 + 4)$$ $$= -\hat{i} - \hat{j} + 9 \hat{k}$$

Thus, the moment of the resultant force about point $A$ is:

$$\overrightarrow{M} = -\hat{i} - \hat{j} + 9 \hat{k}$$

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Key Formulas or Methods Used

• Moment of Force (Torque):

$$\overrightarrow{M} = \overrightarrow{A \vec{P}} \times \overrightarrow{R}$$

The moment is the cross product of the displacement vector $\overrightarrow{A \vec{P}}$ and the resultant force vector $\overrightarrow{R}$.

• Cross Product: The cross product of two vectors $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$ is calculated as:

$$\vec{A} \times \vec{B} = (A_y B_z - A_z B_y) \hat{i} - (A_x B_z - A_z B_x) \hat{j} + (A_x B_y - A_y B_x) \hat{k}$$

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Summary of Steps

1. Calculate the resultant force $\overrightarrow{R}$ by adding the forces $\overrightarrow{F_1}, \overrightarrow{F_2}, \overrightarrow{F_3}$.
2. Find the displacement vector $\overrightarrow{A \vec{P}}$ by subtracting the coordinates of point $A$ from point $P$.
3. Compute the moment by taking the cross product of $\overrightarrow{A \vec{P}}$ and $\overrightarrow{R}$, yielding $\overrightarrow{M} = -\hat{i} - \hat{j} + 9 \hat{k}$.