A force \(\vec F = 3\hat i + 4\;\hat j - 3\;\hat k\) is applied at the point P, whose position vector is \(\vec r = 2\hat i - 2\hat j - 3\hat k\) . What is the magnitude of the moment of the force about the origin?
23 units
Calculating the Magnitude of the Moment of Force
The moment of a force, also known as torque, measures the tendency of a force to rotate an object about a point or axis. When the moment is calculated about the origin, it is given by the cross product of the position vector of the point of application of the force and the force vector itself.
The formula for the moment of force (\(\vec \tau\)) about the origin is:
\(\vec \tau = \vec r \times \vec F\)
where:
- \(\vec r\) is the position vector of the point where the force is applied relative to the origin.
- \(\vec F\) is the force vector.
In this problem, we are given:
- Position vector \(\vec r = 2\hat i - 2\hat j - 3\hat k\)
- Force vector \(\vec F = 3\hat i + 4\;\hat j - 3\;\hat k\)
Calculating the Moment Vector \(\vec \tau\)
We need to compute the cross product \(\vec r \times \vec F\). This can be calculated using a determinant of a 3x3 matrix:
\(\vec \tau = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -2 & -3 \\ 3 & 4 & -3 \end{vmatrix}\)
Expanding the determinant:
\(\vec \tau = \hat i ((-2)(-3) - (-3)(4)) - \hat j ((2)(-3) - (-3)(3)) + \hat k ((2)(4) - (-2)(3))\)
Calculating the terms inside the parentheses:
- For \(\hat i\): \((-2)(-3) - (-3)(4) = 6 - (-12) = 6 + 12 = 18\)
- For \(\hat j\): \((2)(-3) - (-3)(3) = -6 - (-9) = -6 + 9 = 3\)
- For \(\hat k\): \((2)(4) - (-2)(3) = 8 - (-6) = 8 + 6 = 14\)
Substituting these values back into the determinant expansion:
\(\vec \tau = 18\hat i - 3\hat j + 14\hat k\)
So, the moment vector about the origin is \(\vec \tau = 18\hat i - 3\hat j + 14\hat k\).
Finding the Magnitude of the Moment of Force
The magnitude of a vector \(\vec V = a\hat i + b\hat j + c\hat k\) is given by \(|\vec V| = \sqrt{a^2 + b^2 + c^2}\).
For the moment vector \(\vec \tau = 18\hat i - 3\hat j + 14\hat k\), the magnitude is:
\(|\vec \tau| = \sqrt{(18)^2 + (-3)^2 + (14)^2}\)
Calculate the squares:
- \(18^2 = 324\)
- \((-3)^2 = 9\)
- \(14^2 = 196\)
Sum the squares:
\(|\vec \tau| = \sqrt{324 + 9 + 196}\)
\(|\vec \tau| = \sqrt{333 + 196}\)
\(|\vec \tau| = \sqrt{529}\)
To find the square root of 529, we can test perfect squares:
- \(20^2 = 400\)
- \(21^2 = 441\)
- \(22^2 = 484\)
- \(23^2 = 529\)
So, the magnitude of the moment of force is:
\(|\vec \tau| = 23\) units.
| Quantity | Vector Form | Components |
|---|---|---|
| Position Vector (\(\vec r\)) | \(2\hat i - 2\hat j - 3\hat k\) | (2, -2, -3) |
| Force Vector (\(\vec F\)) | \(3\hat i + 4\hat j - 3\hat k\) | (3, 4, -3) |
| Moment Vector (\(\vec \tau = \vec r \times \vec F\)) | \(18\hat i - 3\hat j + 14\hat k\) | (18, -3, 14) |
| Magnitude of Moment (|\(\vec \tau\)|) | N/A | \(\sqrt{18^2 + (-3)^2 + 14^2} = \sqrt{529} = 23\) |
Conclusion on Moment Calculation
The magnitude of the moment of the force \(\vec F = 3\hat i + 4\;\hat j - 3\;\hat k\) applied at the point P with position vector \(\vec r = 2\hat i - 2\hat j - 3\hat k\) about the origin is 23 units.
Revision Table: Key Concepts of Moment of Force
| Concept | Description | Formula (about Origin) |
|---|---|---|
| Moment of Force (\(\vec \tau\)) | Tendency of a force to cause rotation about a point or axis. Also called torque. | \(\vec \tau = \vec r \times \vec F\) |
| Position Vector (\(\vec r\)) | Vector from the reference point (origin) to the point of force application. | N/A |
| Force Vector (\(\vec F\)) | The vector representing the applied force. | N/A |
| Cross Product (\(\times\)) | Vector operation used to find a vector perpendicular to two input vectors; its magnitude is related to the area of the parallelogram they form. | \(\vec A \times \vec B = \begin{vmatrix} \hat i & \hat j & \hat k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}\) |
| Magnitude of a Vector | The length or size of the vector. | \(|\vec V| = \sqrt{V_x^2 + V_y^2 + V_z^2}\) |
Additional Information on Torque and Moment
The moment of force is a fundamental concept in rotational mechanics. It is a vector quantity, and its direction is given by the right-hand rule applied to the cross product \(\vec r \times \vec F\). The unit of moment of force in the SI system is Newton-meter (Nm).
When the moment is calculated about a point other than the origin, say with position vector \(\vec r_0\), the formula becomes \(\vec \tau = (\vec r - \vec r_0) \times \vec F\). Here, \((\vec r - \vec r_0)\) is the position vector of the point of force application relative to the new reference point \(\vec r_0\).
A zero moment of force implies either the force is zero, the position vector relative to the pivot is zero (force acts at the pivot), or the force vector is parallel or antiparallel to the position vector (force acts along the line passing through the pivot). A non-zero moment causes angular acceleration if it's the net moment acting on an object that is free to rotate.
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