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				   give proof for the query in the subject


6 years ago

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										A Couple is a system of forces with a resultant moment but no resultant force. Another term for a couple is a pure moment. Its effect is to create rotation without translation, or more generally without any acceleration of the centre of mass.as its does not move the centre of mass that's why the the moment of force produced by a couple is same about every point.

6 years ago
										The proof of this claim is as follows: Suppose there are a set of force vectors F1, F2, etc. that form a couple, with position vectors (about some origin P) r1, r2, etc., respectively. The moment about P is
$M = \mathbf{r}_1\times \mathbf{F}_1 + \mathbf{r}_2\times \mathbf{F}_2 + \cdots$
Now we pick a new reference point P' that differs from P by the vector r. The new moment is
$M' = (\mathbf{r}_1+\mathbf{r})\times \mathbf{F}_1 + (\mathbf{r}_2+\mathbf{r})\times \mathbf{F}_2 + \cdots$
We can simplify this as follows:
$M' = \mathbf{r}_1\times \mathbf{F}_1+\mathbf{r}\times \mathbf{F}_1 + \mathbf{r}_2\times \mathbf{F}_2+\mathbf{r}\times \mathbf{F}_2 + \cdots$$M' = \left(\mathbf{r}_1\times \mathbf{F}_1 + \mathbf{r}_2\times \mathbf{F}_2 + \cdots\right) + \mathbf{r}\times \left(\mathbf{F}_1 + \mathbf{F}_2 + \cdots \right)$
The definition of a force couple means that
$\mathbf{F}_1 + \mathbf{F}_2 + \cdots = 0$
Therefore,
$M' = \mathbf{r}_1\times \mathbf{F}_1 + \mathbf{r}_2\times \mathbf{F}_2 + \cdots = M$
This proves that the moment is independent of reference point.

6 years ago

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