ගොනුව:බල ඝූර්ණ.jpg

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

${\displaystyle M=\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\cdots }$

Now we pick a new reference point P' that differs from P by the vector r. The new moment is

${\displaystyle M'=(\mathbf {r} _{1}+\mathbf {r} )\times \mathbf {F} _{1}+(\mathbf {r} _{2}+\mathbf {r} )\times \mathbf {F} _{2}+\cdots }$

Now the distributive property of the cross product implies

${\displaystyle M'=\left(\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\cdots \right)+\mathbf {r} \times \left(\mathbf {F} _{1}+\mathbf {F} _{2}+\cdots \right).}$

However, the definition of a force couple means that

${\displaystyle \mathbf {F} _{1}+\mathbf {F} _{2}+\cdots =0.}$

Therefore,

${\displaystyle M'=\mathbf {r} _{1}\times \mathbf {F} _{1}+\mathbf {r} _{2}\times \mathbf {F} _{2}+\cdots =M}$

This proves that the moment is independent of reference point, which is proof that a couple is a free vector.

"http://si.wikipedia.org/wiki/බල_යුග්ම"

12-03-2013

~Ruvini Bandara~

පහත බලන්න.