The line of action of force. â†’F = ˆi - 2ˆj , passes through a point whose position vector is (-ˆj + Ë†k). Find 

(a) the moment of â†’F about the origin, 

(b) the moment of â†’F about the point of which the position vector is Ë†i + Ë†k.


Given:

Force = →F = ˆi - 2ˆj 
Position Vector = →r = -ˆj + Ë†k

To Find:

(a) The moment of F about the origin = â†’r ｘ â†’F 

(b) The moment of F about the point of which the position vector is Ë†i + Ë†k = â†’r ï½˜ â†’F 


Solution: 

(a) The moment of F about the origin = â†’r ｘ â†’F 

→r ｘ→F = (-ˆj + Ë†k ) ｘ( Ë†i - 2ˆj  )

→r ｘ→F = (-1 ｘ 1) Ë†j ï½˜ Ë†i + (-1 ｘ -2) Ë†j ï½˜ Ë†j + (1 ｘ 1) Ë†k ï½˜ Ë†i + (1 ｘ -2) Ë†k ï½˜ Ë†j 

[∴  Ë†j ï½˜ Ë†j  = 0  and
ˆj ï½˜ Ë†i = -`hat {k}` , Ë†k ï½˜ Ë†i = Ë†j, Ë†i ï½˜ Ë†k= -ˆj ]

So by putting these values we have

→r ｘ→F = - (-ˆk)  + 0` + Ë†j -2 (-ˆi )

or

→r ｘ→F = 2 Ë†i + Ë†j + Ë†k ------Ans


(b) The moment of F about the point of which the position vector is Ë†i + Ë†k = â†’r ï½˜ â†’F 

Here we will get the moment arm â†’r by subtracting the new position vector  Ë†i + Ë†k from the origin position vector -ˆj + Ë†k So, 

→r = -ˆj + Ë†k - (ˆi + Ë†k)

→r = -ˆj + Ë†k - ˆi - Ë†k)

→r = -ˆj - Ë†i 

→r = -ˆi - Ë†j 

→r ｘ→F = (-ˆi - Ë†j  ) ｘ( Ë†i - 2ˆj  )

→r ｘ→F = (-1 ｘ 1) Ë†i ï½˜ Ë†i + (-1 ｘ -2) Ë†i ï½˜ Ë†j + (-1 ｘ 1) Ë†j ï½˜ Ë†i + (-1 ｘ -2) Ë†j ï½˜ Ë†j 

[∴ Ë†i ï½˜ Ë†i = ˆj ï½˜ Ë†j  =  = 0
and
ˆi ï½˜ Ë†j = Ë†k, Ë†j ï½˜ Ë†i = -ˆk ]

So by putting these values we have

→r ｘ→F = o + 2 Ë†k - (-ˆk) + 0

or

→r ｘ→F =  3 Ë†k --------Ans



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Numerical Problem 2.13
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List of 11th Physics Numerical Problems of Chapter - 2
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