Numerical Problem No. 2.13 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

The line of action of force. `\vec {F}` = `\hat {i}` - 2`\hat {j}` , passes through a point whose position vector is (-`\hat {j}` + `\hat {k}`). Find 

(a) the moment of `\vec {F}` about the origin, 

(b) the moment of `\vec {F}` about the point of which the position vector is `\hat {i}` + `\hat {k}`.

Given:

Force = `\vec {F}` = `\hat {i}` - 2`\hat {j}` 

Position Vector = `\vec {r}` = -`\hat {j}` + `\hat {k}`

To Find:

(a) The moment of F about the origin = `\vec {r}` ｘ `\vec {F}` 

(b) The moment of F about the point of which the position vector is `\hat {i}` + `\hat {k}` = `\vec {r}` ｘ `\vec {F}` 

Solution: 

(a) The moment of F about the origin = `\vec {r}` ｘ `\vec {F}` 

`\vec {r}` ｘ`\vec {F}` = (-`\hat {j}` + `\hat {k}` ) ｘ( `\hat {i}` - 2`\hat {j}`  )

`\vec {r}` ｘ`\vec {F}` = (-1 ｘ 1) `hat {j}` ｘ `hat {i}` + (-1 ｘ -2) `hat {j}` ｘ `hat {j}` + (1 ｘ 1) `hat {k}` ｘ `hat {i}` + (1 ｘ -2) `hat {k}` ｘ `hat {j}` 

[∴  `hat {j}` ｘ `hat {j}`  = 0  and

`hat {j}` ｘ `hat {i}` = -`hat {k}` , `hat {k}` ｘ `hat {i}` = `hat {j}`, `hat {i}` ｘ `hat {k}`= -`hat {j}` ]

So by putting these values we have

`\vec {r}` ｘ`\vec {F}` = - (-`hat {k}`)  + 0` + `hat {j}` -2 (-`hat {i}` )

or

`\vec {r}` ｘ`\vec {F}` = 2 `hat {i}` + `hat {j}` + `hat {k}` ------Ans

(b) The moment of F about the point of which the position vector is `\hat {i}` + `\hat {k}` = `\vec {r}` ｘ `\vec {F}` 

Here we will get the moment arm `\vec {r}` by subtracting the new position vector  `\hat {i}` + `\hat {k}` from the origin position vector -`\hat {j}` + `\hat {k}` So, 

`\vec {r}` = -`\hat {j}` + `\hat {k}` - (`\hat {i}` + `\hat {k}`)

`\vec {r}` = -`\hat {j}` + `\hat {k}` - `\hat {i}` - `\hat {k})`

`\vec {r}` = -`\hat {j}` - `\hat {i}` 

`\vec {r}` = -`\hat {i}` - `\hat {j}` 

`\vec {r}` ｘ`\vec {F}` = (-`\hat {i}` - `\hat {j}`  ) ｘ( `\hat {i}` - 2`\hat {j}`  )

`\vec {r}` ｘ`\vec {F}` = (-1 ｘ 1) `hat {i}` ｘ `hat {i}` + (-1 ｘ -2) `hat {i}` ｘ `hat {j}` + (-1 ｘ 1) `hat {j}` ｘ `hat {i}` + (-1 ｘ -2) `hat {j}` ｘ `hat {j}` 

[∴ `hat {i}` ｘ `hat {i}` = `hat {j}` ｘ `hat {j}`  =  = 0

and

`hat {i}` ｘ `hat {j}` = `hat {k}`, `hat {j}` ｘ `hat {i}` = -`hat {k}` ]

So by putting these values we have

`\vec {r}` ｘ`\vec {F}` = o + 2 `hat {k}` - (-`hat {k}`) + 0

or

`\vec {r}` ｘ`\vec {F}` =  3 `hat {k}` --------Ans

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Numerical Problem 2.13
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.12  ⇚   11th Physics Main Content List    ⇛ Numerical Problem 2.14

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