The torque or turning effect of force about a given point is given by r x F where r is the vector from the given point to the point of application of F. Consider a force F = - 3 `\hat {i}` + `\hat {j}` +5`\hat {k}` (newton) acting on the point 7`\hat {i}` + 3 `\hat {j}` + `\hat {k}` (m). What is the torque in N m about the origin?


(Ans: 14`\hat {i}` -38 `\hat {j}` +16`\hat {k}` Nm ]

Given:

Force = `\vec {F}` = -3`\hat {i}` + `\hat {j}` + 5`\hat {k}` Newton
Position Vector = `\vec {r}` = 7`\hat {i}` + 3`\hat {j}` + `\hat {k}`  meters

To Find:

Torque = T = ?

Solution: 

We know that 

T = `\vec {r}` ｘ `\vec {F}` 

by putting values

T = `\vec {r}` ｘ `\vec {F}`  = (-3`\hat {i}` + `\hat {j}` + 5`\hat {k}`)ｘ (7`\hat {i}` + 3`\hat {j}` + `\hat {k}`)  Nm

Cross product of two vectors can be solved by writing determinant method and then expanding


T = `\vec {r}` ｘ `\vec {F}`  = `\hat {i}`(15 - 1) - `\hat {j}` (35 + 3) + `\hat {i}`(7 +9)


T = `\vec {r}` ｘ `\vec {F}`  = 14`\hat {i}` - 38`\hat {j}` + 16 `\hat {i}` Nm  ----- Ans



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Numerical Problem 2.12
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List of 11th Physics Numerical Problems of Chapter - 2
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