Question:
A force $F=5 \hat{i}+2\hat{j}-5 \hat{k}$ acts on a particle whose position vector is $r=\hat{i}-2 \hat{j}+\hat{k}$ What is the torque about the origin?
A force $F=5 \hat{i}+2\hat{j}-5 \hat{k}$ acts on a particle whose position vector is $r=\hat{i}-2 \hat{j}+\hat{k}$ What is the torque about the origin?
Updated On: Dec 19, 2024
- $8 \hat{i}+10 \hat{j}+12 \hat{k}$
- $8 \hat{i}+10 \hat{j}-12 \hat{k}$
- $8 \hat{i}-10 \hat{j}-8 \hat{k}$
- $10 \hat{ i }-10 \hat{ j }-\hat{ k }$
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The Correct Option is A
Solution and Explanation
Given,
$F =5 \hat{ i }+2 \hat{ j }-5 \hat{ k }$
and $r =\hat{ i }-2 \hat{ j }+\hat{ k }$
We know that $\tau= r \times F$
So, torque about the origin will be given by,
$=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -2 & 1 \\ 5 & +2 & -5 \end{vmatrix}$
$=\hat{ i }(10-2)-\hat{ j }(-5-5)+\hat{ k }(2+10)$
$=8 \hat{ i }+10 \hat{ j }+12 \hat { k }$
$F =5 \hat{ i }+2 \hat{ j }-5 \hat{ k }$
and $r =\hat{ i }-2 \hat{ j }+\hat{ k }$
We know that $\tau= r \times F$
So, torque about the origin will be given by,
$=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -2 & 1 \\ 5 & +2 & -5 \end{vmatrix}$
$=\hat{ i }(10-2)-\hat{ j }(-5-5)+\hat{ k }(2+10)$
$=8 \hat{ i }+10 \hat{ j }+12 \hat { k }$
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