Application of Scalar Product
Work done
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}\ where\ \overrightarrow{d}= \overrightarrow {OB}- \overrightarrow{OA}\]
PART – B
Example :
\[Find\ the\ work\ done\ by\ the\ force\ 3\overrightarrow{i}+ 5\overrightarrow{j}+ 7\overrightarrow{k},\ when\ the\ displacement\ is\ 2\overrightarrow{i}- \overrightarrow{j} +\overrightarrow{k}\]
Soln:
\[\overrightarrow{F}= 3\overrightarrow{i}+ 5\overrightarrow{j}+7\overrightarrow{k} \]
\[\overrightarrow{d}= 2\overrightarrow{i}- \overrightarrow{j}+\overrightarrow{k} \]
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}= (3\overrightarrow{i}+ 5\overrightarrow{j}+ 7\overrightarrow{k}) .(2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k})\]
= 3 ( 2 ) + 5 ( – 1 ) + 7 ( 1 )
= 6 – 5 + 7
Work done = 8 units
Example :
\[A\ particle\ acted\ on\ by\ the\ forces\ 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} and\ \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k} acting\ on\ the\ particle\]
\[displaces\ the\ particle\ from\ the\ point\ \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k} to\ the\ point\ 3\overrightarrow{i}- 5\overrightarrow{j}+4\overrightarrow{k}. Find\ the\ total\ work\ done\ by\ the\ forces.\]
Soln:
\[\overrightarrow{F_1}= 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} \]
\[\overrightarrow{F_2}= \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k} \]
\[\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} + \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k} \]
\[\overrightarrow{F}= 4\overrightarrow{i}+ 9\overrightarrow{j}+4\overrightarrow{k} \]
\[\overrightarrow{OA}= \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k} \]
\[\overrightarrow{OB}= 3\overrightarrow{i}-5\overrightarrow{j}+ 4\overrightarrow{k} \]
\[\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}\]
\[=3\overrightarrow{i}\ – 5\overrightarrow{j} + 4\overrightarrow{k}- (\overrightarrow{i}\ + 2\overrightarrow{j}+3\overrightarrow{k})\]
\[=\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k}- \overrightarrow{i}\ – 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{d}= 2\overrightarrow{i}- 7\overrightarrow{j}+\overrightarrow{k} \]
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}= (4\overrightarrow{i}+ 9\overrightarrow{j}+ 4\overrightarrow{k}) .(2\overrightarrow{i}-7 \overrightarrow{j}+ 1\overrightarrow{k})\]
= 4 ( 2 ) + 9 ( – 7 ) + 4 ( 1 )
= 8 – 63 + 4
Work done = –51 units
Work done = 51 units (by taking positive value)
Example :
\[A\ particle\ acted\ on\ by\ the\ forces\ 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k} and\ 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\]
is displaced from the point ( 1, 1, 1 ) to the point ( 2, – 3, 5 ). Find the total work done.
Soln:
\[\overrightarrow{F_1}= 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{F_2}= 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\]
\[\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k} + 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\]
\[\overrightarrow{F}= 6\overrightarrow{i}+ 10\overrightarrow{j}-\overrightarrow{k}\]
\[\overrightarrow{OA}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}-3\overrightarrow{j}+ 5\overrightarrow{k}\]
\[\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}\]
\[=2\overrightarrow{i}\ – 3\overrightarrow{j} + 5\overrightarrow{k}- (\overrightarrow{i}\ + \overrightarrow{j}+\overrightarrow{k})\]
\[=2\overrightarrow{i}\ – 3\overrightarrow{j} + 5\overrightarrow{k}- \overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{d}= \overrightarrow{i}- 4\overrightarrow{j}+4\overrightarrow{k}\]
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}= (6\overrightarrow{i}+ 10\overrightarrow{j}-\overrightarrow{k}) .(\overrightarrow{i}- 4\overrightarrow{j}+4\overrightarrow{k})\]
= 6 ( 1 ) + 10 ( -4 ) – 1 ( 4 )
= 6 – 40 – 4
= – 38
Work done = –38 units
Work done = 38 units (by taking positive value)
Example :
\[A\ particle\ is\ displaced\ from\ the\ point\ 5\overrightarrow{i}- 5\overrightarrow{j}- 7\overrightarrow{k}\ to\ the\ point\ 6\overrightarrow{i}+ 2\overrightarrow{j}-2\overrightarrow{k}\]\[under\ the\ action\ of\ forces\ 10\overrightarrow{i}- \overrightarrow{j} + 11\overrightarrow{k},\ 4\overrightarrow{i} + 5\overrightarrow{j} +6\overrightarrow{k}\ and\ -2\overrightarrow{i} + \overrightarrow{j}- 9\overrightarrow{k}.\]\[Calculate\ the\ total\ work\ done\ by\ the\ forces.\]
Soln:
\[\overrightarrow{F_1}= 10\overrightarrow{i}- \overrightarrow{j}+ 11\overrightarrow{k}\]
\[\overrightarrow{F_2}= 4\overrightarrow{i} + 5\overrightarrow{j}+ 6\overrightarrow{k}\]
\[\overrightarrow{F_3}= -2\overrightarrow{i}+ \overrightarrow{j}- 9\overrightarrow{k}\]
\[\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3}\]\[= 10\overrightarrow{i}- \overrightarrow{j}+ 11\overrightarrow{k} + 4\overrightarrow{i} + 5\overrightarrow{j}+ 6\overrightarrow{k}- 2\overrightarrow{i} + \overrightarrow{j}- 9\overrightarrow{k}\]
\[\overrightarrow{F}= 12\overrightarrow{i}+ 5\overrightarrow{j} +8\overrightarrow{k}\]
\[\overrightarrow{OA}= 5\overrightarrow{i}- 5\overrightarrow{j}- 7\overrightarrow{k}\]
\[\overrightarrow{OB}= 6\overrightarrow{i} +2\overrightarrow{j} – 2\overrightarrow{k}\]
\[\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}\]
\[=6\overrightarrow{i} +2\overrightarrow{j} – 2\overrightarrow{k}- (5\overrightarrow{i}- 5\overrightarrow{j}- 7\overrightarrow{k})\]
\[=6\overrightarrow{i} +2\overrightarrow{j} – 2\overrightarrow{k}- 5\overrightarrow{i} + 5\overrightarrow{j}+ 7\overrightarrow{k}\]
\[\overrightarrow{d}= \overrightarrow{i}+ 7\overrightarrow{j}+5\overrightarrow{k}\]
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}= (12\overrightarrow{i}+ 5\overrightarrow{j} +8\overrightarrow{k}) .(\overrightarrow{i}+ 7\overrightarrow{j}+5\overrightarrow{k})\]
\[=12(1) + 5(7) + 8(5)\]
\[= 12 + 35 + 40\]
\[Work\ done\ = 87\ units\]
Application of Vector Product
Moment (or) Torque of a force about a point
\[Let\ A\ be\ any\ point\ and\ \overrightarrow{r}\ be\ the\ position\ vector\ relative\ to\ the\ point\ A\]\[of\ any\ point\ P\ on\ the\ line\ of\ the\ action\ of\ the\ force\ \overrightarrow{F}.\]\[The\ moment\ of\ the\ force\ about\ the\ point\ O\ is\ defined\ as\ \overrightarrow{M}= \overrightarrow{r}× \overrightarrow{F}\]\[where\ \overrightarrow{r}= \overrightarrow{AP}= \overrightarrow{OP}- \overrightarrow{OA}\].
\[ The\ Magnitude \ of\ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]
Example :
\[Find\ the\ moment\ of\ the\ force\ 3\overrightarrow{i}+\overrightarrow{k}\ acting\ through\ the\ point\ \overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k} about\ the\ point\ 2\overrightarrow{i}+ \overrightarrow{j}-2\overrightarrow{k}.\]
Soln:
\[\overrightarrow{F}= 3\overrightarrow{i} + \overrightarrow{k}\]
\[\overrightarrow{OP}= \overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}\]
\[\overrightarrow{OA}= 2\overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}\]
\[\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}\]
\[=\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}- (2\overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k})\]
\[=\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}- 2\overrightarrow{i}-\overrightarrow{j} +2\overrightarrow{k})\]
\[\overrightarrow{r}= -\overrightarrow{i} + \overrightarrow{j} + \overrightarrow{k}\]
\[Moment = \overrightarrow{r}× \overrightarrow{F}\]
\[ =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\
-1 & 1 & 1\\
3 & 0 & 1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 1 – 0) -\overrightarrow{j}(-1 – 3)+\overrightarrow{k}(0 – 3)\]
\[ = \overrightarrow{i}(1) -\overrightarrow{j}(-4)+\overrightarrow{k}(-3)\]
\[\overrightarrow{r}× \overrightarrow{F}= \overrightarrow{i}+ 4\overrightarrow{j}-3\overrightarrow{k}\]
\[ Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]
\[= \sqrt{(1)^2 + (4)^2 + (-3)^2 }=\sqrt{(1 + 16 +9 }=\sqrt{26}\]
\[ Magnitude\ of \ Moment = \sqrt{26}\ units\]
Example :
\[Find\ the\ moment\ of\ the\ force\ 3\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k}\ acting\ through\ the\ point\ \overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k} about\ the\ point\ 4\overrightarrow{i}- 3\overrightarrow{j}+\overrightarrow{k}.\]
Soln:
\[\overrightarrow{F}= 3\overrightarrow{i} +4\overrightarrow{j} + 5\overrightarrow{k}\]
\[\overrightarrow{OP}= \overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{OA}= 4\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k}\]
\[\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}\]
\[=\overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}- (4\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})\]
\[=\overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}- 4\overrightarrow{i} +3\overrightarrow{j} -\overrightarrow{k}\]
\[\overrightarrow{r}= -3\overrightarrow{i} + \overrightarrow{j} + 2\overrightarrow{k}\]
\[Moment = \overrightarrow{r}× \overrightarrow{F}\]
\[ =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
-3 & 1 & 2\\
3 & 4 & 5\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 5 – 8) -\overrightarrow{j}(-15 – 6)+\overrightarrow{k}(-12 – 3)\]
\[\overrightarrow{r}× \overrightarrow{F}= -3\overrightarrow{i}+ 21\overrightarrow{j}-15\overrightarrow{k}\]
\[ Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]
\[= \sqrt{(-3)^2 + (21)^2 + (-15)^2 }=\sqrt{(9 + 441+ 225 }=\sqrt{675}\]
\[ Magnitude\ of \ Moment = \sqrt{675}\ units\]
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