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Question: If $\overrightarrow{F}=3\hat{i}+5\hat{j}-2\hat{k}$ N acting at $7\hat{i}-2\hat{j}+5\hat{k}$, then fi...

If F→=3i^+5j^−2k^\overrightarrow{F}=3\hat{i}+5\hat{j}-2\hat{k} N acting at 7i^−2j^+5k^7\hat{i}-2\hat{j}+5\hat{k}, then find the torque about the point (0,-1,0).

A

−23i^+29j^+38k^-23\hat{i}+29\hat{j}+38\hat{k}

B

−19i^+29j^+44k^-19\hat{i}+29\hat{j}+44\hat{k}

C

−19i^−29j^+44k^-19\hat{i}-29\hat{j}+44\hat{k}

D

−23i^−29j^+38k^-23\hat{i}-29\hat{j}+38\hat{k}

Answer

-23\hat{i}+29\hat{j}+38\hat{k}

Explanation

Solution

To find the torque (τ→\overrightarrow{\tau}) about a point, we use the formula: τ→=r→×F→\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F} where r→\overrightarrow{r} is the position vector from the point about which the torque is calculated to the point where the force acts, and F→\overrightarrow{F} is the force vector.

  1. Identify the force vector (F→\overrightarrow{F}): Given, F→=3i^+5j^−2k^\overrightarrow{F} = 3\hat{i} + 5\hat{j} - 2\hat{k} N

  2. Identify the point where the force acts (P): The force acts at 7i^−2j^+5k^7\hat{i} - 2\hat{j} + 5\hat{k}. So, the position vector of point P is P⃗=7i^−2j^+5k^\vec{P} = 7\hat{i} - 2\hat{j} + 5\hat{k}.

  3. Identify the point about which the torque is calculated (O): The torque is calculated about the point (0,−1,0)(0, -1, 0). So, the position vector of point O is O⃗=0i^−1j^+0k^=−j^\vec{O} = 0\hat{i} - 1\hat{j} + 0\hat{k} = -\hat{j}.

  4. Calculate the position vector (r→\overrightarrow{r}): The position vector r→\overrightarrow{r} is from point O to point P: r→=P⃗−O⃗\overrightarrow{r} = \vec{P} - \vec{O} r→=(7i^−2j^+5k^)−(0i^−1j^+0k^)\overrightarrow{r} = (7\hat{i} - 2\hat{j} + 5\hat{k}) - (0\hat{i} - 1\hat{j} + 0\hat{k}) r→=(7−0)i^+(−2−(−1))j^+(5−0)k^\overrightarrow{r} = (7 - 0)\hat{i} + (-2 - (-1))\hat{j} + (5 - 0)\hat{k} r→=7i^+(−2+1)j^+5k^\overrightarrow{r} = 7\hat{i} + (-2 + 1)\hat{j} + 5\hat{k} r→=7i^−j^+5k^\overrightarrow{r} = 7\hat{i} - \hat{j} + 5\hat{k}

  5. Calculate the torque vector (τ→\overrightarrow{\tau}): Now, calculate the cross product τ→=r→×F→\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}: τ→=(7i^−j^+5k^)×(3i^+5j^−2k^)\overrightarrow{\tau} = (7\hat{i} - \hat{j} + 5\hat{k}) \times (3\hat{i} + 5\hat{j} - 2\hat{k}) This can be calculated using a determinant:

    τ→=∣i^j^k^7−1535−2∣\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -1 & 5 \\ 3 & 5 & -2 \end{vmatrix}

    Expand the determinant:

    τ→=i^((−1)(−2)−(5)(5))−j^((7)(−2)−(3)(5))+k^((7)(5)−(3)(−1))\overrightarrow{\tau} = \hat{i}((-1)(-2) - (5)(5)) - \hat{j}((7)(-2) - (3)(5)) + \hat{k}((7)(5) - (3)(-1)) τ→=i^(2−25)−j^(−14−15)+k^(35−(−3))\overrightarrow{\tau} = \hat{i}(2 - 25) - \hat{j}(-14 - 15) + \hat{k}(35 - (-3)) τ→=i^(−23)−j^(−29)+k^(35+3)\overrightarrow{\tau} = \hat{i}(-23) - \hat{j}(-29) + \hat{k}(35 + 3) τ→=−23i^+29j^+38k^\overrightarrow{\tau} = -23\hat{i} + 29\hat{j} + 38\hat{k}

    The unit of torque is N m.