Question

A force of \(-P\hat k \) acts on the origin of the coordinate system. The torque about the point \((2, -3)\) is \(P(a\hat i+b\hat j)\), The ratio of \(\frac ab\) is \(\frac x2\). The value of x is -

Answer 1

Correct Answer: 3

The torque \( \vec{\tau} \) is given by the cross product of the position vector \( \vec{r} \) and the force \( \vec{F} \): \[ \vec{\tau} = \vec{r} \times \vec{F} \] The position vector \( \vec{r} \) from the origin to the point \( (2, -3) \) is: \[ \vec{r} = (0 - 2) \hat{i} + (0 - (-3)) \hat{j} = -2 \hat{i} + 3 \hat{j} \] The force is given by: \[ \vec{F} = -p \hat{k} \] Now, calculating the cross product: \[ \vec{\tau} = (-2 \hat{i} + 3 \hat{j}) \times (-p \hat{k}) \] Using the cross product identity: \[ \vec{\tau} = (-2 \hat{i} + 3 \hat{j}) \times (-p \hat{k}) = -p(3 \hat{i} + 2 \hat{j}) \] Given that the torque \( \vec{\tau} \) is \( P(a \hat{i} + b \hat{j}) \), comparing the coefficients, we get: \[ a = 3 \quad \text{and} \quad b = 2 \] Thus, the ratio \( \frac{a}{b} = \frac{3}{2} \). We are given that the ratio is \( \frac{x}{2} \), so: \[ \frac{3}{2} = \frac{x}{2} \] Solving for \( x \), we get: \[ x = 3 \] Thus, the value of \( x \) is \( 3 \).