(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:$\vec{\tau} = \vec{r} \times

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$17\hat{i} - 6\hat{j} - 13\hat{k}$

(B) The torque $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:$\vec{\tau} = \vec{r} \times \vec{F}$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix}$Expanding the determinant:$\vec{\tau} = \hat{i} [(2)(4) - (3)(-3)] - \hat{j} [(3)(4) - (3)(2)] + \hat{k} [(3)(-3) - (2)(2)]$$\vec{\tau} = \hat{i} [8 + 9] - \hat{j} [12 - 6] + \hat{k} [-9 - 4]$$\vec{\tau} = 17\hat{i} - 6\hat{j} - 13\hat{k} \text{ N m}$

Class 11 Physics System of Particles and Rotational Motion Torque and Couple

Difficult

The torque of the force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$ acting at the point $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$ about the origin is:

A
$6\hat{i} - 6\hat{j} + 12\hat{k}$
B
$17\hat{i} - 6\hat{j} - 13\hat{k}$
C
$-6\hat{i} + 6\hat{j} - 12\hat{k}$
D
$-17\hat{i} + 6\hat{j} + 13\hat{k}$

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Class 11

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System of Particles and Rotational Motion

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Torque and Couple

$A$ door $1.2 \,m$ wide requires a force of $1 \,N$ to be applied perpendicularly at the free end to open or close it. The perpendicular force required at a point $0.2 \,m$ distant from the hinges for opening or closing the door is: (in $\,N$)

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$A$ wheel of radius $20\, cm$ has forces applied to it as shown in the figure. The net torque produced by the forces $4\, N$ at $A$,$8\, N$ at $B$,$6\, N$ at $C$,and $9\, N$ at $D$ at the angles indicated is:

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Moment of a force of magnitude $20 \, N$ acting along the positive $x$-direction at point $(3 \, m, 0, 0)$ about the point $(0, 2, 0)$ (in $N \cdot m$) is ...........

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The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

MediumJEE MAIN 2019

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The rotational motion of a body is caused by:

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The torque of the force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$ acting at the point $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$ about the origin is:

Similar Questions

$A$ door $1.2 \,m$ wide requires a force of $1 \,N$ to be applied perpendicularly at the free end to open or close it. The perpendicular force required at a point $0.2 \,m$ distant from the hinges for opening or closing the door is: (in $\,N$)

$A$ wheel of radius $20\, cm$ has forces applied to it as shown in the figure. The net torque produced by the forces $4\, N$ at $A$,$8\, N$ at $B$,$6\, N$ at $C$,and $9\, N$ at $D$ at the angles indicated is:

Moment of a force of magnitude $20 \, N$ acting along the positive $x$-direction at point $(3 \, m, 0, 0)$ about the point $(0, 2, 0)$ (in $N \cdot m$) is ...........

The magnitude of torque on a particle of mass $1\,kg$ is $2.5\,Nm$ about the origin. If the force acting on it is $1\,N$,and the distance of the particle from the origin is $5\,m$,the angle between the force and the position vector is (in radians)

The rotational motion of a body is caused by:

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