$ABC$ is an equilateral triangle with $O$ as its centre. $\vec F_1, \vec F_2$ and $\vec F_3$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about $O$ is zero,then the magnitude of $\vec F_3$ is

  • A
    $({F_1} + {F_2})/2$
  • B
    $2({F_1} + {F_2})$
  • C
    $({F_1} + {F_2})$
  • D
    $({F_1} - {F_2})$

Explore More

Similar Questions

$A$ pendulum consists of a bob of mass $m=0.1 \ kg$ and a massless inextensible string of length $L=1.0 \ m$. It is suspended from a fixed point at height $H=0.9 \ m$ above a frictionless horizontal floor. Initially,the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. $A$ horizontal impulse $P=0.2 \ kg \cdot m/s$ is imparted to the bob at some instant. After the bob slides for some distance,the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \ kg \cdot m^2/s$. The kinetic energy of the pendulum just after the lift-off is $K$ Joules. $(1)$ The value of $J$ is. . . . . . $(2)$ The value of $K$ is. . . . . Give the answers of the questions $(1)$ and $(2)$.

$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

$A$ homogeneous disc of mass $2 \ kg$ and radius $15 \ cm$ is rotating about its axis (which is fixed) with an angular velocity $4 \ rad/s$. The linear momentum of the disc is

$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo