Can a body remain in partial equilibrium? Explain with an illustration.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Yes,a body can be in partial equilibrium. This means it may be in translational equilibrium but not in rotational equilibrium,or it may be in rotational equilibrium but not in translational equilibrium.
Case $(a)$: Rotational equilibrium but not translational equilibrium.
Consider a light rod of negligible mass $AB$ of length $2a$. Two parallel forces,both equal in magnitude $F$,are applied perpendicular to the rod at ends $A$ and $B$ in the same direction as shown in figure $(a)$.
Let $C$ be the midpoint of $AB$,so $CA = CB = a$.
The net force on the rod is $\sum \vec{F} = F + F = 2F \neq 0$. Thus,it is not in translational equilibrium.
The net torque about the midpoint $C$ is $\tau = (F \times a) - (F \times a) = 0$. Thus,the rod is in rotational equilibrium.
Case $(b)$: Translational equilibrium but not rotational equilibrium.
Consider the same rod $AB$ of length $2a$. Two equal and opposite forces $\vec{F}$ are applied perpendicular to the rod at ends $A$ and $B$ as shown in figure $(b)$.
The net force on the rod is $\sum \vec{F} = F - F = 0$. Thus,the rod is in translational equilibrium.
The net torque about the midpoint $C$ is $\tau = (F \times a) + (F \times a) = 2Fa \neq 0$. Since the torques are in the same direction (both cause anti-clockwise rotation),the rod is not in rotational equilibrium.

Explore More

Similar Questions

Two men are carrying a uniform bar of length $L$ on their shoulders. The bar is held horizontally such that the younger man gets $(1/4)^{th}$ of the total load. Suppose the younger man is at one end of the bar,what is the distance of the other man from the other end?

Difficult
View Solution

If a street light of mass $M$ is suspended from the end of a uniform rod of length $L$ in different possible patterns as shown in the figure,then

$A$ heavy iron bar of mass $12 \,kg$ has one end on the ground and the other on the shoulder of a man. The rod makes an angle $60^{\circ}$ with the horizontal. The weight experienced by the man is:

$A$ uniform meter scale is balanced at the $30 \ cm$ mark on a knife edge,when weights of $60 \ N$ and $10 \ N$ are suspended at the $10 \ cm$ and $80 \ cm$ marks,respectively. The mass of the meter scale is: $(g = 10 \ m/s^2)$ (in $kg$)

$A$ metal rod of weight $W$ is supported by two parallel knife-edges $A$ and $B$. The rod is in equilibrium in a horizontal position. The distance between the two knife-edges is $r$. The centre of mass of the rod is at a distance $x$ from $A$. The normal reaction on $A$ is

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