Write condition of translational equilibrium of particle.
Write condition of translational equilibrium of particle.
Similar Questions
The $50\,kg$ homogeneous smooth sphere rests on the $30^{\circ}$ incline $A$ and bears against the smooth vertical wall $B$. Calculate the contact forces at $A$ and $B.$
The $50\,kg$ homogeneous smooth sphere rests on the $30^{\circ}$ incline $A$ and bears against the smooth vertical wall $B$. Calculate the contact forces at $A$ and $B.$
A block of mass $1 kg$ is suspended by a string of mass $1 kg$, length $1 m$ as shown in figure. $(g=10\,m /$ $s ^2$ ) Calculate: The tension in string at its lowest point.
A piece of wire is bent in the shape of a parabola $y=k x^2$ ( $y$-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with a constant acceleration $\alpha$. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$-axis is
A piece of wire is bent in the shape of a parabola $y=k x^2$ ( $y$-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with a constant acceleration $\alpha$. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$-axis is
- [IIT 2009]
What is Free body diagram ?
What is Free body diagram ?
A mass of $10 \,kg$ is suspended vertically by a rope of length $5 \,m$ from the roof. A force of $30 \,N$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta=\tan ^{-1}\left(x \times 10^{-1}\right)$. The value of $x$ is ................
$\text { (Given } g =10 \,m / s ^{2} \text { ) }$
A mass of $10 \,kg$ is suspended vertically by a rope of length $5 \,m$ from the roof. A force of $30 \,N$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta=\tan ^{-1}\left(x \times 10^{-1}\right)$. The value of $x$ is ................
$\text { (Given } g =10 \,m / s ^{2} \text { ) }$
- [JEE MAIN 2022]