A mass $M= 40\ kg$ is fixed at the very edge of a long plank of mass $80\ kg$ and length $1\ m$ which is pivoted such that it is in equilibrium. How far (approx.) from the pivot should a mass of $100\ kg$ be attached so that the plank starts rotating with an angular acceleration of $1\ rad/s^2$?
$\frac{1}{{75}}\,m$
$\frac{1}{{25}}\,m$
$\frac{1}{{15}}\,m$
$\frac{1}{{45}}\,m$
A disc of radius $5\, m$ is rotating with angular frequency $10\, rad / sec .$ A block of mass $2\, kg$ to be put on the disc friction coefficient between disc and block is $\mu_{ k }=0.4,$ then find the maximum distance from axis where the block can be placed without sliding (in $cm$)
A uniform rod $AB$ of weight $100\, N$ rests on a rough peg at $C$ and $a$ force $F$ acts at $A$ as shown in figure. If $BC = CM$ and tana $= 4/3$. The minimum coefficient of friction at $C$ is
$A$ thin rod of length $L$ is placed vertically on a frictionless horizontal floor and released with a negligible push to allow it to fall. At any moment, the rod makes an angle $\theta$ with the vertical. If the center of mass has acceleration $= A$, and the rod an angular acceleration $= \alpha$ at initial moment, then
Two particles of mass $m$ each are fixed at the opposite ends of a massless rod of length $5m$ which is oriented vertically on a smooth horizontal surface and released. Find the displacement of the lower mass on the ground when the rod makes an angle of $37^o$ with the vertical. ........ $m$
As shown in Figure the two sides of a step ladder $BA$ and $CA$ are $1.6 m$ long and hinged at $A$. A rope $DE, 0.5 \;m$ is tied half way up. A weight $40\;kg$ is suspended from a point $F , 1.2\; m$ from $B$ along the ladder $BA$. Assuming the floor to be frictionless and neglecting the wetght of the ladder. find the tension in the rope and forces exerted by the floor on the ladder. (Take $g=9.8 \;m / s ^{2}$ )