Two vertical walls are separated by a distance of $2\  m$. Wall $A$ is smooth while wall $B$ is rough with a coefficient of friction $0. 5$. A uniform rod is placed between them as shown. The length of longest rod that can be placed between walls is equal to

820-8

  • A

    $2\  m$

  • B

    $2\sqrt 2\ m$

  • C

    $\sqrt 5\ m$

  • D

    $\frac{{\sqrt {17} }}{2}\ m$

Similar Questions

$3\; m$ long ladder wetghing $20 kg$ leans on a frictionless wall. Its feet rest on the floor $1\; m$ from the wall as shown in Figure Find the reaction forces of the wall and the floor.

$A$ body is in equilibrium under the influence of a number of forces. Each force has a different line of action. The minimum number of forces required is

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 $g=9.8 \;m / s ^{2}$ )

When it is said that body is in mechanical equilibrium ?

A rod of mass $1\ kg$ , length $1\ m$ is suspended horizontally with the help of two ideal strings as shown in figure. First a mass is suspended to the left most end keeping rod horizontal, than a second mass is suspended to the right most end again keeping horizontal orientation. The maximum total mass that can be suspended in that way keeping horizontal orientation of rod ....... $kg.$