A block of mass $m$ is placed on a surface having vertical cross section given by $y=x^2 / 4$. If coefficient of friction is $0.5$ , the maximum height above the ground at which block can be placed without slipping is:
$1 / 4 \mathrm{~m}$
$1 / 2 \mathrm{~m}$
$1 / 6 \mathrm{~m}$
$1 / 3 \mathrm{~m}$
Why coefficient friction is considered as static friction ?
A block of mass $m$ (initially at rest) is sliding up (in vertical direction) against a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta = {\theta _0}t$ where $t$ is time in sec. At $t$ = $0$ , the force is in vertical upward direction and then as time passes its direction is getting along normal, i.e., $\theta = \frac{\pi }{2}$ .The value of $F$ so that the block comes to rest when $\theta = \frac{\pi }{2}$ , is
A $\vec F\,\, = \,\,\hat i\, + \,4\hat j\,$ acts on block shown. The force of friction acting on the block is :
As shown in the figure a block of mass $10\,kg$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$, with horizontal. For $\mu_{ s }=0.25$, the block will just start to move for the value of $F..........\,N$ : $\left[\right.$ Given $\left.g =10\,ms ^{-2}\right]$