Question

A block of mass $1\text{kg}$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.50$. If $g=10{\text{ms}}^{-2}$, then the magnitude of a force (in $\text{N}$) acting upwards at an angle of ${60}^{\circ }$ from the horizontal that will just start the block moving is

Answer 1

From FBD,
$R+P\sin {60}^{\circ }=Mg$
or $R=Mg-P\sin {60}^{\circ }$
$\therefore$ Frictional force,
$F=\mu R$
$=\mu [Mg-P\sin {60}^{\circ }]$
The body just moves when
$P\cos {60}^{\circ }=F$
or $P\cos {60}^{\circ }=\mu [Mg-P\sin {60}^{\circ }]$
or $\frac{P}{2}=\frac{1}{2}[1\times 10-\frac{P\sqrt{3}}{2}]$
or $P=5.36\text{N}$