Question

A block of mass $10Kg$ is placed on a rough horizontal table whose coefficient of static friction ${\mu }_s$ is equal to $0.5$. A Force $F$ acting on the block at angle of ${37}^{\circ }$ is gradually increased from zero. Determine whether the block will first slide or lift up?

Answer 1

There is minimum a magnitude of forces required, both in horizontal and vertical direction, either to slide or lift up the block. The block will first slide or lift up would depend upon which minimum magnitude of force is lesser.
For block to start lifting up, in vertical direction
$Fsin{37}^{\circ }+N-Mg\ge 0$
N becomes zero on just lifting condition.
$F_{lift}\ge \frac{10g}{3/5}$

$\therefore F_{lift}\ge \frac{500}{3}N$
For horizontal direction to start sliding
$Fcos37\ge {\mu }_{s}N$
$Fcos{37}^{\circ }>0.5[10g-Fsin{37}^{\circ }](\because N=10g-Fsin{37}^{\circ })$
Hence $f_{slide}>\frac{50}{cos{37}^{\circ }+0.5sin{37}^{\circ }}$
$f_{slide}>\frac{500}{11}N$
$F_{lift}>\frac{500}{3}N$
$\Rightarrow f_{slide}<F_{lift}$
Therefore the block will begin to slide before lifting.