Question

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 $F$ : [Given $\left.g=10\, ms ^{-2}\right]$ is

• A. $20\, N$
• B. $35.7\, N$
• C. $33.3 \,N$
• D. $25.2\, N$

Answer 1

Answer: (D)

$ N = Mg - F \operatorname{Sin} 30^{\circ}$
$ = mg -\frac{ F }{2}=100-\frac{ F }{2}=\frac{200- F }{2}$
$F \operatorname{Cos} 30^{\circ}=\mu N$
$ \sqrt{3} \frac{ F }{2}=0.25 \times\left(\frac{200- F }{2}\right)$
$ 4 \sqrt{3} F =200- F $
$ F =\frac{200}{4 \sqrt{3}+1}=25.22$