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4-2.Friction
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A block of mass $1\, kg$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.5.$ The magnitude of the force acting upwards at an angle of $60^o$ from the horizontal that will just start the block moving is
A
$5 \,N$
B
$\frac{20}{2 + \sqrt 3}\,N$
C
$\frac{20}{2 - \sqrt 3}\,N$
D
$10 \,N$
Solution
$\mathrm{N}=\mathrm{mg}-\frac{\sqrt{3} \mathrm{F}}{2}$
If block just starts moving
$\mathrm{F} \cos 60^{\circ}=\mu \mathrm{N}$
$\Rightarrow \frac{\mathrm{F}}{2}=0.5\left(\mathrm{mg}-\frac{\sqrt{3}}{2} \mathrm{F}\right)$
$\Rightarrow \mathrm{F}+\frac{\sqrt{3} \mathrm{F}}{2}=10$ or $F=\frac{20}{2+\sqrt{3}}$
Standard 11
Physics
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