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  4. A vertical disc of mass 5 kg and radius 50 cm rests against a step of height 25 cm as shown in figure. What minimum horizontal force applied perpendicular to the axle will make the disc to climb the step? Take g= 10 m/s 2

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Q. A vertical disc of mass 5 $kg$ and radius $50 \,cm$ rests against a step of height $25 \,cm$ as shown in figure. What minimum horizontal force applied perpendicular to the axle will make the disc to climb the step? Take $g=$ $10 \,m/s ^{2}$Physics Question Image

System of Particles and Rotational Motion

Solution:

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$ y=50-25=25 \,cm $
$ x=\sqrt{50^{2}-25^{2}}=25 \sqrt{3} $
$ F y=5 g x $
$\Rightarrow (25)=5 \times 10(25 \sqrt{3}) $
$ \Rightarrow F=50 \sqrt{3} N $