Question 87 of 210

As shown in figure, a $70kg$ garden roller is pushed with a force of $\vec{F}=200N$ at an angle of $30^{\circ}$ with horizontal. The normal reaction on the roller is (Given $g=10m{s}^{-2}$)

A$800\sqrt{2}N$B$600N$C$800N$D$200\sqrt{3}N$

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Answer: (C) $800N$

Component of the force perpendicular to the ground,
$F\sin 30^{\circ}$.
Now for equilibrium along vertical direction, we get
$N=mg+F\sin 30^{\circ} =70\times 10+200\times \dfrac{1}{2} =800N$