A body of mass 10 kg slides along a rough ho | vedclass

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(C) Let $P$ be the force acting at an angle of $30^\circ$ with the horizontal.For the body to be on the verge of motion,the horizontal component of th

Vedclass · Accepted Answer

$50$

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(C) Let $P$ be the force acting at an angle of $30^\circ$ with the horizontal.For the body to be on the verge of motion,the horizontal component of the force must balance the limiting friction: $P \cos 30^\circ = f_k$.Since $f_k = \mu R$,where $R$ is the normal reaction,we have $R = mg - P \sin 30^\circ$.Substituting these into the equation: $P \cos 30^\circ = \mu (mg - P \sin 30^\circ)$.Given $m = 10 \ kg$,$g = 10 \ m/s^2$,$\mu = 1/\sqrt{3}$,and $	heta = 30^\circ$:$P (\sqrt{3}/2) = (1/\sqrt{3}) (100 - P/2)$.Multiplying both sides by $\sqrt{3}$: $P (3/2) = 100 - P/2$.$3P/2 + P/2 = 100$.$2P = 100$.$P = 50 \ N$.

$A$ body of mass $10 \ kg$ slides along a rough horizontal surface. The coefficient of friction is $1/\sqrt{3}$. Taking $g = 10 \ m/s^2$,the least force which acts at an angle of $30^\circ$ to the horizontal is ...... $N$.

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