A body of mass 5 ,kg is placed on a rough hor | vedclass

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Question

(B) Let the mass of the body be $m = 5 \,kg$,the angle of pulling force be $	heta = 45^{\circ}$,and the coefficient of static friction be $\mu = \fra

Vedclass · Accepted Answer

$\frac{25}{\sqrt{2}} \,N$

Vedclass · Answer

(B) Let the mass of the body be $m = 5 \,kg$,the angle of pulling force be $	heta = 45^{\circ}$,and the coefficient of static friction be $\mu = \frac{1}{3}$.Let $F$ be the applied force. The components of $F$ are $F \cos 	heta$ (horizontal) and $F \sin 	heta$ (vertical upward).The normal reaction $N$ is given by $N = mg - F \sin 	heta$.The limiting friction is $f_L = \mu N = \mu(mg - F \sin 	heta)$.For the body to slide,the horizontal component of the force must be equal to the limiting friction: $F \cos 	heta = \mu(mg - F \sin 	heta)$.Rearranging for $F$: $F(\cos 	heta + \mu \sin 	heta) = \mu mg$.$F = \frac{\mu mg}{\cos 	heta + \mu \sin 	heta}$.Substituting the values: $F = \frac{(1/3) 	imes 5 	imes 10}{\cos 45^{\circ} + (1/3) \sin 45^{\circ}} = \frac{50/3}{\frac{1}{\sqrt{2}} + \frac{1}{3\sqrt{2}}} = \frac{50/3}{\frac{3+1}{3\sqrt{2}}} = \frac{50}{3} 	imes \frac{3\sqrt{2}}{4} = \frac{50\sqrt{2}}{4} = \frac{25\sqrt{2}}{2} = \frac{25}{\sqrt{2}} \,N$.

$A$ body of mass $5 \,kg$ is placed on a rough horizontal surface of coefficient of static friction $\frac{1}{3}$. The least pulling force to be applied on the body at an angle $45^{\circ}$ with the horizontal to slide it,is . . . . . . $\left(g=10 \,ms^{-2}\right)$

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