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(D) For pushing at an angle $	heta_A = 30^\circ$ with the horizontal,the minimum force $F_A$ required is given by: $F_A = \frac{\mu mg}{\cos 	heta_A

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

Vedclass · Answer

(D) For pushing at an angle $	heta_A = 30^\circ$ with the horizontal,the minimum force $F_A$ required is given by: $F_A = \frac{\mu mg}{\cos 	heta_A - \mu \sin 	heta_A}$ For pulling at an angle $	heta_B = 60^\circ$ with the horizontal,the minimum force $F_B$ required is given by: $F_B = \frac{\mu mg}{\cos 	heta_B + \mu \sin 	heta_B}$ Given $\mu = \frac{\sqrt{3}}{5}$,$	heta_A = 30^\circ$,and $	heta_B = 60^\circ$: $F_A = \frac{\mu mg}{\cos 30^\circ - \mu \sin 30^\circ} = \frac{\mu mg}{\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{5} \cdot \frac{1}{2}} = \frac{\mu mg}{\frac{\sqrt{3}}{2} (1 - \frac{1}{5})} = \frac{\mu mg}{\frac{\sqrt{3}}{2} \cdot \frac{4}{5}} = \frac{\mu mg}{\frac{2\sqrt{3}}{5}}$ $F_B = \frac{\mu mg}{\cos 60^\circ + \mu \sin 60^\circ} = \frac{\mu mg}{\frac{1}{2} + \frac{\sqrt{3}}{5} \cdot \frac{\sqrt{3}}{2}} = \frac{\mu mg}{\frac{1}{2} + \frac{3}{10}} = \frac{\mu mg}{\frac{5+3}{10}} = \frac{\mu mg}{\frac{8}{10}} = \frac{\mu mg}{\frac{4}{5}}$ Therefore,the ratio is: $\frac{F_A}{F_B} = \frac{\frac{\mu mg}{2\sqrt{3}/5}}{\frac{\mu mg}{4/5}} = \frac{4/5}{2\sqrt{3}/5} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$

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