The tension T in the string shown in the figu | vedclass

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(A) Given: Mass $m = 10\, kg$,angle $	heta = 30^{\circ}$,coefficient of static friction $\mu_s = 0.7$,and acceleration due to gravity $g = 10\, m/s^2

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Zero

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(A) Given: Mass $m = 10\, kg$,angle $	heta = 30^{\circ}$,coefficient of static friction $\mu_s = 0.7$,and acceleration due to gravity $g = 10\, m/s^2$.The component of weight acting down the incline is $F_g = mg \sin 	heta = 10 	imes 10 	imes \sin 30^{\circ} = 100 	imes 0.5 = 50\, N$.The normal force is $N = mg \cos 	heta = 10 	imes 10 	imes \cos 30^{\circ} = 100 	imes \frac{\sqrt{3}}{2} = 50\sqrt{3}\, N$.The maximum static friction force is $f_{s,max} = \mu_s N = 0.7 	imes 50\sqrt{3} = 35\sqrt{3}\, N$.Since $f_{s,max} \approx 35 	imes 1.732 = 60.62\, N$,and the force pulling the block down the incline is $50\, N$,we see that $f_{s,max} > mg \sin 	heta$.Because the static friction is sufficient to prevent the block from sliding down,the tension $T$ in the string is zero.

The tension $T$ in the string shown in the figure is:

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