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

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

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$0$

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(A) Given: Mass $m = 10 \, kg$,coefficient of friction $\mu = 0.7$,angle of inclination $	heta = 30^{\circ}$,and acceleration due to gravity $g = 10 \, m/s^2$.The component of gravitational force acting down the incline is $F_g = mg \sin 	heta = 10 	imes 10 	imes \sin 30^{\circ} = 100 	imes 0.5 = 50 \, N$.The maximum static frictional force acting up the incline is $f_{s,max} = \mu N = \mu mg \cos 	heta = 0.7 	imes 10 	imes 10 	imes \cos 30^{\circ} = 70 	imes \frac{\sqrt{3}}{2} = 35 \sqrt{3} \, N$.Since $35 \sqrt{3} \approx 35 	imes 1.732 = 60.62 \, N$,we see that $f_{s,max} > F_g$.Because the maximum static friction is greater than the force trying to pull the block down the incline,the block remains in equilibrium without the need for tension in the string. Therefore,the tension $T$ in the string is $0 \, N$.

The tension $T$ in the string shown in the figure is ..................... $N$.

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