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Q.
The tension $T$ in the string shown in figure is
Laws of Motion
Solution:
$m g \sin \theta=10(10) \sin 30^{\circ}=50\, N$
Frictional force $=\mu \,mg\, \cos \theta$
$=(0.7)(10)(10) \frac{\sqrt{3}}{2}=35 \sqrt{3} N$
Frictional force is sufficient to oppose gravitational force.
Tension will be zero.
