If S_1, S_2 and S_3 are the tensions at the l | vedclass

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(B) At the point of contact,the forces due to surface tension must be in equilibrium along the surface of the solid.Let $S_2$ be the surface tension a

Vedclass · Accepted Answer

$S_1 \cos 	heta + S_3 = S_2$

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(B) At the point of contact,the forces due to surface tension must be in equilibrium along the surface of the solid.Let $S_2$ be the surface tension at the solid-air interface,$S_3$ be the surface tension at the solid-liquid interface,and $S_1$ be the surface tension at the liquid-air interface.The force $S_1$ acts at an angle $	heta$ with the solid surface.Resolving $S_1$ into horizontal and vertical components,the horizontal component $S_1 \cos 	heta$ acts in the same direction as $S_3$.For equilibrium along the horizontal direction of the solid surface,the forces must balance:$S_2 = S_3 + S_1 \cos 	heta$Thus,the correct relation is $S_1 \cos 	heta + S_3 = S_2$.

If $S_1, S_2$ and $S_3$ are the tensions at the liquid-air,solid-air,and solid-liquid interfaces respectively,and $\theta$ is the angle of contact at the solid-liquid interface,then:

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