Why do we consider the viscous force to be tangential?

Is it that since it is a friction force, it needs to be tangential as other friction forces like kinetic friction and static friction forces are?

• Yes, you got it. – Aaron Stevens Aug 17 at 15:20
• You may think in that way ,as it is the force between the layers of liquid. – yuvraj singh Aug 17 at 15:20

If we orient a Cartesian coordinate system such that the xy plane coincides with the surface of interest, the viscous stress at the surface for a Newtonian fluid is given by: $$\boldsymbol{\tau}=\sigma_{xz}\mathbf{i}+\sigma_{yz}\mathbf{j}+\sigma_{zz}\mathbf{k}\tag{1}$$where $$\sigma_{xz}=\eta\left(\frac{\partial v_x}{\partial z}+\frac{\partial v_z}{\partial x}\right)\tag{2}$$ $$\sigma_{yz}=\eta\left(\frac{\partial v_y}{\partial z}+\frac{\partial v_z}{\partial y}\right)\tag{3}$$$$\sigma_{zz}=2\eta\frac{\partial v_z}{\partial z}-\left(\frac{2}{3}\eta-\kappa\right)\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)\tag{4}$$where $$\eta$$ is the shear viscosity and $$\kappa$$ is the "dilatational viscosity."
From the non-slip boundary condition, we know that $$v_x=v_y=v_z=0$$ at the surface. This enables us to reduce Eqns. 2-4 at the surface to: $$\sigma_{xz}=\eta\frac{\partial v_x}{\partial z}\tag{5}$$ $$\sigma_{yz}=\eta\frac{\partial v_y}{\partial z}\tag{6}$$$$\sigma_{zz}=\left(\frac{4}{3}\eta+\kappa\right)\frac{\partial v_z}{\partial z}\tag{7}$$In general, since $$\sigma_{zz}$$ is not generally equal to zero, the viscous stress will not generally be tangent to the surface. However, for the special case of an incompressible fluid, we know that $$\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}=0$$And then, from the zero slip boundary condition, it follows that $$\frac{\partial v_z}{\partial z}=0$$. This means that for the special case of an incompressible fluid, $$\sigma_{zz}=0$$ Under these circumstances, the viscous stress will be tangent to the surface.