In the attached picture, can I say that there is a lift force in the Y-direction, and a drag force too in the same Y-direction?

$F_L$ proportional to $V_fx^2$ ?

$F_D$ proportional to $V_py^2$ ?

Is this equation of motion correct here: $ma = -mg -F_D$ + V$\rho$$g+$ $F_L$ ?

The lift force is because of the horizontal flow which is in fact lifting the particle, and the drag force is the resistance to the particle's upward motion. Is this correct to say.

Please assume that the picture is correct, and such an observation was made.

enter image description here

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    $\begingroup$ I think that you should consider what causes drag. $\endgroup$ – don_Gunner94 May 2 '16 at 12:44


I apologise : as you suggest, there might be a lift force on the sphere if there is a shear flow in the fluid (see Discussion in 1st Link). However, this force is likely to be much smaller than the drag on the particle.




I do not see any reason for there to be a lift force since the particles do not appear to have an aerodynamic shape. Even if they did, the fact that they are free to rotate means that the 'angle of attack' could be reduced to zero, which would reduce lift to zero.

Possibly the particles could spin which would introduce a force due to the Magnus Effect (see video link). However, if the particles are quite small, viscous drag would quickly wipe out the rotation.


Your buoyancy term should use the difference in density between the particles and the fluid.

Drag is friction which opposes relative motion. Assuming the particles are denser than the fluid the drag force should probably be in the direction of fluid flow around the particles, which may not be upwards. This direction depends on position in relation to the step, and would be difficult to predict, even when the flow is 'laminar' rather than 'turbulent'. The particles themselves might also disrupt the pattern of fluid flow. If the particles are initially on the bottom of the step, they might be 'dragged' into the corner but probably not upwards - unless they are almost floating and/or there is a lot of turbulence.

Remember that your equation of motion is a vector equation : forces ma and Fd both have x components as well as y.

  • $\begingroup$ Hi, many thanks for your reply. I was thinking that the lift force is because there is zero fluid velocity at particle bottom, but non zero velocity at particle top, may be some small velocity gradient around the particle due to turbulence. Ok lets assume then is lift force from the Magnus effect then. Ya the buoyancy term should be edited, you are right. I think the drag should be in the direction of relative particle motion and since flow velocity in vertical direction is zero, hence I took took drag as downwards. $\endgroup$ – user2617526 May 2 '16 at 13:48
  • $\begingroup$ Many thanks for your response and adding even more details. $\endgroup$ – user2617526 May 4 '16 at 3:07
  • $\begingroup$ Hi, when I am using the lift force in the y-direction equation of motion, will be it proportional to (V_fx - V_py)$^2$ or (V_fx$^2$ - V_py$^2$) or something else. I guess this will be the relative velocity for the particle right but not very sure which term is correct in this case? I am sure for drag in the horizontal direction it will be (V_fx - V_px)$^2$ in the motion equation. Thanks a lot for your help. $\endgroup$ – user2617526 May 5 '16 at 16:02

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