A plate of area 10m^(2) is made to move horizontally with a speed of 0.5 ms^(-1) by applying a force over the surface a liquid. If the depth of the liquid is 2.5m, coefficient of viscosity of liquid is 10^(-3)Nsm^(-2), then calculate the horizontal force needed to move the plate.

A plate of area 10m^(2) is made to move horizontally with a speed of 0

1.	A plate of area 10m^(2) is made to move horizontally with a speed of 0.5 ms^(-1) by applying a force over the surface a liquid. If the depth of the liquid is 2.5m, coefficient of viscosity of liquid is 10^(-3)Nsm^(-2), then calculate the horizontal force needed to move the plate.
Answer» Solution :Given `Deltav=0.5 MS^(-1)` `Deltax=2.5m` velocity gradient `=(Deltav)/(Deltax)=(0.5)/(2.5)=0.2(ms^(-1))/(m)` We KNOW that \|F\| `=etaA((Deltav)/(Deltax))=10^(-3) xx 10^(2) xx 10^(-2)=10^(-3)N` Hence `10^(-3)N` of force is needed to be applied, to KEEP the PLATE moving.