A sliding gate 3m and 1.5m high situated in a vertical plane has a coefficient of friction between itself and guide of 0.18 if the gate weight is 19n and its upper edge is at a depth of 9m, what vertical force is required to raise it? Neglect buoyancy force on gate. -

Width of the gate, b=3 m

Depth/Height of the gate, d=1.5m

Area of the gate, $A=bxd=3x1.5=4.5m^2$

Weight of the gate 19KN

Coefficient of friction $μ=0.18$

Vertical force required to raise the gate:

Depth of c.g. of the gate from water surface,

$$\bar{x}=9+1.5/2=9.75 m$$

Pressure force on the gate ,

$$P=wA\bar{x}=9.81×4.5×9.75=430.4kN$$

Force required to raise the gate = frictional force+ weight of the gate

$$=μP+W=0.18×430.4+19=96.472 kN……….(ans)$$