The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:
Pressure = density x g x height
In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:
P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa
Answer:
The 50N weight be hung at 23 cm to maintain equilibrium
Explanation:
Given;
length of the uniform rod = 30 cm
center of the uniform rod = 15 cm
weight of 40N is hung at 5 cm mark
weight of 50 N will be hung at ?
0------5cm-----------------15cm-------------P---------30cm
↓ 10cm Δ xcm ↓
40N 50N
Take moment about the pivot point and apply the principle of moment
50N (x cm) = 40N (10 cm)
x = (400) / 50
x = 8cm
P = x cm + 15 cm
P = 8 cm + 15 cm
P = 23 cm
Therefore, the 50N weight be hung at 23 cm to maintain equilibrium