Find the force on the bottom of a 47.5-cm-tall, cylindrical beaker with a diamet

Question

Find the force on the bottom of a 47.5-cm-tall, cylindrical beaker with a diameter of 5.29 cm. The beaker is filled to its brim with water.

So far I have done:

Density x Volume = mass
Volume = h(pi(r^2))
= (.475m)pi(.02645)^2
= .001 m^3
Density of water = 1000 kg/m^3

mass = (.001 m^3)(1000 kg/m^3) = 1 kg

F = mg = (1 kg)(9.81 m/s^2) = 9.81 N

However this is incorrect. I have also tried a few other methods and come up with he wrong answer. Help would be appreciated. I think I am missing a step somewhere.

Explanation / Answer

Height of the cylinder h = 47.5 cm = 0.475 m Radius of the cylinder r = 0.02645 m Density of water = 1000 kg/m3 Atmospheric pressure P = 1.013*105 Pa -------------------------------------------------------------- Now the pressure acting on the bottom of the cylinder is                 P = P0 + gh                    = 1.013*105 Pa + (1000 kg/m3 )(9.8 m/s2)(0.475 m)                    = 105955 Pa The pressure P = F/A therefore the force on bottom                       F = PA = (105955 Pa) (r2)                                    = (105955 Pa) (0.02645 m)2                                    = 232.87 N Atmospheric pressure P = 1.013*105 Pa -------------------------------------------------------------- Now the pressure acting on the bottom of the cylinder is                 P = P0 + gh                    = 1.013*105 Pa + (1000 kg/m3 )(9.8 m/s2)(0.475 m)                    = 105955 Pa The pressure P = F/A therefore the force on bottom                       F = PA = (105955 Pa) (r2)                                    = (105955 Pa) (0.02645 m)2                                    = 232.87 N                                    = 232.87 N

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