# Solve this physics problem

Find the pressure exerted on the ground by a three legged milking stool if the stool has a mass of
3 kg, the mass of the milker is 93 kg and the legs are circular each with a diameter of 3 cm

hi

pressure = force / area

Force = mg + Mg = 3 * 9,8 + 93 * 9,8 = 941,76 N
Area one leg = pi * r² = 3 * (0,03/2)² = 7 * 10⁻⁴
Area 3 legs = 3 * 7 * 10⁻⁴ = 21 * 10⁻⁴

Pressure = 960 / (21 * 10⁻⁴) = about 444126 Pa = 444 kPa

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To find the pressure exerted on the ground by the three-legged milking stool, we can use the formula:
Pressure=Force/Area

The total force exerted by the stool and the milker is given by:
Force=(Mass of stool+Mass of milker)×Acceleration due to gravity

where the mass of the stool is 3 kg, the mass of the milker is 93 kg, and the acceleration due to gravity is 9.81 m/s29.81m/s2.

So, the total force is:
Force=(3 kg+93 kg)×9.81 m/s2=96 kg×9.81 m/s2=941.76 N

Now we need to find the total area of the stool legs in contact with the ground.

Each leg is circular with a diameter of 3 cm, so the radius is 32 cm=1.5 cm=0.015 m23​cm=1.5cm=0.015m.

The area AA of one leg in contact with the ground is given by πr2:
A=π(0.015 m)2=π×0.000225 m2=0.00070686 m2 (approximately)

Since there are three legs, the total area is 3×A3×A:
Atotal=3×0.00070686 m2=0.0021206 m2

Finally, the pressure PP exerted on the ground is:
P=941.76 N0.0021206 m2≈444,126 Pa≈444 kPa

So the pressure exerted on the ground by the three-legged milking stool is approximately 444kPa.

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