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 m23cm=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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