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A 73 kg man in a 7.4 kg chair tilts back so that all his weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area of radius 1.5 cm. Find the pressure exerted on the floor by each leg. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.A 73 kg man in a 7.4 kg chair tilts back so that all his weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area of radius 1.5 cm. Find the pressure exerted on the floor by each leg. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.

Respuesta :

Answer:

55738.539 Pa

Explanation:

Gvien:

The mass of the man  = 73 kg

Mass of the chair = 7.4 kg

radius of the leg of the chair = 1.5 cm = 0.015 m

Now,

the force of gravity due to both the masses, F = (73+7.4) kg x 9.8 = 787.92 N

Also,

Pressure = force / area

Area of a circle of leg = πr² = π x 0.015² = 7.068 x10⁻⁴ m2

Now,

there are 2 legs so the force will be  divided evenly in each leg

thus,

F' =  [tex]\frac{787.92}{2}=393.96N[/tex]

Hence, the pressure on each leg will be

Pressure = [tex]\frac{393.96N}{7.068\times 10^{-4}}=55738.539 Pa[/tex]

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