Question

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.

Answer 1

55738.539 Pa

Step-by-step 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' =
`(787.92)/(2)=393.96N`

Hence, the pressure on each leg will be

Pressure =
`(393.96N)/(7.068* 10^(-4))=55738.539 Pa`