Question

A 72 kg person sits on a 3.6 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.4 cm in diameter. Find the pressure exerted on the floor by each leg of the chair, assuming the weight is evenly distributed. (in Pa)

Answer 1

Weight of person: 79 × g = 79 × 9.8 = 774.2 Weight of chair: 3.5 × g = 3.5 × 9.8 = 34.3 So the total force exerted on the floor by the chair and person together is 774.2 + 34.3 = 808.5 N This force is evenly distributed over four legs, so the force on the floor at each leg is 808.5 ÷ 4 = 202.125 N Pressure is force ÷ area. The contact area of each leg is πr^2, where the radius r is 0.006m π × (0.006)^2 = 1.13 × 10^-4 P = F/A = 202.125/(1.13 × 10^-4) = 1.79 × 10^6 Pa = 1.79 MPa The key formulae here were F=ma (which became F=mg where g is gravitational acceleration) to calculate weight and P=F/A.