Question

Find the pressure exerted by the water bed on the floor when the bed rests in its normal position. Assume the entire lower surface of the bed makes contact with the floor. SOLUTION When the water bed is in its normal position, the area in contact with the floor is 4.00 m2. Use the the definition of pressure to find the pressure (in Pa). P = Mg 4.00 m2 = Pa EXERCISE A carpenter decides to build a very sturdy bedframe to support the water bed. The frame has four legs, each with a square bottom with side length 7.0 inches. The bedframe weighs 110 kg. What is the pressure (in Pa) exerted by the water bed on the floor when it is supported by the bedframe?

Answer 1

A water bed is 2.00m on a side and 30.0cm deep. Find the pressure that the water bed exerts on the floor. Assume that the entire lower surface of the bed makes contact with the floor.

Density of water, ρ=1000kg/m^3 g=10m/s

This is the missing part of the question.

Answer:

102343.75 Pa

Explanation:

Lenght of bed = 2 m on each side

Depth = 30 cm = 0.3 m

Volume of bed = 2 x 2 x 0.3 = 1.2 m^3

Weight of the bed = pgv

Where p = density of water

g is acceleration due to gravity 10 m/s^2.

V is the volume

Weight of bed = 1000 x 10 x 1.2 = 12000 N

Mass of bed frame = 110 kg

Weight of bed frame = 110 x 10 = 1100 N

Total weight of bed frame and bed = 12000 + 1100 = 13100 N

Side lenght of bed leg = 7 in = 0.1778 m

Area of each bed frame leg = 0.1778 x 0.1778 = 0.032 m2 (legs are square faced)

Total surface area = 0.032. X 4 = 0.128 m2.

Pressure exerted on floor = total weight ÷ area

= 13100/0.128

= 102343.75 Pa