Question

Part A) A car misses a turn and sinks into a shallow lake to a depth of 6.6m. If the area of the csr door is 0.46m^2, what id the force exerted on the outside of the door by the water? Note: 1 atm=101.325kPa.

Part B) What is the force exerted on the inside of the door by air, assuming the inside of the car is at atmospheric pressure? Think about what the occupant should do to get the door open.

i got 76.4N for part A which is incorrect and 46.6N for part B which is also incorrect

Answer 1

part A)

for the pressure at this depth

pressure at this depth, p = 1.013 *10^5 + 1000 * 9.8 * 6.6

P = 165980 Pa

for the force acting on the car door outside

force exerted on the door = P * Area

force exerted on the door = 165980 * 0.46

force exerted on the door = 76351 N

part B) for the force exerted on the inner car door

force exerted on the inner car door = pressure inside * area

force exerted on the inner car door = 1.013 *10^5 * 0.46

force exerted on the inner car door = 46600 N

the force exerted on the inner car door is 46600 N