A car misses a turn and sinks into a shallow lake to a depth of 8.9 m. If the ar

Question

A car misses a turn and sinks into a shallow lake to a depth of 8.9 m. If the area of the car door is 0.49 m^2, what is the force exerted on the outside of the door by the water assuming that the inside of the car is at atmospheric pressure? Part B What is the force exerted on the inside of the door by the air, assuming the inside of the car is at atmospheric pressure? Part C If the mass of the driver is 61.0 kg, what is the ratio of the net force on the door to the weight of the driver? Think about what the driver should do to get the door open.

Explanation / Answer

(a) Pressure at the depth of 8.9 m
= pgh = 1000*9.81*8.9 = 87.309 kPa (gauge pressure )
Absolute pressure = 101.325+87.309 = 188.634 kPa
Force on the door = pressure*area = 188.634*0.49 = 92.431 kN
(b) Force from inside = 101.325*0.49 = 49.64925 kN
(c) Fnet = 92.431 - 49.64925 = 42.782 kN
Ratio = Fnet / weight = 42.782*103 /(61*9.81) = 71.4923
Driver have to apply the force more than the Fnet i.e 42.782 kN to open the door.

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