Answer:
The force is [tex]F =102.34 \ N [/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m_c = 1500\ kg[/tex]
The top radius of the pipe is [tex]r_t = 24 \ cm =0.24 \ m[/tex]
The bottom radius is [tex]r_b = 2 \ cm = 0.02 \ m[/tex]
Generally the area of the small end of the pipe is
[tex]A_2 = \pi r_b^2[/tex]
=> [tex]A_2 = 3.142 * 0.02 ^2[/tex]
=> [tex]A_2 = 0.00126\ m^2 [/tex]
Generally the pressure in the pipe is mathematically represented as
[tex]P = \frac{F}{A_1}[/tex]
Here [tex]A_1[/tex] is the area of the larger opening of the pipe and this is mathematically represented as
[tex]A_1 = \pi * r^2_t[/tex]
=> [tex]A_1 = 3.142 * 0.24^2[/tex]
=> [tex]A_1 = 0.18098 \ m^2[/tex]
Also F is the force in the at the larger opening(the top) of the pipe which mathematically evaluated as
[tex]F = m * g[/tex]
=> [tex]F = 1500 * 9.8[/tex]
=> [tex]F = 14700 \ N [/tex]
So
[tex]P = \frac{14700}{0.18098}[/tex]
[tex]P = 81224.44 \ N / m^2[/tex]
Generally the force at the smaller opening of the pipe is
[tex]F = P * A_2[/tex]
=> [tex]F = 81224.44 *0.00126[/tex]
=> [tex]F =102.34 \ N [/tex]
