Question

Two cylinders each with a 60 cm diameter, thatare closed at one end, open at the other, are joined to form asingle cylinder, then the air inside is removed.
How much force does the atmosphere exert onthe flat end of each cylinder?
Suppose one cylinder is bolted to a sturdy ceiling. How many 90 kg football players would need to hang from the lower cylinder to pull the two cylinders apart

Answer 1

Answer:

a

The force is  $F = 2864561.4 \ N$

b

The number is $N = 3248 \ players$

Explanation:

From the question we are told that

    The  of each cylinder is  $d = 60 \ cm = 6 \ m$

     The mass of the players is  $m = 90 \ kg$

Generally the cross-sectional  area of the cylinder is mathematically represented as

     $A = \pi * \frac{d^2}{4}$

=>  $A = 28.3 \ m^2$

Generally force exerted on the flat end of each cylinder is mathematically represented as

      $F = A * P$

Here P  is the atmospheric pressure with value  $P = 101300 \ Pa$

So

       $F = 28.3 * 101300$

=>    $F = 2864561.4 \ N$

Generally the weight of a single football player is  

        $W = m * g$

=>     $W = 90 * 9.8$

=>     $W = 882\ N$

Generally the number of player required to pull the two cylinders apart is mathematically represented as

       $N = \frac{ F }{W}$

=>     $N = \frac{ 2864561.4 }{882}$

=>     $N = 3248 \ players$