The drawing shows a cylinder fitted with a piston that has a mass mi of 0.760 kg

Question

The drawing shows a cylinder fitted with a piston that has a mass mi of 0.760 kg and a radius of 2.50 times 10-2 m. The top of the piston is open to the atmosphere. The pressure beneath the piston is maintained at a reduced (but constant) value by means of the pump. As shown, a rope of negligible mass is attached to the piston and passes over two massless pulleys. The other end of the rope is attached to a block that has a mass of m2 = 7.10 kg. The block falls from rest down through a distance of 1.75 m in 1.82 s. Ignoring friction, find the absolute pressure beneath the piston.

Explanation / Answer

Given,

m1 = 0.760 kg ; r = 2.5 x 10-2 m ; m2 = 7.10 kg ; d = 1.75 m ; t = 1.82 s

we need to find the absolute pressure beneath the piston.

Let T be the tension in the cord. So we can write;

m2g - T = m2a

for piston: T -( m1g + p x pi r2 ) = m1a (p is the pressure and a is acc.)

adding the above two equations

m2g - (m1g + p x pi r2 ) = (m1 + m2) a (1)

for the moving block; using third equation of motion:

1.75 = 1/2 a (1.82)2

a = 1.06 m/s2

Now putting the given values and the value of acceleration in equation 1 we get

m2g - (m1g + p x pi r2 ) = (m1 + m2) a

7.1 x 9.8 - (0.76 x 9.8 + p x 3.14 x 0.025 x 0.025 ) = (7.1 + 0.76) x 1.06

69.58 - 7.45 - p(0.00196) = 8.33

p = 27449 Pa

So the pressure beneath the piston = 101325 - 27449 = 73876 Pa

Hence, P = 73876 Pa.

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