The drawing shows a cylinder fitted with a piston that has a mass m_1 of 0.760 k

Question

The drawing shows a cylinder fitted with a piston that has a mass m_1 of 0.760 kg and a radius of 2.50 x 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 m_2 = 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

Let T = tension
In second block,
m2g - T = m2 a, here a = acceleration of the block
For the piston,
T - (m1g - pxxr2) = m1 a; p = pressure difference across the piston.
Eliminating T from the two eqns.,
m2g - (m1g - p*r^2) = (m1 + m2) a ... ( 1 )
For the motion of the second block,
1.75 = (1/2) a (1.8)2
=> a = 1.75 / (1.8)2 = 0.54 m/s2
Plugging this value of a and values of m1 and m2 and r in eqn. ( 1 ),
(7.1 - 0.76) x 9.8 - p x 3.14 (0.025)2 = (7.6 + 0.76)x (0.54)
62.132 - 1.962x10-3p = 4.5144 `
p = (62.132 -4.5144) / (1.962x10-3) = 29367 N/m2

atmospheric pressure (p_a = 101325Pa as standard value)

absolute pressure beneath the piston = p-a -p
= 101325 - 29367 N/m2
71958N/m2.

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