Question

A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T

Answer 1

The force acting on the shaft is 1324.75 N

Step-by-step explanation:

Given that,

Cross sectional area of shaft
`A_(sh)=0.8\ cm^(2)`

Gas pressure
`P_(gas)=3\ bar=3*10^5\ Pa`

Total mass M=24.5+0.5=25 kg

Diameter of piston, d=10 cm

We need to calculate the cross section area of piston

Using formula of area

`A_(p)=\pi*(d^2)/(4)`

Put the value into the formula

`A_(p)=\pi*(0.1^2)/(4)`

`A_(p)=0.00785\ m^(2)`

We need to calculate the weight of the piston and shaft

Using formula of weight

`W=mg`

Put the value into the formula

`W=25*9.81`

`W=245.25\ N`

We need to calculate the force due to gas pressure

Using formula of force

`F_(gas)=P_(gas)* A_(p)`

Put the value into the formula

`F_(gas)=3*10^(5)*0.00785`

`F_(gas)=2355\ N`

We need to calculate the force due to atmospheric pressure,

Using formula of force

`F_(atm)=P_(atm)* A_(p)`

Put the value into the formula

`F_(atm)=1*10^5*0.00785`

`F_(atm)=785\ N`

We need to calculate the force acting on the shaft

from free body diagram

`F_(gas)=F+F_(atm)+W`

Put the value into the formula

`2355=F+785+245.25`

`F=2355-785-245.25`

`F=1324.75\ N`

Hence, The force acting on the shaft is 1324.75 N

Answer 2

The complete question is :

The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.

Solution :

Given

The cross sectional area of the shaft,
`$A_(s) = 0.8 \ cm^2$`

Gas pressure,
`$P_(gas) = 3 \ bar = 3 * 10^5 \ Pa$`

The total mass, m = 24.5 + 0.5

= 25 kg

Diameter of the piston, d = 10 cm

The cross sectional area of the piston,
`$A_(p) = (\pi)/(4) * (0.1)^2$`

Weight of the piston and shaft, W = mg

= 25 x 9.81

= 245.25 N

Force due to the gas pressure,

`$F_(gas) = P_(gas) * A_(p)$`

`$F_(gas) = 3 * 10^5 * 0.00785 $`

= 2355 N

Force due to atmospheric pressure,

`$F_(atm) = P_(atm) * A_(p)$`

`$= 1 * 10^5 * 0.00785$`

= 785 N

Now
`$F_(gas) = F + F_(atm) + W$`

⇒ 2355 = F + 785 + 245.25

⇒ F = 1324.75 N