Question #db318

The pressure of steam injected into the lower part of a cylinder raises the steam hammer which drops under gravity when the pressure is released.

1. Let us calculate what is the velocity free falling hammer of mass `m` as it strikes the iron piece. We use the kinematic equation

   `v^2-u^2=2as` .......(1)
   `u and v` are initial and final velocities respectively, `s` is displacement and `a` is acceleration. For a free falling hammer `a=g=9.81m^-2`, acceleration due to gravity. Inserting given values in SI units we get
   `v^2-0^2=2xx9.81xx0.6`
   `=>v=sqrt11.772ms^-1`

2. Once the hammer strikes the iron piece it is stopped through an average indentation of `6mm`, as hammer decelerates.
   Using equation (1) we get
   `0^2-(sqrt11.772)^2=2axx0.006`
   `=>a=-11.772/(2xx0.006)=-981ms^-2`
3. Using Newton's Second law of motion we get
   Average Force `F=ma`
   `F=360xx(-981)=-353160N`
   `-ve` sign shows that it is a stopping force.

The pressure of steam injected into the lower part of a cylinder raises the steam hammer which drops under gravity when the pressure is released.

Following assumptions are made for simplicity.

1. Due to mechanical construction of steam hammer, it is assumed that stroke includes the average indentation produced in the hot piece of iron.
2. All the mechanical energy of hammer is expanded towards the creation of indentation and none is wasted due to friction or production of heat light and sound due to impact.
3. The net average force during the period production of indentation is constant and always acts in the direction of displacement vector of movement of hammer/tup.
4. The net average force acts in direction displacement of hammer.

Let us calculate total potential energy of the hammer of mass `m`.
At a of height `h` of hammer `PE=mgh`
where `g=9.81ms^-2` is acceleration due to gravity.
Maximum `PE` of the hammer is at the top of stroke. Which is
`PE_max=360xx9.81xx600/1000=2118.96J` ........(1)

In the next step:
Lets assume that net average force in the blow is `vecF`.

Work `W` done by this force though a displacement which is equal to the indentation produced`=vecFcdotvecd`
`W=|vecF|xx6/1000cos0^@J` .....(2)
From Law of Conservation of energy, `PE=W`.

Equating (1) and (2) we get
`|vecF|xx6/1000=2118.96`

`=>|vecF|=2118.96xx1000/6`
`=>|vecF|=353160N`

For starters, you know that the tup will fall for

`600 color(red)(cancel(color(black)("mm"))) * "1 m"/(10^3color(red)(cancel(color(black)("mm")))) = "0.6 m"`

under gravity, so if you take `v` to be the velocity with which it hits the piece of iron, you can say that

`v^2 = v_i^2 + 2 * g * h`

Here

• `v_i` is the initial velocity of the tup, equal to `"0 m s"^(-1)`
• `g` is the gravitational acceleration, equal to `"9.81 m s"^(-2)`
• `h` is the height of the fall, equal to `"0.6 m"`

The above equation will thus be equivalent to

`v^2 = 2 * g * h" " " "color(orange)("(*)")`

Now, after the tup hits the piece of iron, it travels for another

`6 color(red)(cancel(color(black)("mm"))) * "1 m"/(10^3color(red)(cancel(color(black)("mm")))) = "0.006 m"`

before coming to a complete stop, i.e. its final velocity will be `"0 m s"^(-2)`. You can thus say that for the penetrating distance, you have

`0^2 = v^2 - 2 * a * d`

Here

• `a` is the acceleration required to stop the tup (it opposes the gravitational acceleration)
• `d` is the penetrating distance, equal to `"0.006 m"`

The above equation can be rearranged as

`v^2 = 2 * a * d`

Plug this back into equation `color(orange)("(*)")` to find that

`2 * g * h = 2 * a * d`

which is equivalent to

`a = g * (h/d)`

Now, you know that after the tup hits the piece of iron, a force `F` that opposes gravity is required to bring the tup to a complete stop.

If you take `m` to be the mass of the tup, which we know to be equal to `"360 kg"`, you can say that

`overbrace(F - m * g)^(color(blue)("the net force felt by the tup")) = m * a`

Plug in the expression you have for `a` to find that

`F - m * g = m * overbrace(g * (h/d))^(color(blue)(=a))`

which is equivalent to

`F = m * g * (1 + h/d)`

Finally, plug in your values to find

`F = "360 kg" * "9.81 m s"^(-2) * overbrace(( (0.6 color(red)(cancel(color(black)("m"))))/(0.006color(red)(cancel(color(black)("m")))) + 1))^(color(blue)(=101))`

`F = "356,691.6 N" = color(darkgreen)(ul(color(black)("357,000 N")))`

I'll leave the answer rounded to three significant figures, but keep in mind that you only have one significant figure for the height of the fall and for the penetrating distance.