# Question #db318

##### 3 Answers

#### Answer:

(

This solution has serious problems with respect to the Law of Conservation of Energy. Hence Alternate Solution as below has been proposed.

#### Explanation:

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.

- 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# - 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# - 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.

#### Answer:

Previous solution had serious problems with respect to the Law of Conservation of Energy. Hence Alternate Solution as below has been proposed.

#### Explanation:

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.

- Due to mechanical construction of steam hammer, it is assumed that stroke includes the average indentation produced in the hot piece of iron.
- 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.
- 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.
- The net average force acts in direction displacement of hammer.

Let us calculate total potential energy of the hammer of mass

At a of height

where

Maximum

In the next step:

Lets assume that net average force in the blow is

Work

From Law of Conservation of energy,

Equating (1) and (2) we get

#### Answer:

Here's what I got.

#### Explanation:

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 **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

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

Here

#a# is theaccelerationrequired to stop the tup (itopposesthe 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

#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 **opposes gravity** is required to bring the tup to a complete stop.

If you take

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

Plug in the expression you have for

#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.