Download The head of a drop hammer is raised 2m and then falls

The head of a drop hammer is raised 2m and then falls freely to strike a billet of metal. If the
mass of the hammer is 300 kg and the billet is reduced in size by 15 mm, in a time of 0.0048
s, what is the average force applied by hammer?
Solve the problem using D’Alembert’s principle and the conservation of energy method and
compare and contrast the methods as suitable for solving the particular problem.
Names of variables used:
m = 300 kg - the mass of the hammer
h = 2 m - the initial height of the hammer
s = 15 mm - the billet size reduction
t = 0.0048 s - the time when the hammer’s force was applied
F = ??? - the average force applied by the hammer.
Please note: I’m writing this as ”me”, not an unknown ”one” because it is easier. I hope the
lack of “scientific” language won’t disturb you (smile).
There is one problem with this problem: too many known variables! Because the distance s
and the time t are both given one may calculate from kinetics the hammer’s deceleration and,
multiplying it by the hammer mass calculate the force F. Also multiplying the deceleration and
the time t one may find the initial speed v of the hammer when it struck the billet.
From the other side this speed is fully determined by the height h, or rather h is determined
by v. One of the given variables is redundant.
Fortunately I calculated the hammer speed using kinetics i.e using the formula:
p
√
m
v = 2 g h = 2 · 9.81 · 2 = 6.264
s
(1)
and from the distance s and time t (also using kinetics)
v=
2s
2 · 0.015
m
=
= 6.25
t
0.0048
s
(2)
Both results are very similar (their difference is about 0.2%) so I assume I can use as many
known variables as needed and forget about the redundant one.
If you are not familiar with formulas in equations (1,2) here is how to get them:
1
1
2 g h = 2 g · g t2 = 2 g · g
2
2
and
µ ¶2
v
= v2
g
(3)
1 2 1v 2 1
at =
t = vt
(4)
2
2 t
2
Equation (4) means that in an uniformly decelerated movement with the initial speed v and
the final speed of zero the distance is the average speed multiplied by time.
s=
Please see the next page for solutions.
Solution - using the energy conservation principle. This method (in case of the current
problem) is much simpler than the other. The potential energy of the hammer raised on the
height h changes into the hammer’s kinetic energy and then into the work W of the billet
resistive force F on the distance s. The energy conservation equation is very simple:
mgh = F s
(5)
From eq. (5) I calculate the force F
mgh
(6)
s
The units of [F] are newtons but The distance s should be expressed in meters, s = 0.015 m.
Let’s calculate the numeric value of F:
300 · 9.81 · 2
F =
≈ 392000 N = 392 kN
(7)
0.015
F =
In one of the previous problems I added the distance s to h when calculating the potential
energy difference. Yes, I should do it here too, but 0.015 m is a small height comparing to 2 m,
so it may be dropped from the calculations.
Solution - using the d’Alambert’s principle. This principle says that in a moving reference
frame the sum of all acting forces, real (which are the gravity and the bullet resistance) and
the inertia force should reduce to zero.
The inertia force is equal m a, where a is the deceleration. It acts downwards, in the opposite
way as the billet resistance. (Please NOTE ! The inertia force acts in a direction opposite to
−
→
the vector F . It tries to hold the current speed of the hammer head. That’s the reason it is
called an “inertia” force). The gravity force m g acts downwards too. So the equation for forces
looks as follows:
F − ma − mg = 0
(8)
The deceleration may be obtained from the distance s and time t (compare the discussion in
the beginning of my text).
1 2
2s
at
therefore
a= 2
2
t
I’m interested in the value of a, so I calculate it before substituting into equation (8)
s=
(9)
2 · 0.015
m
≈ 1302 2
(10)
2
0.0048
s
The deceleration is about 130 times stronger than the Earth gravity and in the equation (8)
the m g part might be dropped. It is the same thing as forgetting the small difference in the
potential energy on the distance s. From eq. (8) I calculate F:
a=
F = −m (a + g) = −300 · (1302 + 10) = 393.6 kN
(11)
The result is very similar to that calculated form eq. (7). Remember that previously I underestimated the change in the potential energy so the results are even closer. The other sources
of difference are the time t and distance s which are not exactly proper for a hammer head
dropped from h = 2 m.
As you can see the second method took more calculations but results are the same (they would
be exactly the same if the problems data were correct).
Best greetings - Antek