1. A 2 200-kg pile driver is used to drive a steel I-beam into the ground. The p

Question

1. A 2 200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.20 m before coming into contact with the top of the beam, and it drives the beam 14.8 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

_______ magnitude

2. As shown in the figure below, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m1, a 3.60 kg block, originally at rest on the horizontal table at a height 1.21 m above the floor, to m2, a hanging 2 kg block originally a distance d = 0.950 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m1 is projected horizontally after reaching the edge of the table. The hanging block m2 stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system.

(a) Find the speed at which m1 leaves the edge of the table. (Assume m2 hits the ground before m1 leaves the table.)
m/s

(b) Find the impact speed of m1 on the floor.
m/s

(c) What is the shortest length of the string so that it does not go taut while m1 is in flight?
m

Explanation / Answer

1.
m = mass of the pile driver = 2200 kg
h = distance fallen to contact = 3.2 m
d = distance into the ground = 0.148 m
g = acceleration by gravity = 9.8 m/s2

F = force exerted by I-beam on the pile driver = ?

The TOTAL distance the pile driver falls is
D = h + d
D = (3.2 m) + (0.148 m) = 3.348 m

The difference in potential energy of the pile driver between initial and final position is:
PE = m x g x D
PE = (2200 kg) x (9.8 m/s2) x (3.48 m)
PE = 75028.8 J

That is also how much work that has to be done by the force exerted by the beam on the pile driver, so
PE = F x d
m x g x (h + d) = F x d
(75028.8 J) = F x (0.148 m)
F = 506951.35 N
The pile driver is falling down. So to stop it, the I-beam must exert a force in the opposite directen.
F is upward

2.
(a)
Tension T pulling m1 to the right gives
T = m1 * a............(1)
Forces acting on m2 moving downward give
T + (-m2 * g) = m2 * (-a)
T = m2 * g - m2 * a.........(2)
Substituting (1) in (2) gives
m1 * a = m2 * g - m2 * a
(m1 + m2) * a = m2 * g
a = (m2 * g) / (m1 + m2)...........(3)
m2 moving downward from rest with acceleration given in (2) yields
0.950 m = d = 1/2 * a * (t^2)
(t^2) = 2 * d * (m1 + m2) / (m2 * g)
t = sqrt [2 * d * (m1 + m2) / (m2 * g)]..........(4)
t = sqrt [2 * 0.95 * (3.6 + 2) / (2 * 9.8)]
t = 0.74 sec

m1 moving to the right from rest with acceleration in (3) and time given
given in (4) attains final speed of
vx = a * t = (m2 * g) / (m1 + m2) * 0.74
vx = 2 * 9.8 / (3.6+2) * 0.74
vx = 2.59 m/s

(b)
Without any acceleration along the horizontal, the x-component of velocity of m1 during its fall remains constant at 2.59 m/s.
The y-component of velocity increases from zero to some value vy given by
vy^2 = 2 * g * h
where h = 1.21 m is the height of the table top.
vy^2 = 2 * 9.8 * 1.21 = 23.716
vy = 4.87 m/s
The magnitude of the velocity with which m1 hits the ground is therefore
v = sqrt [vx^2 + vy^2]
v = sqrt [2.59^2 + 23.716]
v = 5.52 m/s

(c)
Since the y-component of velocity of m1 is zero when it leaves the table top, the time of fall of m1 can be obtained from
1.21 m = h = 1/2 * g * (t^2)
t = sqrt [2h/g] = sqrt(2*1.21/9.8)
t = 0.5 sec
The horizontal distance traveled by m1 in that time is
S = vx * t = (2.59) * (0.5) = 1.29 m

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