Mechanical Engineering Principles

140 MECHANICAL ENGINEERING PRINCIPLES

Problem 8. The average force exerted on

the workpiece of a press-tool operation is

150 kN, and the tool is in contact with the

workpiece for 50 ms. Determine the change

in momentum.

From above, change of linear momentum

=applied force×time(=impulse)

Hence,change in momentum of workpiece

= 150 × 103 N× 50 × 10 −^3 s

=7500 kg m/s(since 1 N=1kgm/s^2 )

Problem 9. A force of 15 N acts on a body

of mass 4 kg for 0.2 s. Determine the change

in velocity.

Impulse=applied force×time=change in linear
momentum

i.e. 15 N× 0 .2s=mass×change in velocity

=4kg×change in velocity

from which,change in velocity

=

15 N× 0 .2s

4kg

= 0 .75 m/s

(since 1 N=1kgm/s^2 )

Problem 10. A mass of 8 kg is dropped

vertically on to a fixed horizontal plane and

has an impact velocity of 10 m/s. The mass

rebounds with a velocity of 6 m/s. If the

mass-plane contact time is 40 ms, calculate

(a) the impulse, and (b) the average value of

the impulsive force on the plane.

(a) Impulse=change in momentum=m(u 1 −v 1 )

where u 1 = impact velocity = 10 m/s and

v 1 =rebound velocity=−6m/s

(v 1 is negative since it acts in the opposite

direction tou 1 )

Thus,impulse=m(u 1 −v 1 )

=(8kg)( 10 −− 6 )m/s

= 8 × 16 =128 kg m/s

(b) Impulsive force=

impulse

time

=

128 kg m/s

40 × 10 −^3 s

=3200 Nor3.2 kN

Problem 11. The hammer of a pile-driver

of mass 1 t falls a distance of 1.5 m on to a

pile. The blow takes place in 25 ms and the

hammer does not rebound.

Determine the average applied force exerted

on the pile by the hammer.

Initial velocity,u=0, acceleration due to gravity,

g= 9 .81 m/s^2 and distance,s= 1. 5 m.

Using the equation of motion:

v^2 =u^2 + 2 gs

Gives: v^2 = 02 + 2 ( 9. 81 )( 1. 5 )

from which, impact velocity,

v=

√

( 2 )( 9. 81 )( 1. 5 )= 5 .425 m/s

Neglecting the small distance moved by the pile and

hammer after impact,

momentum lost by hammer

=the change of momentum

=mv=1000 kg× 5 .425 m/s

Rate of change of momentum

=

change of momentum

change of time

=

1000 × 5. 425

25 × 10 −^3

=217000 N

Since the impulsive force is the rate of change of

momentum,the average force exerted on the pile

is 217 kN.

Problem 12. A mass of 40 g having a

velocity of 15 m/s collides with a rigid

surface and rebounds with a velocity of

5 m/s. The duration of the impact is 0.20 ms.

Determine (a) the impulse, and (b) the

impulsive force at the surface.