ME 232 Homework Problems

Problem 3/124 Page 170, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A pulley system consisting of masses m1 and m2 on the left and a mass m3 on the right.  The system is released from rest.

Find: The speed after masses m1 and m2 have dropped through a distance L.  Reconsider the problem if mass m1  is replaced by a constant downward force acting on m2 of magnitude m1 g.

Parameter Values:
m1 = 10 kg
m2 = 20 kg
m3 = 25 kg
L = 2 m

1. Mechanical Systems: Masses m1 and m2 and the string connecting them as they fall through the distance L is system 1.  Mass m3 as it moves upward through the distance L is system 2.

2. Free Body Diagrams: Not included.  System 1 would show masses m1 and m2, the connecting string, and the weights of masses m1 and m2 acting vertically downward and would show a tension T acting vertically upward.  System 2 would show mass m3 and the weight of mass m3 acting vertically downward and the same tension T acting vertically upward.

3. Equations
Note that system 2 goes up the same distance system 1 goes down.  Thus their speeds are equal in magnitude.

System 1 Work = -(m1 + m2 ) g Dh1 (gravity) + T(-L)  (constant force)
                         = (m1 + m2) g L - T L
System 1 D Kinetic Energy = 1/2 (m1 + m2) (v2 - 0)

System 2 Work = -m3 g Dh2 (gravity) + T L (constant force)
                         = -m3 g L + T L
System 2 D Kinetic Energy = 1/2 m3 (v2 - 0 )

4. Solve
(m1 + m2) g L - T L = 1/2 (m1+m2) v2
-m3 g L + T L          = 1/2 m3 v2

Adding the equations:

(m1 + m2 - m3) g L = 1/2 (m1 + m2 + m3) v2

v = {2(m1 + m2 - m3) g L / (m1 + m2 + m3) }1/2 = 1.889 m/s

Now considering the modified problem:

1. Mechanical Systems: Mass m2 as it falls through the distance L is system 1.  Mass m3 as it moves upward through the distance L is system 2.

2. Free Body Diagrams: Not included.  System 1 would show mass m2 and the weight of m2 acting vertically downward, the constant force m1 g vertically downward,  and would show a tension T acting vertically upward.  System 2 would show mass m3 and the weight of mass m3 acting vertically downward and the same tension T acting vertically upward.

3. Equations
Note that system 2 goes up the same distance system 1 goes down.  Thus their speeds are equal in magnitude.

System 1 Work = -m2  g Dh1 (gravity) + m1 g L (constant force) + T(-L)  (constant force)
                         = (m1 + m2) g L - T L
System 1 D Kinetic Energy = 1/2 m2 (v2 - 0)

System 2 Work = -m3 g Dh2 (gravity) + T L (constant force)
                         = -m3 g L + T L
System 2 D Kinetic Energy = 1/2 m3 (v2 - 0 )

4. Solve
(m1 + m2) g L - T L = 1/2 m2 v2
-m3 g L + T L          = 1/2 m3 v2

Adding the equations:

(m1 + m2 - m3) g L = 1/2 (m2 + m3) v2

v = {2(m1 + m2 - m3) g L / (m2 + m3) }1/2 = 2.09 m/s