Physics Solutions Manual

Chapter 24 continued
Challenge Problem
page 656
The figure shows two identical motors with a
common shaft. For simplicity, the commutators
are not shown. Each armature coil consists of
48 turns of wire with rectangular dimensions of
17 cm wide by 35 cm deep. The armature
resistance is 12 . The red wire travels to the left
(along half the width) and then back to the rear
of the motor (along the depth). The magnetic
field is 0.21 T. The diameter of the pulley is
7.2 cm. A rope fixed to the pulley and the floor
prevents the motor shaft from turning.
Pulley
S
Spring
scale
S 1
S
Shaft
N
120 V
1. Given F ILB, derive an equation for the
torque on the armature for the position
shown.
Torque is defined as the product of the
force and the lever arm. In the case of
the motor armature position shown, the
lever arm is equal to half of the width of
the armature coil (the shaft is the center
of rotation and is at the midpoint of the
coil width). The length of the wire acted
upon by the field is equal to the depth
of the coil. This length is effectively
increased by n, the number of turns in
the coil. Finally, the torque is doubled
because as one side is pushed up by
the magnetic field, the other side is
pushed down according to the third
right-hand rule.
2nBI(depth)(width/2)
S 2
N
35 V
Simplifying and replacing (depth)(width)
with area, A, gives:
nBIA
The torque produced by the motor
armature, in the position shown, is
equal to the number of turns times the
field strength times the armature current
times the area of the armature coil.
2. With S 1 closed and S 2 open, determine the
torque on the shaft and the force on the
spring scale.
nBIA
(48)(0.21 T)1 20
V
(0.35 m)(0.17 m)
12
6.0 Nm
Because the shaft cannot turn, the system
is in equilibrium and the force on
the spring scale is found by considering
half the pulley diameter:
6.
0 Nm
Fspring scale 170 N
0.
036
m
3. With both switches closed, determine the
torque on the shaft and the force on the
spring scale.
Both motors produce counterclockwise
torque:
1 (48)(0.21 T)1 20
V
(0.35 m)(0.17 m)
12
6.0 Nm
2 (48)(0.21 T) 35
V
(0.35 m)(0.17 m)
12
1.7 Nm
net 7.7 Nm counterclockwise
7.
7 Nm
Fspring scale 210 N
0.
036
m
4. What happens to torque if the armature is
in a different position?
The torque is reduced when there is
any rotation from the position shown
because the lever arm is reduced. With
90° rotation, the force on the armature
will be up and down (canceling) with
the effective lever arm equal to zero.
With the shown position as 0°:
nBIA cos
500 Solutions Manual Physics: Principles and Problems
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