Cutting off AC current to motor with thermistor
Here's one way to do this with a few transistors and resistors:



R1 is the thermistor, which goes lower in resistance as its temperature rises.  When that happens, Q1 is turned on, which turns on Q2, which turns on the relay, which turns off the motor.  The normally closed contact of the relay (pins 1 and 4 in this example) are in series with the motor on the hot side of the AC line.

R5 provides a little positive feedback to get snap action instead of fading the relay on and off.  This is better for the relay and for overall operation.

R1 and R2 form a voltage divider.  The relay will be energized (motor turned off) when the output of this divider gets to about 600 mV.  In this example, that means the thermistor trip resistance is about 73 k&Omega;.  You will need to adjust R2 to set the trip point for whatever thermistor you actually get.

Much lower thermistor values, with accordingly lower R2 values, will work, but be aware of self-heating.  At the trip point, the thermistor will have about 4.4 V across it.  The dissipation of the thermistor is therefore

&nbsp;&nbsp;W = 19.4 V2 / R

where R is the resistance of the thermistor in Ohms, and W its dissipation in Watts.  You want to keep W below a few 10s of mW.  Working backwards using 25 mW as the maximum dissipation, that means the thermistor must be 776 &Omega; or more at the trip point.  Fortunately thermistors of a few k&Omega; are easy to find.

As I said above, R5 sets the hysteresis size.  You probably need to experiment to find what provides snap action without causing a overly large hysteresis band.  I'd start with the current thru R5 about 1% that of R2.

As others have said, use a cheap wall wart that runs from the same power the motor is driven with to get the 5 V.