Choosing voltage divider resistor values
In the situation you outline, the rule of thumb is to eliminate R2 and determine the value of the voltage required at the \$ R1Rl \$ junction by using Ohm's law:

$$ E_{J} = IR = 0.01A \times 100\Omega = 1 \text{volt} $$

Since the current in a series circuit is everywhere the same, the 0.01 amperes through the load resistor must also go through R1.

Then since the supply voltage, Vcc, is 5 volts and you want the motor to see 1 volt, Ohm's law dictates that the value of R1 must be:

$$R1 = \frac{Vcc -  V_{J}}{I} -\frac{4V}{0.01A} = 400 \text{ohms}  $$