Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Your simulation completely ignores the current drawn by the fan motor.
A simple resistive voltage divider is not a good solution for adjusting the speed of a motor.
What is the power/current rating of the motor?
As a simple modification for a fairly low current load, you could add an NPN darlington transistor as an emitter follower after your potentiometer circuit, to give it higher current capability.
The pot wiper would connect to the base, +10V to the collector and the emitter connects to motor positive.
Also connect a diode such as a 1N4000 series (4002, 4004, 4007 etc) across the motor output from the circuit, with cathode (band) to the transistor emitter and anode to 0V. That protects the transistor from any inductive spikes (back EMF) that the motor may produce).
The output voltage will be around 1.2V - 1.5V lower than the pot voltage, so you may need to compensate with the supply voltage.
The transistor will get warm (or hot) and needs a heatsink.
A TIP120 is readily available NPN darlington which should be suitable.
That pretty much gives you the same circuit as used in many simple model train speed controllers - eg.
Below is a circuit, similar to what rjenkinsgb suggested, except it uses a Sziklai pair instead of a Darlington to reduce the output drop voltage at the maximum pot setting from two base-emitter drops to one.
The simulation shows a maximum output of 9.29V with a TIP32 transistor.