## Power Screw

Consider a square thread power screw uses 1,000 N∙m of torque to lift a 25 kN load. If the lead is 0.02 m and the thread coefficient of friction is 0.08, what is the power screw’s overall efficiency?

Hint
The power screw’s efficiency is:
$$\eta=\frac{Fl}{2\pi T}$$$where $$F$$ is the load, $$l$$ is the lead, and $$T$$ is the torque. Hint 2 The friction coefficient is not needed to solve the problem. It only exists to cause confusion. The power screw’s efficiency is: $$\eta=\frac{Fl}{2\pi T}$$$
where $$F$$ is the load, $$l$$ is the lead, and $$T$$ is the torque.
Notice how the friction coefficient is not needed to solve the problem. It only exists to cause confusion. Thus,
$$\eta=\frac{(25,000N)(0.02m)}{2\pi (1,000N\cdot m)}=\frac{500}{6,280}=0.08$$\$
0.08