## Power Screw Torque

Consider a square thread power screw can lift a 20 kN load with a 70% efficiency. If the lead is 0.01 m and the thread coefficient of friction is 0.05, how much torque (N∙m) is needed?

Expand 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. Solving for torque:
$$0.7=\frac{(20,000N)(0.01m)}{2\pi \cdot T}$$$$$T=\frac{(20,000N)(0.01m)}{2\pi(0.7)}=\frac{200N\cdot m}{4.396}=45.5\:N \cdot m$$$
45.5 N∙m