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?
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. Thus,
$$$\eta=\frac{(25,000N)(0.02m)}{2\pi (1,000N\cdot m)}=\frac{500}{6,280}=0.08$$$
0.08
Time Analysis
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