Answer:
a) 49.95 watts
b) The self locking condition is satisfied
Explanation:
Given data
weight of the square-thread power screw ( w ) = 100 kg = 1000 N
diameter (d) = 20 mm ,
pitch (p) = 2 mm
friction coefficient of steel parts ( f ) = 0.1
Gravity constant ( g ) = 10 N/kg
Rotation of electric power screwdrivers = 300 rpm
A ) Determine the power needed to raise to the basket board
first we have to calculate T
T = Wtan (∝ + Ф ) * [tex]\frac{Dm}{2}[/tex] ------------- equation 1
Dm = d - 0.5 ( 2) = 19mm
Tan ∝ = [tex]\frac{L}{\pi Dm}[/tex] where L = 2*2 = 4
hence ∝ = 3.83⁰
given f = 0.1 , Tan Ф = 0.1. hence Ф = 5.71⁰
insert all the values into equation 1
T = 1.59 Nm
Determine the power needed using this equation
[tex]\frac{2\pi NT }{60}[/tex] = [tex]\frac{2\pi * 300 * 1.59}{60}[/tex]
= 49.95 watts
B) checking if the self-locking condition of the power screw is satisfied
Ф > ∝ hence it is self locking condition is satisfied