Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch ca
1 answer:
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Answer:
power = 49.95 W
and it is self locking screw
Explanation:
given data
weight W = 100 kg = 1000 N
diameter d = 20mm
pitch p = 2mm
friction coefficient of steel f = 0.1
Gravity constant is g = 10 N/kg
solution
we know T is
T = w tan(α + φ ) ...................1
here dm is = do - 0.5 P
dm = 20 - 1
dm = 19 mm
and
tan(α) = ...............2
here lead L = n × p
so tan(α) =
α = 3.83°
and
f = 0.1
so tanφ = 0.1
so that φ = 5.71°
and now we will put all value in equation 1 we get
T = 1000 × tan(3.83 + 5.71 )
T = 1.59 Nm
so
power = .................3
put here value
power =
power = 49.95 W
and
as φ > α
so it is self locking screw