Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most tr
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Answer:
i) The torque required to raise the load is 15.85 N*m
ii) The torque required to lower the load is 6.91 N*m
iii) The minimum coefficient of friction is -0.016
Explanation:
Given:
dm = mean diameter = 0.03 m
p = pitch = 0.004 m
n = number of starts = 1
The lead is:
L = n * p = 1 * 0.004 = 0.004 m
F = load = 7000 N
dc = collar diameter = 0.035 m
u = 0.05
i) The helix angle is:
The torque is:
ii) The torque to lowering the load is:
iii)
Clearing u:
u = -0.016
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) / 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
=3.335 *10^-4
ε(avg) =150 *10^-6
orientation of γmax
θ = 31.71 or -58.29
To determine the direction of γmax
= 1.67 *10^-4