Answer:
0.243
Explanation:
Step 1: Identify the given parameters
Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,
collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09
Frictional diameter =45mm
Step 2: calculate the torque required to raise the load
[tex]T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]
[tex]T_{R}[/tex] = (9.66 + 6.75)N.m
[tex]T_{R}[/tex] = 16.41 N.m
Step 3: calculate the torque required to lower the load
[tex]T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]
[tex]T_{L}[/tex] = (1.64 + 6.75)N.m
[tex]T_{L}[/tex] = 8.39 N.m
Since the torque required to lower the thread is positive, the thread is self-locking.
The overall efficiency = [tex]\frac{F(L)}{2\pi(T_{R})}[/tex]
= [tex]\frac{5(5)}{2\pi(16.41})}[/tex]
= 0.243
