Question

2. A load of 2.5 kN is to be raised by a screw jack with mean diameter of 75 mm and pitch of 12
mm. Find the efficiency of the screw jack, if the coefficient of friction between the screw and nut
is 0.075.​

Answer 1

Answer:

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Answer 2

Explanation:

Given  load of 2.5 kN is to be raised by a screw jack with mean diameter of 75 mm and pitch of 12 mm. Find the efficiency of the screw jack, if the coefficient of friction between the screw and nut  is 0.075.

• We have load W = 2.5 kN
• Mean diameter of screw d = 75 mm
• Pitch p = 12 mm
• Coefficient of friction between screw and nut = tan φ = 0.075
• So we have  
•              tan α = p / π d
•                       = 12 / 3.14 x 75
•                       = 0.0509
• So efficiency of the screw jack ƞ = tan (α + φ)
•                                                     = tan α / tan α + tan φ / 1 – tan α tan φ
•                                                      = 0.051 / 0.051 + 0.075 / 1 – (0.051 x 0.075)
•                                                      = 0.051 / 0.1265
•                                                        = 0.403
•                                                        = 40.3 %
• So efficiency of the screw jack is 40.3%

Reference link will be

https://brainly.in/question/13758377