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A jack shown in figure 1, uses a single square-thread screw to raise a known load. The major diameter and pitch of the screw and the thrust collar mean diameter were known. (a) Determine the thread depth [3 Marks) (b) Determine helix angle. [4 Marks] (c) Estimate the torque required to raise the load. [6 Marks] (d) Estimate the torque required to lower the load. [6 Marks] (e) Estimate the efficiency of the jack for raising the load [6 Marks)

Question-Answer › Category: Machine Design › A jack shown in figure 1, uses a single square-thread screw to raise a known load. The major diameter and pitch of the screw and the thrust collar mean diameter were known. (a) Determine the thread depth [3 Marks) (b) Determine helix angle. [4 Marks] (c) Estimate the torque required to raise the load. [6 Marks] (d) Estimate the torque required to lower the load. [6 Marks] (e) Estimate the efficiency of the jack for raising the load [6 Marks)
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Cornwell asked 1 year ago

A jack shown in figure 1, uses a single square-thread screw to raise a known load. The major diameter and pitch of the screw and the thrust collar mean diameter were known. (a) Determine the thread depth [3 Marks) (b) Determine helix angle. [4 Marks] (c) Estimate the torque required to raise the load. [6 Marks] (d) Estimate the torque required to lower the load. [6 Marks] (e) Estimate the efficiency of the jack for raising the load [6 Marks)

QUESTION 3 (25 MARKSI A jack shown in figure 1, uses a single square-thread screw to raise a known load. The major diameter a

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R cotwell answered 1 year ago

Answers
 

  1. The above problem is related to Machine Design.
  2. A screw jack is given for the analysis.
  3. All the necessary steps are clearly mentioned and easy to understand.
  4. Hope the answer will help you. Thank you for your support.
  5. Lin Solutions A gack shown single square thread screw 1 figure raise to uses al a known loed. Load thread screw W = 50 kn ft(c) 2 2 Now, Torque required to raise the load T= Wam (findmth W fede I dm- fL (sorod) (0.033)(-0:24 (01033) +0.006 (0,033 33= 300-N-m. Work output during one revolution W.p = (50000) (0.006) Efficiency of the Jack for raising the load Worch Output =
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