Tek-Tips is the largest IT community on the Internet today!

Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!

  • Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Dynamics application 1

Status
Not open for further replies.

891879

Electrical
Joined
Jan 24, 2003
Messages
3
Location
US
I have a table which needs to be tilted. The current unbalanced load is 1886 lbs to the left of the pivot at 35" from the pivot (66,000 in-lbs of resultant torque on the shaft). I would like to use a electro mechanical jack to lift the table. The jack is positioned below the table to the left of the pivot at a 27 edgree angle from horizontal. The jack attaches to the table at 49.4" from the pivot. The perpendicular distance from the jack centerline to the pivot is 11.6". Using Torque analysis, I determined that I need a 5,689 lb force to balance the table (66,000/11.6).

Using a vector analysis it seems that I need a 2942 lb force pushing at the 27 degree angle to balance the load at 49.4 inches out from the pivot (load at 49.4" = 66,000 in-lbs/49.4 = 1336, 1336/sin 27 = 2942 lbs.

Which analysis is correct? Which is incorrect and why?
 
Either way should give you the same answer, so either is more correct than the other. When I work your problem, the moment arm for the jack force I calculate is 22.43" (49.4sin27). To find the force due to this would be 1886*35/22.43 which yeilds about 2943 lbs, the same as that found when you used a vector approach. I would verify how you calculated themoment arm for your force of 11.6" I hope this helps.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top