Question

For all problems, make sure you define a force equilibrium and a moment equilibrium equation. Use vectors wherever possible.

Problem 2 Determine the force F needed to lift the load, and find the distance x required to place the hook to maintain equilibrium. Each pulley has a radius of r=100 mm. The beam has a variable mass along the length as given by the function How does the distance x change if the mass along the beam is constant and equal to ρ=100 kg/m?

EGM 2511 Statics University of Florida Subject: Recitation 5.2 Frame Analysis Name: Date: Problem 2 d=0.9 m Determine the force F needed to lift the load, and find the distance x required to place the hook to maintain equilibrium. Each pulley has a radius of r=100 mm. The beam has a variable mass along the length as given by the function m(s)=c{1+2 C =100 here How does the distance x change if the mass along the beam is constant and equal to p=100 kg/m? Answer: x = 0.702 m (for the variable self weight) x = 0.754 m (for the constant self weight) 16 = 0.3 m ko SC P = 3 kN

Answer 1

Case - variable mass of beam along length. -d-oogm. VA BL 0.3m - > . b=0:3 sel P-3KN

Assume pulleys Assume all pullies are so, FBD 7 Beam. friction -less with r=o. 1m 0.9+0) + 0.3m- & mg V P- 3 KN long & me) = c(+ mis): 100 ( 1+ ) where ca 100kg/m. when s=o, 1.6m m(o) = 100 kg/m. when soul m(L) = 300 kg/m 300 1 20 100 center of mass workort 3 beam = 100 X16X0.8+ +*200*1,6 3 x16 x (100+300) x 1.6 = 0.93m. (from Right.) to tal mass beam = + X16X 1100 +300) - 320kg let g= 9.8 m/s² a total weight of beam = max mg (N) - 320 x 9.8) e X103 KN - 3. 14 KN @ 0.93m from right

For equilibrium. (Force equilibrium) Ety =0 2F +F = 3+ 3.14 → F = 2.047KN moment equilibrium Eme=0 (EMB=0 SPOxNzret 2F XI - px x - mg x 0.633 = 0 → 2X2.047 - 3x2 - 3.14.40633 =0 5x = 0.702m] mam case - For a constant self weight & weight of beam = mga = 100 x 106 x 9.87X1073 = 1.5696 KN e 0.8 from right. for force eq. Efyzo @ 3F = 3+ 1.5696 > F = 1.5232 KN I moment equilibrium EMB=0 afx 1 - PxPC) - mg x 0.5=0 2X1.5232 - 3x x - 1.56 96 *0.5 =0 x = 0.758 m from