Question

4. Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.15 and that 0=45°, determine the smallest force P required to raise block A. (20 pts) 700 N A B e P

Answer 1

# Solution 1700N given data waight of pipe column = 700N А B is = 0.15 & 0-45° P Note : To lift the block A upword the friction force will act on block A in downword direction 700N FBD of Block Net Horizontal force on Block A Efp = 0 (Normal Rex") NA A N NA fs. Cas 45 Ncasus =0 ISA2 (friction force) NA Us N casus Ncasus = o ISAL NA Ncasus ( Ms + !) = o net vertical force Block A E fuo on ESAI 700 + FSA, Sinus N sinus Ms NA + 700 M N sin us - Nşin us=o • 15 NA + 700 (ons sinus - sinus) N = 0 2

Solve D Equation and to get the value of N & NA N= 1461:18 N & NA 1188.188 N F BD of Block В fsAz (friction force) (Normal Rexn) 45 -P net force in Horizontal Direction EFH = 0 45° i Ne Is B (friction force) N N casus + FSA casus - Р P + f fse O Ncasus + Ms N casus P & Ms NB = o

14618 Casus 0.15 X 1461.18 casus - Pt 0015 NB =0 3 * + 0.1S NG 1188.1918 net force in vertical Direction E fu + fsA. Sinus NB N sinus 2 1461 18 sinus + Mg N Sinus-o Ng = 878.2287 Ni 3 to to get the put the value of No in equation 3 value of P 1188.1918 - P + 0.15 X 878.2287 Pmin 1319.926 N