Question

10 The block shown in figure below is acted upon rts weight W 200 kN a horizontal force Q=600 kN, and the pressure P exerted by the inclined plane. The resultant R of these forces Is up and parallel to the incline thereby sliding the block up it. Determine P and R Hint: Take one axis parallel to the incline. W-200 kN R Q-600 kN P 30 Solution: ΣFy -0 P Cos 15-600 Sin 30-200 Cos 30=0 P Cos 15 300 173.20 473.20 P = Cos 15 490 kN R2 = (Fy)2+ (Fx2 R2-6 ΣΕ? + y P 15 30° R

why summation of forces Fy is considered 0? (see photo) thank you!

Answer 1

W 200KN 6001N o ne axis Paralle to the incline 30

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case GOO KN F2 =W ニ 200 H. Resultant FasLe P P 1 V R- Hence by consideing is taken as static equiiboium condition Pooof F2 FR FI Force R Resultaot Equilibbium Fosce causes unbalance distibuting Resultant acts aS a in the syslem. Equilibrium Fosce '- the effect of cancels in which It is the Fore resultant Faxce the system stable Makes force This resullant a s equilibriam foce clbe Same Valu e of is For ce dorection of equilibum for ce, but the Tesultant Foce. opposite to lets awnsso R the will be the aesultant. Forces Equilibaium
To solve this problem, Newton’s third first law of motion can be referred , by using static equilibrium condition. And hence, summation of all the vertical forces Fy is taken as zero, as the law states itself a body is said to be in equilibrium if the object is at rest or in motion with constant velocity, by using Newtons first law of motion, we can calculate the forces acting on the system.

Hence F = 0 To calculate the