2. Find the horizontal force P to start the motion of any part of the system of three blocks resting upon oneanother as shown. The weight of the blocks are W-300 KN, Ws= 100 KN and Wc =200 KN. Between A and Bμ=0.30, between C and B, μ =0.20 and between C and the ground, p=0.10.1sni brod nog02.0-4BC05.0-4Solution: tamos quoy worla bre sosiaasquar005300 en of jells400N₁-300F3N-300-N₂=300Assuming B and C will not move.F₁ = 300 (0.30)F₁ = 90 kNne à bns A zelboß 1P = 90 kNes ensla berbhebnu 02.0- y elin 05.0= q Vin Celpris orllanimetsa MX 008Assuming C will not move: al 1stF2=400 (0.20) = 80 kNOTEBm ribirteF₂vicaAssuming the three blocks move:F3 = 0.10 (600)F3 = 60 kNP=60 KNLeast force is 60 kN.

Friction:- Friction is the force which resists the relative motion. It is given that WA= 300 KN,…

Answered: Find the horizontal force P to start…

Find the horizontal force P to start the motion of any part of the system of three blocks resting upon one other as shown The uni

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

2. Find the horizontal force P to start the motion of any part of the system of three blocks resting upon one
another as shown. The weight of the blocks are W-300 KN, Ws= 100 KN and Wc =200 KN. Between A and B
μ=0.30, between C and B, μ =0.20 and between C and the ground, p=0.10.
1
sni brod nog
02.0-4
B
C
05.0-4
Solution: tamos quoy worla bre sosia
asquar
005
300 en of jells400
N₁-300
F3
N-300-
N₂=300
Assuming B and C will not move.
F₁ = 300 (0.30)
F₁ = 90 kN
ne à bns A zelboß 1
P = 90 kN
es ensla berb
hebnu 02.0- y elin 05.0= q V
in Celpris orllanimetsa MX 008
Assuming C will not move: al 1st
F2=400 (0.20) = 80 kN
OTE
Bm ribirte
F₂
vica
Assuming the three blocks move:
F3 = 0.10 (600)
F3 = 60 kN
P=60 KN
Least force is 60 kN.

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