Question

Five forces acting

at a point o are in equilibrium. Four of them whose magnitudes are 3,4, 4, 3 kg. wt., act along the coplanar lines OA, OB, OC, OD respectively, such thatZAOB = 15°, BOC = 60°, ZCOD = 15°;Find the magnitude and direction of the fifth force.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer: Fifth force magnitude should be 7.706 kg-wt making 225° with OA

Step-by-step explanation:

All five forces at O are in equlibrium therefore,

Net Force at O = 0

Fifth force = - ( sum of other 4 forces )

OA = 3 i

OB = 4cos(15°) i + 4sin(15°) j

OC = 4cos(75°) i + 4sin(75°) j

OD = 3  j

OA+OB+OC +OD = 3i +(4cos(15°) i + 4sin(15°)j) +(4cos(75°)i +4sin(75°) j) +3j

OA+ OB+ OC + OD = (3 + 4cos(15°) + 4cos(75°)) i  + (4sin(15°) + 4sin(75°)+3)j

OA + OB+ OC + OD =  (3 + 4(cos(15°) + sin(15°))) (i + j)

OA + OB+ OC + OD =  (3 + 2√6) (i + j)  

|OA + OB+ OC + OD| =  ( 3√2 + 2√3 ) = 7.706

Magnitude of fifth force is 7.706 kg-wt

Phase of resultant = 45°

Therefore, phase of 5th force = 225°

Therefore, fifth force magnitude should be 7.706 kg-wt making 225° with OA

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