Question

find the magnitude and direction of the resultant of following forces impressed on a particle f1 = 3 root 2 kilometre per force due north east and forced to this route to kgf due south east and f3 is equal to root 2 kgf northwest

Answer 1

Answer:

Explanation:

f1 = 3√2  North east

=> North = 3√2Sin45  = 3

& East = 3√2Cos45 = 3

f1 = 3 East  + 3 North

f2 = 1   South East

f2 = 1/√2 East  + 1/√2  South

f3 = 2  North West

f3 = √2 West   + √2 North

Total = 3 East  + 1/√2 East + √2 West    + 3 North + 1/√2  South + √2 North

= 3 East  + 1/√2 East - √2 East    + 3 North - 1/√2  North + √2 North

= (3 - 1/√2) East + (3 + 1/√2) North

Magnitude = √ (3 - 1/√2)² +  (3 + 1/√2)² = √ 19

Tanα = (3 + 1/√2)/(3 - 1/√2)  => α = 58.26 °

58.26 ° North of East    or  31.74 East of North

Answer 2

Explanation:

Answer will be

F (resultant force) will be 8.246N and it acts 14.04 degree South-West.

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