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two forces F1=500 N due east and F2=250 N due north have their common initial point. F2-F1 is and angle is.. Note:their is an arrow on F1 and F2..

two forces F1=500 N due east and F2=250 N due north have their common initial point. F2-F1 is and angle is..


Note:their is an arrow on F1 and F2..

Grade:12

4 Answers

nirmit jain
18 Points
7 years ago

let

f = f1 + f2

v= f1 - f2

there fore on adding both we get

f+v = 2 f1

and f can be calculated by vetor sum of addition = 559.07N towards north east

and therefore

v = 724.06 N towards 22.5 degree north

Bevkoof Singh
43 Points
7 years ago

since both are in two diff direction so

the angle between them is 900

so l F2vector+(-F1vector) l = √f12+f22-2f1f2(cos90).......................cos90=0

by puting the value of f1  f2

we get 

250√5 N and angle is 

Θ=tan-1(500/250) in north of west

Θ=tan-1(2)

SOURAV MISHRA
37 Points
7 years ago

MAGNITUDE OF RESULTANT FORCE IS CLEARLY 250√5N AND THE ANGLE BETWEEN THE RESULTANT AND F1 IS tan-1(1/2).

Sachin
21 Points
2 years ago
F2- F1=√F1^2+F2^2+2F1.F2.cos 90°
F2-F1= √62500+250000+0
          =√312500
          =250√5
tanA = 250/500
A = tan^-1 (1/2)

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