Two forces 3 N & 2N are at an angle theta such that the resultant is R. The 1st force is now increased to 6 N & the resultant become 2R. Find the value of theta.
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Answered by
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Formula used
R^2 = A^2 + B^2 + 2AB Cos theta
A= 3 N
B= 2 N
R^2 = 13 + 12 Cos theta
Case 2
4R^2 = 40 + 24 Cos theta
put the value of R^2
here we got
4 ( 13 + 12 Cos theta ) = 40 + 24 Cos theta
52 + 48 Cos theta = 40 + 24 Cos theta
Now 24 Cos theta = 40-52
Cos theta = -12 / 24
Cos theta = -1/2
Cos 120 degree = -1 /2
Theta = 120 degree
Formula used
R^2 = A^2 + B^2 + 2AB Cos theta
A= 3 N
B= 2 N
R^2 = 13 + 12 Cos theta
Case 2
4R^2 = 40 + 24 Cos theta
put the value of R^2
here we got
4 ( 13 + 12 Cos theta ) = 40 + 24 Cos theta
52 + 48 Cos theta = 40 + 24 Cos theta
Now 24 Cos theta = 40-52
Cos theta = -12 / 24
Cos theta = -1/2
Cos 120 degree = -1 /2
Theta = 120 degree
falgunichapekar:
Can anyone ans it plz
sorry
i thought there is a increase of 6 in 3 N and i taken it as 9 N
the formula i used i R^2 = A^2 + B^2 + 2AB
Plz try it again
That's √ a² + b² + 2ab cos theta
yeap
sorry for lots of mistakes
Check now
OK thanks
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