Two forces 3n and 2n are acting at an angle theta

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Motilal National Institute of Technology

Here 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

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Text Solution

`30^@``60^@``90^@``120^@`

Answer : D

Solution : `A=3N,B=2N` then `R=sqrt(A^(2)+B^(2)+2ABcostheta)` <br> `R=sqrt(9+4+12costheta)` <br> Now `A=6N,B=2N` then …(i) <br> `2Rsqrt(36+4+24costheta)` .(ii) <br> from (i) and (ii) we get `costheta=-(1)/(2)becausetheta=120^@`