Question

The resultant of two forces 3p and 2p is r. if the first force is doubled then the resultant then the resultant is also doubled. the angle between te two forces

Answer 1

Hey.

Here is the answer.

Resultant,R =√(a^2+b^2+2abcosx)

As, r = √{(3p)^2+(2p)^2+2×3p×2p×cosx}

or, r = √(9p^2+4p^2+12p^2cosx)

or, r = √(13p^2+12p^2cosx)

According to question;

2r = √{(6p)^2+(2p)^2+2×6p×2p×cosx}

or, 2r =√(36p^2+4p^2+24p^2cosx)

or, 2r = √(40p^2+24p^2cosx)

or, 2{√(13p^2+12p^2cosx)}
= √(40p^2+24p^2cosx)

or, 4(13p^2+12p^2cosx) = (40p^2+24p^2cosx)

or, 52p^2 + 48p^2cosx= 40p^2 + 24p^2cosx

or, 12 p^2 = - 24 p^2cosx

or, cosx = - 1/2

or, x = cos^-1(-1/2)

or, x = 120°

So, angle between the forces is 120°.

Thanks.

Answer 2

In 1st case, R2=(3P)2+(2P)2+2*3*2*P2*cosΘ = 9P2+4P2+12P2cosΘ = 13P2+12P2cosΘ ----->(1)

In 2nd case, (2R)2=(6P)2+(2P)2+2*6*2*P2*cosΘ ; 4R2=36P2+4P2+24P2cosΘ = 40P2+24P2cosΘ ------->(2)

Multiply (1) by 4; 4R2=52P2+48P2cosΘ ------>(3)

(3)-(2) gives, 0=12P2+24P2cosΘ

Therefore, cosΘ = (-1/2) => Θ = cos-1(-1/2) = 120°

Hope it helps u mark as brainliest please