A particle is subjected to two equal forces along two different directions. If one of them is halved, the angle which the resultant makes with the other is also halved, The angle between the forces is halved. The angle between the forces is ?

Suppose the forces are P and Q. Let the angle between the forces be α.

|P| = |Q|

Let the resultant be R and the resultant makes an angle β with Q.

Now,

tanβ = (P sinα)/(Q + P cosα)

=> tanβ = (P sinα)/(P + P cosα)

=> tanβ = (sinα)/(1 + cosα) = tan(α/2)

=> β = α/2 …………………..(1)

Again,

tan(β/2) = (P/2)(sinα)/[Q + (P/2)cosα]

=> tan(β/2) = (P/2)(sinα)/[Q + (P/2)cosα]

=> tan(β/2) = (1/2)(sinα)/[1 + (1/2)cosα]

=> tan(β/2) = (sinα)/[2 + cosα]

(1) => tan(α/4) = (sinα)/[2 + cosα]

=> [sin(α/2)]/[1 + cos(α/2)] = (sinα)/[2 + cosα]

=> [sin(α/2)]/[1 + cos(α/2)] = [2sin(α/2)cos(α/2)]/[2 + {2cos2(α/2) - 1}]

=> 1/[1 + cos(α/2)] = [2cos(α/2)]/[2cos2(α/2) + 1]

cos(α/2) = x

So,

1/(1 + x) = 2x/(2x2 + 1)

=> 2x2 + 1 = 2x + 2x2

=> x = ½

So,

cos(α/2) = ½ = cos60

=> α = 120o

 This is the angle between the forces.

  • 59

Sorry the revised question is : A particle is subjected to two equal forces along two different directions. If one of them is halved, the angle which the resultant makes with the other is also halved. What is the angle between the forces?

  • 4
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