The resultant of two forces 3P and 2P is R. If the first force is doubled, the resultant is also doubled. What is the angle between the forces?

 The resultant of two forces 3P and 2P is R. If the first force is doubled, the resultant is also doubled. What is the angle between the forces?

Solution

The magnitude R of the resultant R of two forces 3 P and 2 P is given by

R = [ (3P)² + (2P)² + 2 × 3 P × 2 P Cos ß]½

= [ 9 P² + 4 P² + 12 P² Cos ß]½

= [13 P² + 12 P² Cos ß]½

= P [ 13 + 12 Cos ß]½,

where ß is the angle between the two forces and 0 ≤ ß ≤ 180°.

When the first force is doubled, the magnitude of the new resultant R' (= 2 R) is given by

R' = [ (6 P)² + (2 P)² + 2 × 6 P × 2 P Cos ß]½

= [ 36 P² + 4 P² + 24 P² Cos ß]½

= [ 40 P² + 24 P² Cos ß]½ = P [ 40 + 24 Cos ß]½

According to statement of the question,

R' = 2 R

=> P [ 40 + 24 Cos ß]½ = 2 × P [ 13 + 12 Cos ß]½

=> [ 40 + 24 Cos ß]½ = 2 × [ 13 + 12 Cos ß]½ —(1)

Squaring both sides

[40 + 24 Cos ß] = 4 [ 13 +12 Cos ß]

=> 40 + 24 Cos ß = 52 +48 Cos ß

=> 48 Cos ß - 24 Cos ß = 40 - 52

=> 24 Cos ß = -12,

Cos ß = - 1/2

ß = arc Cos ( -1/2) = 120°

The angle between the forces is 120°.