Question

A) the forces 20 n, 30 n, 40 n, 50 n and 60 n are acting at one of the angular points of a regular hexagon, towards the other five angular points, taken in order. Find the magnitude and direction of the resultant force.

Answer 1

Answer:

Magnitude of the resultant force

resolving all the forces horizontally,

∑ H= 20 cos 0 °+ 30 cos 30° + 40 cos 60° + 50 cos 90° + 60 cos 120°  N

  = (20*1)+ (30*0.866)+(40*0.5)+(50*0)+ 60(-0.5)  N

= 36.0 N

resolving all the forces vertically,

∑ V= 20 sin 0° + 30 sin 30° + 40 sin 60° + 50 sin  90 °+ 60 sin 120°  N

 =(20*0)+ (30*0.5)+(40*0.866)+(50*1)+ (60*0.866)  N

 = 151.6 N

The Magnitude of the resultant forces:

R = √(∑ H)^2+(∑ V^2)

 =√(36.0)^2+(151.6)^2

 = 155.8  N

Direction of the resultant force:

θ = Angle which the resultant force makes with horizontal

we know that

tanθ  = ∑ V/∑ H

       = 151.6/ 36.0

      = 4.211

or  θ  = 76.6°

Hence the magnitude is 155.8 N

and the direction is 76.6°