Three forces act on particle:

Three forces act on particle 'O' as shown in the figure given below. Determine the value of 'P' such that resultant of these three forces is horizontal. Find magnitude and direction of the fourth force which when acting along with the given forces, will keep 'O' in equilibrium.

Sol.: As resultant(R) is horizontal so the vertical component of resultant is zero, that is,

ΣV = 0, ΣH = R

ΣV = 200sin10° + Psin50° + 500sin150° = 0

On solving,      ΣP = -371.68N                                                                                                                      ...(i)

ΣH = 200cos10° + Pcos50° + 500cos150° = 0

By putting value of 'P', we get

ΣH = -474.96N                                                                                                                     ...(ii)

Let Unknown force be 'Q' and subtends an angle of ? With horizontal x axis. Additional force makes system in equilibrium now,

ΣH = Qcos   -474.96N = 0 that is,          Qcosθ   = 474.96N------(3)

As ΣV is already zero, on addition of force Q, the body is in equilibrium so again            V is zero.

ΣV = 200sin10° -371.68sin50° + 500sin150° + Qsinθ = 0

But 200sin10° - 371.68sin500 + 500sin1500 = 0 by equation (1) So,    Qsinθ = 0, that means Q = 0or sinθ   = 0,

ΣQ is not zero so sinθ = 0,  Q= 0

By putting  = 0 in equation (iii),

Q = 474.96N,     = 0º                                  .......ANS