Maths M1-Help Needed!!

[ashutoshamatya]

9. A particle P rests in equilibrium on a smooth horizontal surface under the action of the four horizontal forces shown in the diagram. The angles between the forces of magnitudes 1.2 N and 2 N, and between the forces of magnitudes X N and Y N is, are each 90 degrees. The angle between the forces of magnitudes 2 N and Y N is 130 degrees. Find the values of X and Y.

The force of magnitude 1.2 newton is now removed. Given that the mass of P is 0.4 kg, find the magnitude of the acceleration with which P starts to move, and show clearly on the diagram the direction of this acceleration.

 

[usman]

I guess Force Resolution wd b the most appropriate here:
R(in vertical direction):- Y = 2*cos(180-130) + 1.2*cos(180-90-50) = 2cos(50) + 1.2cos(40)
R(in horizontal direction):- X = 2*sin(180-130) - 1.2*sin(180-90-50) = 2sin(50) - 1.2sin(40)

We know that in a system of coplanar forces that are in equilibrium, the magnitude of the resultant force in any direction from the origin is zero. Using this fact, we can deduce that when the 1.2N force is removed, there will b a resultant force of equal magnitude (1.2N) in the opposite direction (to the 1.2N force). Then, applying Newton's 2nd Law, we can say that there will b an acceleration in this very direction (of resultant force), with the magnitude force/mass i.e.
a = 1.2/0.4 = 3ms^-2