Question

3 forces p,q,r acting along a particle in a plane the angle between p and q is 150 and q and r is 120 then the equilibrium forces p,q,r are in ratio

Answer 1

Answer:

At equilibrium, the forces P, Q, R are in the ratio √3 : 2: 1

Explanation:

Let the three forces P, Q, R act at point O in the plane, such ∠POQ = 150° and ∠QOR = 120°.

Also, the vector OP is along X axis, OQ along Y axis and OR along Z axis.

We know that,  

∠POQ +  ∠QOR   + ∠ROP  = 360°

==> ∠ROP = 360 - (150+120) = 90°

Since Vector OP and OR are perpendicular, we can represent OR as PS, such that OR is parallel to PS.

So, OPS forms a triangle and according to the law of triangle, OS = Force P + Force R.

At equilibrium, OS should be equal and opposite to the force Q, so that the particle on the plane at O is at equilibrium. ( QOS is a straight line)

Now, ∠QOS = ∠QOP + ∠POS = 180°

∠POS = 180° - 150° = 30°

Now, Sine rule states that, in a triangle, side a/ Sin A = side b /  Sin B = side c / Sin C

By applying Sine rule for the triangle OPS, ( Forces are P , Q and R  represent a triangle and opposite  angles are 60°, 90° and 30°)

| P | / Sin 60°  =  | Q | / Sin 90°  =  | R | / Sin 30°

| P | / (√3/2) =  | Q | / 1  =  | R | / (1/2)

| P | / (√3) =  | Q | / 2  =  | R | / 1

Ratio is √3 : 2: 1