## Lesson 3. SOLVING NUMERICALS

 Module 1. Statics and Dynamics Lesson 3SOLVING NUMERICALS 3.1 Problem (OR) Adding equations (1) and (2), we get (P+Q) 2 = P2 + Q2 + 2PQ = 10+6 = 16 P+Q = 4 → (3) Similarly, (P-Q) 2 = P2 + Q2 - 2PQ = 10 - 6 = 4 P-Q = 2 → (4) (OR) Solving the equations 3 & 4 P = 3 N Q = 1N 3.2 Problem Two forces act at an angle of 1200. The bigger force is of 40 N and the resultant is perpendicular to the smaller one. Find the smaller force. Fig. 3.3 Sol. Bigger force P = 40 N Angle between the resultant and the Smaller force = 900 Angle between the two forces θ = 1200 Angle between the bigger force and the resultant, Α=120-90=300 Let Q = Smaller force 3.3 Problem A body of mass100 N is suspended by two strings of 4 m & 3m lengths attached at the same horizontal level of 5 m a part. Find the tensions in strings. Fig. 3.4 3.4 Problem A component of weight of 50 N is hauled along a rough horizontal plane, by a pull of 18 N acting at an angle of 140 with the horizontal. Find co-efficient of friction. Fig. 3.5 Sol. Weight W = 50 N Inclination of force θ = 140 Force P = 18 N Let R = Normal reaction µ = co-efficient of friction F = Force of friction Resolving the forces at right angles to the plane R= 50-18 Sin 140 = 45.65 N → (1) Now resolving the force along the plane F = 18 Cos 140 = 17. 46 N → (2) We, know that F = µ R 3.5 Problem A force of 40 N pulls a component of weight 60 N up an inclined of the plane, to the horizontal, is 300, calculate the co-efficient of friction. Fig. 3.6 Sol. Given weight, W = 60 N Force P = 40 N Inclination θ = 300 Let R = Normal reaction µ = co-efficient of friction F = Force of friction Resolving forces along the inclined plane F = 40-60 Sin 300 = 10 N → (1) Resolving force at right angles to the plane R = 60 Cos 300 = 51.26 N → (2)