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## Lesson 3. SOLVING NUMERICALS

Module 1. Statics and Dynamics Lesson 3 SOLVING NUMERICALS
Adding equations (1) and (2), we get (P+Q) P+Q = 4 → (3) Similarly, (P-Q) P-Q = 2 → (4) (OR) Solving the equations 3 & 4 P = 3 N Q = 1N
Two forces act at an angle of 120
Sol. Bigger force P = 40 N Angle between the resultant and the Smaller force = 90 Angle between the two forces θ = 120 Angle between the bigger force and the resultant, Α=120-90=30 Let Q = Smaller force
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.
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 14
Sol. Weight W = 50 N Inclination of force θ = 14 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 14 Now resolving the force along the plane F = 18 Cos 14 We, know that F = µ R
A force of 40 N pulls a component of weight 60 N up an inclined of the plane, to the horizontal, is 30
Sol. Given weight, W = 60 N Force P = 40 N Inclination θ = 30 Let R = Normal reaction µ = co-efficient of friction F = Force of friction Resolving forces along the inclined plane F = 40-60 Sin 30 Resolving force at right angles to the plane R = 60 Cos 30 |