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Resolving Forces & Newton’s Second Law

Resolving Forces & Newton’s Second Law. F 2. F 1. Forces are vector quantities. Using vector addition we see that forces can be split into 2 perpendicular forces. j. F. . i. ‘i’ direction (Horizontal). ‘j’ direction (Vertical). cos  = adj/hyp = F 1 / F.

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Resolving Forces & Newton’s Second Law

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  1. Resolving Forces&Newton’s Second Law

  2. F2 F1 Forces are vector quantities. Using vector addition we see that forces can be split into 2 perpendicular forces j F  i ‘i’ direction (Horizontal) ‘j’ direction (Vertical) cos  = adj/hyp = F1 / F sin  = opp/hyp = F2 / F F1 = F cos  F2 = F sin  This process is known as “Resolving Forces”

  3. Resolving Forces and N2L 0.5 ms-2 R 30 N 25o resistance F 40kg N A sled (40kg) being towed by a rope (tension 30N) at 25o, it accelerates at 0.5 ms-2. Find the resistance (F) and the normal reaction. N2L : F = ma weight = 40kg = 40 x 9.8 = 392 N 30 N Draw a Force diagram The resultant horizontal force provides the acceleration N2L : F = ma Resultant = 40 x 0.5 = 20N Resolve horizontally Resolve vertically Resultant = 30 cos 25 - F = 20N = 27.2 - F = 20 F = 27.2 - 20 = 7.2N R + 30 sin 25 = 392 R = 392 - 12.7 = 379.3 N

  4. Resolving Forces and N2L a ms-2 R 50 N 25o 20 N 50g N A sled (50kg) being towed by a rope (tension 50N) at 25o, has resistive force of 20N. Find the acceleration N2L : F = ma weight = 50g = 50 x 9.8 = 490 N 50 N Draw a Force diagram Resolve horizontally Resultant = 50 cos 25 - 20 = 45.3 - 20 = 25.3 N N2L : F = ma Resultant = 50 x a 25.3 = 50a a = 25.3 / 50 = 0.51 ms-2

  5. Newton’s Laws of Motion- and slopes

  6. A Force Question - Reminder Remember the Normal reaction is always perpendicular to the slope rope R Tension T 30o In questions with slopes, it always works out easiest to resolve forces parallel and perpendicular to the slope. 30o 60o 30o 20 N weight Parallel to slope Perpendicular to slope R = 20 cos 30 = 17.3 N T = 20 sin 30 = 10 N

  7. A slope and N2L problem (1) First: Draw a Force diagram An object of mass 2.8kg is pulled up a smooth slope inclined at 40o by a light string parallel to the slope. The object is accelerating at 0.5 ms-2. Find : the tension in the string and the normal reaction 0.5 ms-2 R 2.8 kg T 0.5 ms-2 40o N2L : F = ma weight = 2.8g = 2.8 x 9.8 = 27.44 N 40o 27.44 N

  8. A slope and N2L problem (2) Force diagram An object of mass 2.8kg is pulled up a smooth slope inclined at 40o by a light string parallel to the slope. The object is accelerating at 0.5 ms-2. Find : the tension in the string and the normal reaction Resolve Parallel to slope A resultant force must exist to make the object accelerate R T 0.5 ms-2 Resultant force = T- 27.44 cos 50 = (T - 17.64) N 50o N2L : F = ma Resultant = 2.8 x 0.5 = 1.4 N 40o 27.44 N So, T - 17.64 = 1.4 T = 1.4 + 17.64 = 19.0 N

  9. A slope and N2L problem (3) Force diagram An object of mass 2.8kg is pulled up a smooth slope inclined at 40o by a light string parallel to the slope. The object is accelerating at 0.5 ms-2. Find : the tension in the string and the normal reaction Resolve Perpendicular to slope No resultant force - acceleration is parallel to slope R 19 N 0.5 ms-2 R = 27.44 sin 50 = 21.0 N 50o 40o 27.44 N

  10. KEY POINT FROM THIS LESSON If an object is accelerating in a force question then you need to used N2L to work out the resultant force providing the acceleration

  11. Activity • Mechanics 1 book • Exercise G [page 97] • Q 1 , 2, 3

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