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A Level Mechanics

A Level Mechanics. Bodies in Equilibrium. Bodies in Equilibrium. We have used 2 methods for solving problems with 3 forces in equilibrium:. using a triangle of forces,. finding the components of the forces by resolving. However, if

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A Level Mechanics

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  1. A Level Mechanics Bodies in Equilibrium

  2. Bodies in Equilibrium We have used 2 methods for solving problems with 3 forces in equilibrium: • using a triangle of forces, • finding the components of the forces by resolving. However, if we have more than 3 forces, or the forces are not in equilibrium, we cannot use a triangle of forces, so we then resolve the forces. Since the method of resolving forces can be applied to any of these problems, we’ll use it in the following examples.

  3. e.g.1. The diagram shows a particle of weight of 2 newtons that is tied to a light inextensible string attached to a wall. The particle is held in equilibrium, as shown, by a horizontal force of magnitude P newtons. Find the tension in the string and the value of P. 40 Solution: The first step is to show the forces acting on the particle. T P 2

  4. Find T and P. Equilibrium. 2 across the angle means cos cos40  P = Tsin40 P =2·61sin40   P=1·68 newtons ( 3 s.f. ) Resolving: 40 40 = 0 Tcos40 - 2 Decide with your partner what the component of T is, without drawing a separate diagram. We should never miss out this stage as doing so leads to errors in later problems.  Tcos40 = 2 T P T= = 2·61 ( 3 s.f. ) 2 The tension is 2·61newtons. P - Tsin40 = 0

  5. 35 e.g.2. A particle of weight 10 newtons rests on a smooth plane inclined at 35 to the horizontal. The particle is supported, in equilibrium, by a light inextensible string parallel to the slope. Find the magnitudes of the tension in the string and the contact force between the particle and plane.

  6. Equilibrium Smooth plane 35 R T Weight 10 newtons. 10 Find Tand R. Solution: The plane is smooth so there is no friction. Ans: There are 2 reasons for preferring parallel and perpendicular to the plane: • We will only need to resolve 1 force, the weight. • T will appear in one equation and R in the other, so we won’t have to solve simultaneous equations.

  7. R T 35 10 35 35 55 To find the components of the weight, we need an angle. the 3rd angle is Using this right angled triangle . . . 90 - 35 =55 and using the right angle between the slope and the perpendicular, 90 - 55 =35

  8. R T 55 35 10 35 35 We can just use 35 ( the angle of the slope ) without needing to subtract.

  9. R T 10 35 Find Tand R. Solution: 10cos35 35 35 10 10sin35 Resolving: T - 10sin35 =0 T =10sin35  =5·74newtons ( 3 s.f. ) - 10cos35 =0 R  R =10cos35 =8·19newtons ( 3 s.f. )

  10. W a Tip: Lots of problems you will meet in M1 involve objects on slopes so it is well worth remembering the component of the weight down the slope: Component of W down the slope: Wsina

  11. P 14 25 e.g.3 A box of weight 10 newtons is being pushed at a constant speed in a straight line across a horizontal surface by a force of magnitude P newtons at 25 to the surface. There is a constant resisting force of magnitude 14 newtons. Find P and the magnitude of the normal reaction. Solution: R Constant velocity  equilibrium Can you see what the horizontal component of the pushing force is, without drawing a separate diagram? 10 Pcos25 Ans:

  12. P 14 25 14 cos25 e.g.3 A box of weight 10 newtons is being pushed at a constant speed in a straight line across a horizontal surface by a force of magnitude P newtons at 25 to the surface. There is a constant resisting force of magnitude 14 newtons. Find P and the magnitude of the normal reaction. Solution: R Resolving: Pcos25 - 14 = 0  P= 10 P = 15·4 ( 3 s.f. ) -Psin25 R - 10 = 0 R = 10 + 15·4sin25  = 16·5 newtons ( 3 s.f. )

  13. SUMMARY • A body in equilibrium is either at rest or moving with a constant velocity. • To solve equilibrium problems we resolve the forces and form equation(s) using X= 0 and/or Y= 0. • For bodies on a slope we usually resolve parallel and perpendicular to the slope. • The component of the weight, W, down a slope is Wsina where a is the angle of the slope.

  14. P 2 42 EXERCISE 1. A particle of weight 2newtons rests on a smooth plane inclined at 42 to the horizontal. It is supported by a force of magnitude Pnewtons acting parallel to the slope. Find the value of P and the magnitude of the normal reaction.

  15. P 2 42 EXERCISE Solution: R Resolving: P - 2sin42 =0  P =2sin42 =1·34 ( 3 s.f. ) - 2cos42 =0 R  R =2cos42 =1·49newtons ( 3 s.f. )

  16. A B 60 30 C EXERCISE • A particle of weight • W is held in equilibrium by two inextensible strings AC and BC at 60 and 30 to the horizontal as shown in the diagram. If the tension in BC is 1 newton, find the value of W and the tension in AC.

  17. A B cos30 cos60 60 30 T = 1·73 ( 3 s.f. )  C + 1·73sin60 1sin30  W = 2 EXERCISE Solution: Method 1: Resolving horizontally and vertically 1cos30 = 0 - Tcos60 T 1  -Tcos60 = -cos30 T =  W = 0 + Tsin60 1sin30 - W = W  The weight is 2 newtons and the tension in AC is 1·73newtons.

  18. A B T 1 1 cos60 60 30 C W  W = 2  W = 1·73 ( 3 s.f. ) EXERCISE Solution: Method 2: Resolving parallel and perpendicular to CB = 0 1 - Wcos60 = Wcos60  1 = W  60 - Wsin60 = 0 T  T= 2sin60 The weight is 2 newtons and the tension in AC is 1·73newtons.

  19. BODIES IN EQUILIBRIUM Summary • A body in equilibrium is either at rest or moving with a constant velocity. • To solve equilibrium problems we resolve the forces and form equation(s) using X= 0 and/or Y= 0. • For bodies on a slope we usually resolve parallel and perpendicular to the slope. • The component of the weight, W, down a slope is Wsin a, where a is the angle of the slope.

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