1. A Level Mechanics Bodies in Equilibrium

2. Since the method of resolving forces can be applied to any of these problems, we’ll use it in the following examples.  using a triangle of forces, We have used 2 methods for solving problems with 3 forces in equilibrium: 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.  finding the components of the forces by resolving. Bodies in Equilibrium

3. 40  P 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. 2 Find the tension in the string and the value of P. Solution: The first step is to show the forces acting on the particle. T

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

5. 35  Find the magnitudes of the tension in the string and the contact force between the particle and plane. 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.

6. 35  Weight 10 newtons. Solution: 10 T R Equilibrium Smooth plane 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. Find T and R.

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

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

9. 35  10 T R  10 cos 35  R  R  10 cos 35  T  8·19 newtons ( 3 s.f. ) Resolving: 10 35  10sin 35  10cos 35   0 0 Solution: Find T and R.  5·74 newtons ( 3 s.f. )  T  10 sin 35   10 sin 35   0 0 35 

10. 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:  W Component of W down the slope : W sin 

11. P 14 25  10 R Solution: Can you see what the horizontal component of the pushing force is, without drawing a separate diagram ? Ans: P cos 25  Constant velocity  equilibrium 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.

12. 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: Resolving: P cos 25   0   14 P  14 cos 25  P  15·4 ( 3 s.f. ) R  R   16·5 newtons ( 3 s.f. )  P sin 25   10  0 10  15·4 sin 25  10 R

13.  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 W sin  where  is the angle of the slope. SUMMARY

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

15. 42  2 P R Solution:  2 cos 42  R  R  2 cos 42  P  1·49 newtons ( 3 s.f. )  P  2 sin 42   1·34 ( 3 s.f. ) Resolving:  2 sin 42   0 0  0 0 EXERCISE

16. 2.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. 60  30  A C B EXERCISE

17. 60  30  A B C Solution: W T 1 Method 1: Resolving horizontally and vertically 1 cos 30   0   T cos 60   0 1 sin 30   T sin 60   W  cos 30  T cos 60   T  cos 30  cos 60   T  1·73 ( 3 s.f. ) 1 sin 30   1·73 sin 60   W  2  W  The weight is 2 newtons and the tension in AC is 1·73 newtons. EXERCISE

18. 60  30  A B C W T 1 Solution: 1  0   W cos 60   0 T  W sin 60  T  2 sin 60   W cos 60  1   W  W  2  W  1·73 ( 3 s.f. )  The weight is 2 newtons and the tension in AC is 1·73 newtons. EXERCISE Method 2: Resolving parallel and perpendicular to CB 60  1 cos 60 

19. Summary BODIES IN EQUILIBRIUM  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 W sin  , where  is the angle of the slope.