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Teach A Level Maths Bodies in Equilibrium

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"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" © Christine Crisp Volume 4: Mechanics 1 Bodies in Equilibrium

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Since the method of resolving forces can be applied to any of these problems, well 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.

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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

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P 2 T Equilibrium. Find T and P. 40 T cos 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

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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.

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35 Weight 10 newtons. Solution: 10 T R Equilibrium Smooth plane The plane is smooth so there is no friction. Agree with your partner which directions you would use to resolve the forces. 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 wont have to solve simultaneous equations. Find T and R.

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

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

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

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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

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P 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.

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P 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 P cos 25 P 15·4 ( 3 s.f. ) R R 16·5 newtons ( 3 s.f. ) P sin ·4 sin R

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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

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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 P Find the value of P and the magnitude of the normal reaction.

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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 EXERCISE

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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 A C B EXERCISE

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60 30 A B C Solution: W T 1 Method 1: Resolving horizontally and vertically 1 cos 30 0 T cos 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

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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

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The following page contains the summary in a form suitable for photocopying.

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Summary BODIES IN EQUILIBRIUM TEACH A LEVEL MATHS – MECHANICS 1 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.

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