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Teach A Level Maths Bodies in Equilibrium

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**Volume 4: Mechanics 1 Bodies in Equilibrium**

"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

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**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.

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**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: T The first step is to show the forces acting on the particle. P 2

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**across the angle means cos**

Equilibrium. Find T and P. Resolving: 40 40 Tcos 40 - 2 = 0 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. Tcos 40 = 2 T P 2 cos 40 T = = 2·61 ( 3 s.f. ) across the angle means cos 2 The tension is 2·61 newtons. P - Tsin 40 = 0 P = Tsin 40 P = 2·61sin 40 P = 1·68 newtons ( 3 s.f. )

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**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.

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35 R T Equilibrium Smooth plane Weight 10 newtons. 10 Find T and R. Solution: Agree with your partner which directions you would use to resolve the forces. 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.

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

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

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**10 T R 10 T - 10sin 35 = 0 T = 10sin 35 = 5·74 newtons ( 3 s.f. )**

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

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**Component of W down the slope :**

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: a W Component of W down the slope : W sin a

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

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**P 14 R Pcos 25 - 14 = 0 14 P = 10 P = 15·4 ( 3 s.f. ) R - 10**

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: P 14 25 R Resolving: Pcos 25 - 14 = 0 14 cos 25 P = 10 P = 15·4 ( 3 s.f. ) R - 10 - Psin 25 = 0 R = ·4sin 25 = 16·5 newtons ( 3 s.f. )

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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 W sin a where a is the angle of the slope.

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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. 42 2 P Find the value of P and the magnitude of the normal reaction.

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**42 2 P R P - 2sin 42 = 0 P = 2sin 42 = 1·34 ( 3 s.f. ) R**

EXERCISE Solution: 42 2 P R Resolving: P - 2sin 42 = 0 P = 2sin 42 = 1·34 ( 3 s.f. ) R - 2cos 42 = 0 R = 2cos 42 = 1·49 newtons ( 3 s.f. )

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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. 60 30 A C B If the tension in BC is 1 newton, find the value of W and the tension in AC.

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**The weight is 2 newtons and the tension in AC is 1·73 newtons.**

EXERCISE Solution: 60 30 A B C Method 1: Resolving horizontally and vertically T 1 cos 30 - T cos 60 = 0 1 -T cos 60 = -cos 30 cos 30 cos 60 T = W T = 1·73 ( 3 s.f. ) 1 sin 30 + T sin 60 - W = 0 1 sin 30 + 1·73 sin 60 = W W = 2 The weight is 2 newtons and the tension in AC is 1·73 newtons.

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**The weight is 2 newtons and the tension in AC is 1·73 newtons.**

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

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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 a , where a is the angle of the slope.

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