Equilibrium of rigid bodies in contact
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Equilibrium of Rigid Bodies in Contact
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This subject is a further extension of that branch of mechanics that is called statics and is concerned with the equilibrium of static objects - objects that are not moving. he theory that is assumed, apart from a general understanding of algebra, is: 1. The process of resolving forces into components - often horizontal and vertical components - and the principal that the sum of forces that are in static equilibrium is zero. 2. For a body in static equilibrium the sum of all moments about any point or axis is zero. 3. Basic knowledge about centres of mass. There is no new theory involved. This topic is concerned with an extension of the given theory of statics to more complicated examples. 4. Newton's Third Law: every action has an equal and opposite reaction. 5. Knowledge of the coefficient of friction and understanding of toppling and sliding. What is new is that success in problem-solving in this area requires (1) the development of a certain way of viewing objects - the ability to see objects in part to whole relationships - and (2) the development of physical intuition and understanding of forces. Success in this topic requires a clear grasp of existing theory and visual and physical insight. To illustrate these ideas we consider an asymmetrical step-ladder. [Diagram goes here - download the original to see it.] The ladder comprises two metal pieces X and Y, hinged together and standing on a smooth floor, held together by a rope, Z. Y is shorter than X. Since the floor is smooth there are no frictional forces at the floor, consequently there are only the normal reactions to consider at the floor. In this illustration we shall also suppose that both vertical pieces, X and Y, have the same weight W. The spatial insight required is illustrated as follows. The ladder can be viewed either as a WHOLE or as three separate PARTS. The whole is shown above, but as parts, the object is a composite of: [Diagram goes here - download the original to see it.] When we look at the object as a whole we consider only the external forces acting on it as a whole. These are just the weight of each and the reactions at the floor. Because the rope is "light" we shall assume it has no weight. Since the floor is "smooth" there are no frictional forces on the floor. Thus, the forces acting on the ladder as a whole are: [Diagram goes here - download the original to see it.] We ignore all the internal forces such as the tension in the rope - these all cancel out when the object is viewed as a whole. However, a force that is internal to an object viewed as a whole may be external to PART of the object. The forces acting on each point of the ladder are: [Diagram goes here - download the original to see it.] The metal piece X is being acted upon by a tension in the rope pulling in the direction of Y. At the hinge the piece Y is pushing against X. This contact force has been resolved into two parts - a horizontal component and a vertical component. Physical intuition is required to see which way the forces are acting. Piece Y is "pushing" piece X upwards. To see why observe that since Y is shorter than X it will be bearing none of the common weight, hence, it must be pulling X upwards. The contact forces acting on X due to Y must be equal in size to the contact forces acting on Y due to X. This is because the object as a whole is not moving, and is an application of Newton's Third Law. That law also tells us about the direction of the forces - the contact forces between two objects must be equal in size and opposite in direction. [Diagram goes here - download the original to see it.] The contact forces of Y on X must be equal in size and opposite in direction to the contact forces of X on Y. We will now develop this into a full example. Example (1) [Diagram goes here - download the original to see it.] A step-ladder is made of two metal pieces, AB and BC. The length of AB is 3.9m and that of BC is 3.75m, when open the height of the ladder is 3.6m above the ground. The metal pieces are modelled as uniform rods and they are smoothly hinged at B. The floor is smooth and the distance AC = 2.55m. The two metal parts are joined by a light horizontal rope of length 1.7m. The weight of AB is twice that BC. If the tension in the rope is 12.25N find W. Firstly, we mark on a diagram the forces acting on the ladder as a whole: [Diagram goes here - download the original to see it.] These are the weights of the two sections acting on their centres of mass situated half way up the ladder. Resolution of the problem will depend on the geometry of the ladder, and hence a separate diagram for this is needed. [Diagram goes here - download the original to see it.] Since we will be taking moments experience tells us that we will need to know the perpendicular distances DE, EF, BE, EG, AG, GC so we proceed to calculate those using ratios and Pythagoras's theorem. If we know the perpendicular distances we will not, in fact, need to know the angles a and b. It is easier to work with perpendicular distances than with angles, so we use these wherever possible. [Diagram goes here - download the original to see it.] [Equation goes here - download the original to see it.] [Diagram goes here - download the original to see it.] [Equation goes here - download the original to see it.] [Diagram goes here - download the original to see it.] The completed diagram is [Diagram goes here - download the original to see it.] Recall that the external forces on the ladder are: [Diagram goes here - download the original to see it.] From this, by resolving vertically, we obtain the equation: [Equation goes here - download the original to see it.] [Example continues - download the original to see it.]
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Contents of
Equilibrium of rigid bodies in contact
1 Equilibrium of Rigid Bodies in Contact
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