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Unformatted text preview: COLLEGE PHYSICS, Part I Chapter 4 (Second Part): Applications of Newton’s Laws Equilibrium of a Particle Applications of Newton’s Laws for Nonequilibrium Condiitons Contact Forces and Friction Gravitational Forces Forces of Nature This chapter teaches to solve problems that include relationships between forces and motion. Equilibrium of a Particle An object is in equilibrium when it is at rest or moves with constant velocity. According to Newton’s first law, the vector sum of all forces acting on an object in equilibrium is zero, and therefore, according to Newton’s second law, the acceleration is zero. When R = Σ F = , the object is in equilibrium . This definition is sufficient only if the object can be treated as a particle . Σ F x = , Σ F y = (the object is in equilibrium) SET UP: 1. Draw a sketch showing all dimensions and angles 2. Identify the object or objects in equilibrium that you will consider 3. Draw a freebody diagram for each object a) present the object as a particle and avoid including other objects, such as the surface the object may be resting on or the rope pulling on it. b) Identify and draw all force vectors acting on the object (do not include forces exerted by the object ) c) Label each force (e.g., w for weight, n for the normal force, T for a tension force, etc.) 1. Choose a coordinate system and represent each force by its components along the x and y axes. At your choice, the coordinate system may by tilted, for example in problems where the object is sliding on a tilted surface. Problem solving strategy: Objects in equilibrium SOLVE: 1. Apply the principle of Σ F x =0 and Σ F y = 0 , 2. If two or more objects interact with each other, use Newton’s third law to relate the forces they exert on each other. 3. Identify at least as many independent equations as the number of unknown parameters in the problem, and use them to find the unknowns. Example: A gymnast has just begun climbing up a rope hanging from a gymnasium ceiling. Her weight is 500 N and the weight of the rope is 100 N. Analyze the forces on the gymnast and on the rope. The figure shows: a) A diagram of the whole problem b) Forces acting on the gymnast c) Forces acting on the rope Note that we chose the y axis to be vertical (upward) and the x axis horizontal. There are no x components of force. Part b) The gymnast is motionless, so the sum of the forces acting on her is zero ( Σ F y = 0). 500 N + T 1 = 0, so T 1 = 500 N (These two forces are not an actionreaction pair) Part c) At the top of the rope, the gymnast’s weight and the ropes own weight are equilibrated by an upward force T 2 : T 2 – 500 N – 100 N = 0, so T 2 = 600 N Example: Twodimensional equilibrium A car engine with weight w hangs from a chain that is linked at point O to two other chains, one fastened to the ceiling at a 60 o angle and horizontally to the wall. Find the tension in each of the chains, assuming w is known and weights of the chains can be neglected. From this diagram (b), we immediately find that T...
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This note was uploaded on 09/21/2011 for the course PHY 2053 taught by Professor Vellaris during the Spring '08 term at University of Central Florida.
 Spring '08
 Vellaris
 Physics, Force, Friction

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