Plan for Today: Dry Friction Belt Friction Friction ▪ Depends on the characteristics of the contacting surfaces. ▪ Acts in a direction that opposes motion. ▪ In Statics we further specify that friction opposes impending motion ▪ Acts parallel to the contacting surfaces. CHARACTERISTICS OF DRY FRICTION Experiments show that the friction force F is directly proportional to force (F = P) until a certain point at which slip occurs. That point is the point of maximum static fraction and is given as: Fs = s N where s is the coefficient of static friction and N is the normal force. After the point of maximum static friction is surpassed, motion occurs and friction force is given as F = k N where k is the coefficient of kinetic friction. DETERMING s EXPERIMENTALLY A block with weight W is placed on an inclined plane. The plane is slowly tilted until the block just begins to slip. The inclination, s, is noted. Analysis of the block just before it begins to move gives (using Fs = s N): + Fy = N – W cos s = 0 + FX = S N – W sin s = 0 Using these two equations, we get s = (W sin s ) / (W cos s ) = tan s This simple experiment allows us to find the S between two materials in contact. Will the body tip or slip? At rest (A), the normal force is uniformly distributed. As a pushing force is applied (B), the distributed normal force is redistributed, moving the equivalent point load to the right. If pushing force becomes large enough (C), the moment couple exerted by gravity and normal force will be unable to counter the couple exerted by the pushing force and friction force If the moment couple formed by gravity and the normal force is less than the moment couple formed by the friction force and pushing force, tipping will occur. In this, class we will only be dealing with problems for which tipping does not occur. Decision Process for Solving Dry Friction Problems (Is friction µN?) PROBLEMS INVOLVING DRY FRICTION Steps for solving equilibrium problems involving dry friction: 1. Draw necessary free body diagrams. Make sure that you show the friction force in the correct direction (it always opposes the motion or impending motion). 2. Determine the number of unknowns. Do not assume that F = Fs = S N unless the motion is impending. 3. Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns. Note: If you are not told that motion is impending, solve for the friction force with equilibrium equations and then check to see if it is greater than Fs. If it is, the body is slipping and F is actually k N . If it isn’t, then F is just equal to the F that you found. BELT FRICTION When there is no friction, T1 = T2 as we have previously discussed. However, now we need to consider what happens when friction is of concern. Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to radians. If the belt slips or is just about to slip, then T2 must be larger than T1 and the direction of motion resisting friction forces. Hence, T2 must be greater than T1. Detailed analysis (see textbook) shows that T2 = T1 e where is the coefficient of static friction between the belt and the surface. Be sure to use radians when using this formula!! Example The coefficient of static friction between the 50-lb box and the inclined surface is 0.10. The coefficient of static friction between the rope and the fixed cylinder is 0.05. Determine the force the woman must exert on the rope to cause the box to start moving up the inclined surface. Example: The maximum tension that can be developed in the cord is 500 N. If the pulley at A is free to rotate and the coefficient of static friction at the fixed drums B and C is 0.25, determine the largest mass of the cylinder that can be lifted by the cord. Start by looking at drum B: When trying to determine wrap angle, one helpful fact to remember is that the angle between the line tangent to the point where the rope contacts the drum and radial line is a right angle. Example: A 180-lb farmer tries to restrain the cow from escaping by wrapping the rope two turns around the tree trunk as shown. If the cow exerts a force of 250 lb on the rope, determine if the farmer can successfully restrain the cow. The coefficient of static friction between the rope and the tree trunk is µs = 0.15, and between the farmer’s shoes and the ground is µs = 0.3.