Unit Dynamics Module The Forces of Friction page 1 of 2 Problems on Friction and Inclines A block on an incline with friction remains motionless until the incline reaches a critical angle Beyond this angle the block accelerates down the ramp The magnitude of static friction depends on the magnitude of other forces As the angle of the incline increases so does fs The static frictional force of an object on an incline is given by the formula s tan The acceleration of a block down an incline is a k g cos g sin On a frictionless incline a block will slide down with an acceleration of g sin m s2 If there is friction between the block and the incline the block will remain motionless until the incline reaches a critical angle At angles greater than the critical angle the block will slide down the ramp Consider a stationary block on an incline that is tilted at the critical angle You would like to understand the relationship between the critical angle and the coefficient of static friction Use Newton s second law You have already focused on one object the block Now consider all the forces on the block The weight of the block is downward The normal force is perpendicular to the surface of the incline The force of static friction is parallel to the incline in the direction away from the motion of the block Since you are considering acceleration down the incline choose a coordinate system so that the x axis is parallel to the incline and the y axis is perpendicular to the incline First focus on the y coordinates The forces are the normal force in the positive y direction and the ycomponent of weight in the negative y direction The y component of weight is mg cos There is no acceleration in the y direction so the right hand side of Newton s law is zero www thinkwell com info thinkwell com Copyright 2001 Thinkwell Corp All Rights Reserved 1885 doc rev 03 29 2001 Unit Dynamics Module The Forces of Friction page 2 of 2 Problems on Friction and Inclines Now focus on the x components The forces are the force of static friction in the negative y direction and the x component of weight in the positive ydirection The y component of the weight is mg sin Since the block is stationary the right hand side of Newton s second law is zero Since the angle of the incline is the critical angle the force of static friction is maximal You can rewrite Newton s second law in the y direction as fs max mg sin Recall that the force of static friction is maximal when it is equal to the product of the coefficient of static friction and the normal force fs max s N If you set the two equations for fs max equal to each other you can solve for the coefficient of static friction in terms of the critical angle Now consider the acceleration of the block when the angle of the incline increases beyond the critical angle The acceleration is in the positive x direction so consider the x components The force of kinetic friction is in the negative x direction and the xcomponent of weight is in the positive x direction Recall that the force of kinetic friction is the product of the coefficient of kinetic friction k and the normal force Since there is no acceleration in the y direction the normal force is equal in magnitude and opposite in direction to the y component of weight N Wy mg cos In this example there is acceleration so the right hand side of Newton s second law is max Solve for acceleration of the block ax k g cos g sin www thinkwell com info thinkwell com Copyright 2001 Thinkwell Corp All Rights Reserved 1885 doc rev 03 29 2001