 Ramps - Study Guides Practice Problems are at the BOTTOM of this page MY VIDEO Introduction to Ramp Problems (Friction is ABSENT) is HERE MY VIDEO Introduction to Ramp Problems (Friction is PRESENT) is HERE (re-uploaded at 3:00 PM on Sunday, ScreenCastify is "Processing") The basis for ramp problems is the same for ALL of our forces problems: If an object experiences unbalanced forces in the "x" or "horizontal" direction, it will accelerate in the direction of the unbalanced force. If an object experiences unbalanced forces in the "y" or " vertical" direction, it will accelerate in the direction of the unbalanced force. If an object is at rest in the "x" or "horizontal" direction the forces in that direction MUST be balanced or absent. If an object is at rest in the "y" or "vertical" direction the forces in that direction MUST be balanced or absent. THE VERY BIG difference in RAMP problems is that don't really have an easily defined horizontal (x) or vertical (y). I recomend that we change our perspective: Our two dimensions are: down the ramp (dtr). normal to the ramp (N) Also, there is ONE and ONLY one force that is responsible for every aspect of our conversations of ramps.... weight (mg) As always, we mathematically we express those situations as: ∑Fdtr = madtr Which is very helpful. Keeping with convention we'll define down the ramp to be negative and up the ramp to be positive. Any object sliding down a ramp is NOT being launched into space by the normal force of the ramp. Similarly, any object sliding down a ramp is NOT crashing through the ramp. Therefore the sum of the forces in the Normal (perpindicular to the ramp) direction are balanced, and we only really care about those when we calculate friction!   TO WIT: RAMP PROBLEMS REVIEW ► As noted above, the weight of an object posed to slide down a ramp is the entire source of all the forces present that we'll discuss when working on ramp problems. The force pulling the object down the ramp in the "dtr" dimension is NOT simply the weight of the force. It is the weight of the force in the "dtr" direction: mgsinΘ The frictional force is generated in preportion to the Normal force of the weight (mg) of the object. The deciding factor here is the frictional coefficient μ: f = μN Since the Normal force itself is dependent on the weight (mg) of the object, we can express friction in terms of the weight of the object by substituting for N as follows: f = μmgcosΘ ═══════════════════════════ Now that is all well and good... IF you know how to get the dtr and N components of the objects weight (mg). The first step is to sketch and label the object on the ramp with the weight (mg) vector pointing stright down to the center of the Earth. Next, sketch the Normal force of the ramp pushing up on the box perpindicular to the surface of the ramp The Normal force is resisted by the (dtr) component of the the weight (mg) of the object, so sketch that next. Label that mgcosΘ The last step is connect your two vectors with a vector that is paralell to the direction of the ramp itself. That is the component of the force pulling the object down the ramp. Label that mgsinΘ ► If Fdtr (the force down the ramp) is greater than f (friction) then the object will slide. If they are equal, it won't. Remember, friction ALWAYS works against motion. Also, the frictional force can NEVER be larger than Fdtr otherwise an object could be pushed up the ramp by friction, which is clearly impossible! Practice Problems: ► Kahn Academy has some helpful quizzes and worked problems HERE ► Check out problems #22, #23 & #24 on the physics classroom website (I'd recomend creating your own force diagrams and ignore the diagrams there, they are a bit confusing) ► Here's an intriguing problem combining what we've learned about forces AND motion