Question

IlI. A SMOOTH ACCELERATING PLANE An smooth plane is inclined from the horizantal at angle of 0 as shown. (Assume for now that 0- 70 degrees.) A mass m is placed at the top of the inclined plane and allowed to slide downwards (friction is negligible). At some point during the mass's trip. however, the plane begins to accelerate to the right at a rate of a, causing the mass to stop sliding on the plane. How/why does it stop sliding? That's the question. Only at ane particular acceleration will the block stop sliding. Our goal is to find that acceleration. A. Draw a picture of the situation and a system schema. B. Draw a pure FBD of the mass C. Choose a coordinate system in which the x-axis is parallel to the direction of acceleration. Hint: This may prove simpler than you first think; if it looks simple at first, that's probably a good sign Remember, the mass stopped sliding D. Create a component FBD of the mass E Create NII equations for the mass in both directions. F. Look at your component FBD & your NIl equations. In 1-2 sentences, explain how the mass could have stopped sliding down the plane. What force was causing it to slide downwards before the plane started accelerating? What force (or what component of what force) balanced that force? G. Use the NIl equation in the y-direction to compute the magnitude of the Normal force, if the block is not sliding on the plane H. Use the NII equation in the x-direction to compute the acceleration a that the plane must be accelerating at, if the block is not sliding on the plane. Some reflections: answer each question below in I sentence of English. No numbers required What will happen to the block if the acceleration drops below the a you found in step H? J What will happen to the block if the acceleration goes above the a you found in step H? K. What will happen to the block if the acceleration is exactly a? (Not a trick question. An "are you with me?" question) The GENERAL CASE: This looks scarier, but it's really just exactly what you did in steps D-H all over again, with one tiny change. Now, instead of assuming that = 70, assume that 0 is some unknown angle. Recreate the steps you did above, leaving as 0. So, for example, where you had sin( 70) before, you'll know have sin(). Of course, you can't put sin(0) into your calculator, so you'll just leave it as sin(e). Ditto with cos(e). L.

Answer 1

- A before acceleration After aceleration ♡ -a. N. (normal) ма - (pseudoforced + ya Nuoso Insino MO

(6) As Mass stogs Slieling. Asino = ma Allos = ma la direction) ly direction) (f) before acceleration 'a'. meals is sliding down due to component of force of gravity applying along the plane ie masino mgloss holingsino. When acceleration is given to inclined plone, a pseudo fone start acting on block in opposite direction of acceleration which balonce the normal force component. (b) for o= 70° News 70' :ma No ma cos 70

(1) a = nsino m mq.sin 70 los 70'. m = 9f an 70 acceleration, as 9.8*2.75 . a = 26.95m/s² ril ll acceleration drops below 'a', Nsino will be greater than ma, block will trip off inclined. y acceleration goes above 'a' Nsino will be less than ma, block will trip back. Block will move together with inclined. it all s to all f is G. same for Noso = ma Msino = ma. tono a (az gdano