MAV doesn't like to make much sense

[Mao]

I think anyone who went to MAV lecture would agree with me that the degree of difficulty is WAYY over the top.

Here's a few questions I cannot seem to understand:

Sketch SOLVED
took a while. Letting z=x+yi, I've managed to show that for , and definitely not for
but I can't seem to show that for x<0 and -1<y<1 that expression is false...
[btw, this region describes an arc, as the angles subtended by a chord on any point on the circumference are equal]



The wedge is not fixed on the ground, and the ground is smooth. Find an expression for acceleration in terms of SOLVED

So, the block exerts a force perpendicular to the inclined plane,
resolving this force into components parallel to the ground: , that can then be used to calculate the acceleration of the wedge.
but apparently that is wrong. there has to be a NET force acting perpendicular to the inclined plane, i.e.
what o.O


it then goes on to ask if the wedge is now placed on a rough plane, what is the minimum coefficient of friction in terms of .SOLVED

what the hell is a minimum coefficient of friction? would that be the coefficient of kinetic friction?
apparently the minimum occurs at the point of sliding, but this is where the maximum friction force occurs... what o.O


what are the forces acting on the smaller mass?SOLVED
From checkpoints, only and are. I assumed it was written with respect to VCAA assessment reports, specifically Question 27 (MCQ, part 1) in VCAA 2003 SM exam 1. According to the Assessment Report, acts downwards on the lower mass also... what o.O



When the pulling force is 10g N, the block is on the point of sliding down. When the force is 12g N, the block is on the point of sliding up. Find two possible angles of inclination and two coefficients of friction

resolving into components perpendicular and parallel:
10g N, parallel to plane:
10g N, perpendicular to plane:
12g N, parallel to plane:
12g N, perpendicular to plane:

now, the solution suggests that since friction force is maximum in both cases, the normal reaction force R must be equal. does that imply 5=6?!
[it then goes on to show that one of the set of solutions give and , ]

okay, that's all for now, any help is appreciated.

[shinny]

For the 2nd question, you forgot to take into account the 2kg mass of the wedge in calculating acceleration, so divide that answer by 2 (I don't get whats wrong overall with that answer though o.O). When they're asking for the minimum coefficient, I think they mean the minimum coefficient of friction so that the wedge does not move. I mean, what other application could they be asking for o_O For the 3rd question, VCAA is correct because of Newton's Law about for every reaction, there is an equal and opposite reaction. It makes logical sense anyway; how could chucking a mass on top of something have no effect on its forces...

EDIT: For part 2, does anyone know whether the friction is just coefficient of friction*normal of the wedge? or normal of the wedge AND the block? It physically makes sense for the normal of the block to come into account as well, but I've never been taught this...

2nd EDIT: Disregard all that working out then if my N1 was wrong from the very beginning o__o" I'll try the first part properly later...

3rd EDIT @_@: Removed the images because they're huge and misleading since they're wrong =P

[Mao]

For the 2nd question, you forgot to take into account the 2kg mass of the wedge in calculating acceleration, so divide that answer by 2.

no, that will still be wrong according to the set of solutions MAV gave us.
apparently, the 1kg block has a net acceleration perpendicular to the wedge, and the acceleration (after MUCH mucking around) is

When they're asking for the minimum coefficient, I think they mean the minimum coefficient of friction so that the wedge does not move. I mean, what other application could they be asking for o_O

VERY NICE. why didn't I think of that!!

For the 3rd question, VCAA is correct because of Newton's Law about for every reaction, there is an equal and opposite reaction. It makes logical sense anyway; how could chucking a mass on top of something have no effect on its forces...

yeah, I thought it made sense too. Who would like to send Checkpoints an email?

(Y) thanks man, good work

[doboman]

Shinjitsuzx has girl handwriting...lol

[Mao]

shinjitsuzx: really appreciate your efforts, but your expression for is the expression for acceleration (first part). their

also, how you said "the block is not moving", i don't think that's true... it's a smooth surface between the wedge and the block, and there is gravity... so the block HAS to be moving, and moving under constant acceleration.

EDIT: AHHH MAJOR THING I DIDNT THINK ABOUT
IF THE WEDGE IS ACCELERATING TO THE LEFT, THE INCLINED PLANE IS "SLIPPING AWAY" FROM UNDER THE BLOCK, HENCE THE BLOCK MUST ALSO BE ACCELERATING IN THE SAME DIRECTION (KIND OF, WITH RESPECT TO GROUND), THAT IS, IT MUST KEEP ACCELERATING "DOWNWARDS" PERPENDICULAR TO THE INCLINED PLANE TO REMAIN ON THE PLANE, HENCE THE REACTION FORCE IS NOT g*cos(theta), ITS LIKE STANDING IN A LIFT ACCELERATING DOWNWARDS.

okay, i'm calmed down now. the block is accelerating in two directions, parallel to the plane (this is fairly straight forward), and perpendicular to the plane. If you add these two vector accelerations together, the net downwards acceleration will be greater (than just parallel no perpendicular), which makes sense because the wedge is moving away AND it is slipping down.
When the system has downward acceleration (think a lift lowering, and you are inside), the normal reaction force R from the wedge acting on the block will be less (standing on a scale in a lowering lift will let you instantly lose weight). Since the sidewards force of the wedge is , the rest comes fairly good.

[Collin Li]

Shinjitsuzx has girl handwriting...lol

Haha, yeah! He's gonna get a good English score (cue aidansteele to agree).

[Mao]

Self-solve: 1st question

for x<0 and -1<y<1, z+i lies in the second quadrant, and z-i lies in the third quadrant.

hence, arg(z+i) is positive, and arg(z-i) is negative, and , let the conjugate of z-i be w, also lies in the second quadrant.



now, the sum of two angles in the second quadrant is bounded within and , and hence will not equal to , the relationship cannot exist for x<0 and -1<y<1


that leaves... the last question, which would be easily acceptable if I declare the solutions to be wrong in its assumption that

[shinny]

Yeh I realised I got pretty much everything entirely wrong just after you mentioned that the block itself is moving in the j direction but I had no time to correct it, was going out =P I'll try again now.

[ice_blockie]

Mao, who was the lecturer at the MAV lecture? I presume you went to the Glen Waverly ones...

[Mao]

yes, Glen Waverley... His name escapes me...

[gabrielle__]

The dude at the brunswick lecture was such a cock wank.
"Lets spend 20 minutes defining a bound reference book. And for the rest of the time, here's some question's i stole out of checkpoints. ENJOY!"

Waste of a day and $33.
lol Didn't even get a summary of the course.

[bigtick]

Self-solve: 1st question

for x<0 and -1<y<1, z+i lies in the second quadrant, and z-i lies in the third quadrant.

hence, arg(z+i) is positive, and arg(z-i) is negative, and , let the conjugate of z-i be w, also lies in the second quadrant.



now, the sum of two angles in the second quadrant is bounded within and , and hence will not equal to , the relationship cannot exist for x<0 and -1<y<1

argument of z is not unique, any two arguments of z differ by an integer multiple of 2pi.

[shinny]

Shinjitsuzx has girl handwriting...lol

Haha, yeah! He's gonna get a good English score (cue aidansteele to agree).

Since when was there a direct correlation between good handwriting and English o_o Besides, mine gets almost illegible at the speed I write at in an English exam <.< It's bad enough that it's been written on my trial exam comments by my tutor and teacher =P

[shinny]

Oh and as for the last question, Ru shouldn't equal Rd...can't see why it would. I mean, if you're lifting it diagonally upwards with more force, the force that the block exerts on the slope would be less? So ignoring their assumption, I eventually got (after much CAS lag) theta=23.45 degrees...Anyone else want to go for a shot at this and verify it before I finish off the question? o_O

[ice_blockie]

The dude at the brunswick lecture was such a cock wank.
"Lets spend 20 minutes defining a bound reference book. And for the rest of the time, here's some question's i stole out of checkpoints. ENJOY!"

Waste of a day and $33.
lol Didn't even get a summary of the course.

...should sooooo complain about that guy...