CE PHY...

abc666

當一件物件放在一條斜坡上
normal reaction on 物件by斜坡是(angle to horizontal=X)
R=mg cosX
還是
R=mg / cosX

當物件不動,或向下滑動時
或斜坡有friction或沒friction
或物件在preform circular motion時(AL)
weight 和 normal reaction的關係有沒有分別的?

Ivanhy92

原帖由 小霞的fans 於 28-4-2009 16:08 發表

                               
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話說我唔記得左mgsin θ呢舊野係搵咩-_- "

Weight component downward the inclined plane...

Friction-compensated slope:

f = mgsin θ

小霞的fans

話說我唔記得左mgsin θ呢舊野係搵咩-_- "

abc666

咦...
我之前看時竟把perpendicular和 alone掉轉了...
希望考試時不會又看錯...

超夢夢的夢境

原帖由 abc666 於 8/4/2009 06:49 PM 發表

                               
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一直以為主要是計和錢銀有關的東東...

完全沒關係
paper I 是 mechanics
paper II 是 probability,statistics,numerical analysis(就是只重視數字上的答案,而不是實際數值),differential equation等

原帖由 abc666 於 8/4/2009 06:49 PM 發表

                               
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祝你拿A~!

Applied?機會不大,我想大概兩成吧
B很穩就是

Physics?機會更小,只有mechanics比較好

無論如何,謝謝

p.s.上面改了一下

abc666

謝了~
diagram不用了,夠功力看
只是concept有點弄不清
主要是因為書有些不是circular motion的圖的normal reaction條線比mg長...
圖中又沒解說,平時又懶看文字...
導致我一直以為一件物件停在斜坡上時R cosX=mg ...
今天早上做exercise時發覺ans有點奇怪
現在一問之下才發現...

applied math 這麼多phy囧?
一直以為主要是計和錢銀有關的東東...

祝你拿A~!

超夢夢的夢境

原帖由 abc666 於 8/4/2009 10:35 AM 發表

                               
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當一件物件放在一條斜坡上
normal reaction on 物件by斜坡是(angle to horizontal=X)
R=mg cosX
還是
R=mg / cosX

當物件不動,或向下滑動時
或斜坡有friction或沒friction
或物件在preform circular motion ...

R = mg cosX
in the cases that the inclined plane is smooth or rough, no matter the object is sliding down or not.

Note that there are only 3 forces exerted on the object (weight, normal reaction and friction)
Resolve mg respectively into components along and perpendicular to the inclined plane
Components along the plane=mg cos X
Components perpendicular to the plane=mg sinX
Components along the plane is balanced.Therefore, R = mg cosX
EDIT:匆忙間打錯了
Components perpendicular to the plane=mg cos X
Components along the plane=mg sinX
Components perpendicular to the plane is balanced.Therefore, R = mg cosX

When the object is performing HORIZONTAL circular motion without friction,
components along the plane is NOT balanced.
(The reason is that the resultant force is the centripetal force pointing towards the centre,which can be resolve into components along and perpendicular to the inclined plane)
So R= mg cosX does not hold.

Consider the vertical components of forces.
Note that the object is moving on a horizontal plane.
Therefore, vertical components of forces is balanced.
R cosX = mg
R = mg / cos X

In addition, solving the velocity can be done by the following steps.
Vertical components of forces contribute to centripetal force.R sinX = mv^2/r
v=sqrt(gr tanX)   (sqrt=square root)
p.s. circular motion 這部分在今天的 Applied paper I 價值兩分

When the object is performing HORIZONTAL circular motion WITH friction,
Consider the vertical and horizontal components of forces.
EDIT:For v<designated speed,i.e. friction f is pointing upwards
R cosX = mg - f sinX
R sinX - f cosX = mv^2/r
EDIT:For v>designated speed,i.e. friction f is pointing downwards
R cosX = mg + f sinX
R sinX + f cosX = mv^2/r
Obviously,R changes.

Solving the velocity 是我今天不夠時間想的那部分
想到了,原來今早被自己局限了,概念沒錯可是想不下去(淚目*2036)
Only consider the friction point downwards.
For the case surface in contact is small

                               
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For the case a plane is in contact

                               
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Then take moment about centre of gravity.
Moment about centre of gravity = 0

[ 本帖最後由 超夢夢的夢境 於 8/4/2009 09:01 PM 編輯 ]