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菊花仔 2012-11-11 03:18 PM

物理 力學一條....

各位ching   呢題小弟諗左好耐都諗唔清
我有諗過呢個方法
對a受力分析 求出a既加速度 跟住以a為參考系 對b受力分析 從而計出b相對於a滑到底的時間  再用呢個時間計返a行左幾遠 但個答案仲有好多未知量係到....

動量方面我都諗過水平方向系統動量受恒  豎直方向不守恒 . 有冇ching可以指點下

[[i] 本帖最後由 菊花仔 於 2012-11-11 04:46 PM 編輯 [/i]]

Zzlaz 2012-11-11 04:30 PM

work done???

菊花仔 2012-11-11 04:31 PM

[quote]原帖由 [i]Zzlaz[/i] 於 2012-11-11 04:30 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347175616&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
work done??? [/quote]
??

↗亢龍無悔↙ 2012-11-11 04:47 PM

唔知啱唔啱,我試下做嫁詐~~~

Consider free body diagram for mass B,
Weight = m g
Normal reaction = m g cosθ
Net force = m g sinθ = m a, => a = g sinθ

Consider free body diagram for mass A,
Weight = 3 m g
Action force from mass B = m g sinθ
Normal reaction = 3 m g + m g cosθ cosθ
Net force = m g cosθ sinθ = 3 m a', => a' = 1/3 g sinθ cosθ

Consider mass B and apply equation of motion,
(a - b) / cosθ = (1/2) (g sinθ) t*2, => t*2 = 2 (a - b) / (g sinθ cosθ)
Consider mass A and apply equation of motion,
s = (1/2) (1/3 g sinθ cosθ) 2 (a - b) / (g sinθ cosθ)
s = 1/3 (a - b)

Note that, when we look at B (for the observer in mass A), the track of B will be along the plane A
When we are being an isolated observer, the track will be a [color=red]sloping plane steeper than the plane A[/color]
[color=black]The concept is like~[/color]
[color=black]For an object performing uniform circular motion, we say that is the centripetal force keep the object in motion[/color]
When we are located in the object, we should say it is the centrifugal force keep the object in motion instead.

[[i] 本帖最後由 ↗亢龍無悔↙ 於 2012-11-11 05:15 PM 編輯 [/i]]

↗亢龍無悔↙ 2012-11-11 05:04 PM

呢啲數,我地唔會用momentum去計
正如你話要拆x, y component,因為momentum係vector
Consider y direction and law of conservation of momentum,
Mass B momentum increases from 0 to m Vy
咁你有啲咩野可以無啦啦將B嘅momentum由0增加到m Vy先?
由於mass A無vertical motion,即係你要consider the "earth" itself
咁你會愈整愈多unknown

菊花仔 2012-11-11 05:22 PM

[quote]原帖由 [i]↗亢龍無悔↙[/i] 於 2012-11-11 04:47 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347176547&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
唔知啱唔啱,我試下做嫁詐~~~

Consider free body diagram for mass B,
Weight = m g
Normal reaction = m g cosθ
[color=red]Net force = m g sinθ[/color] = m a, => a = g sinθ  [color=red] 題目只係話A與地面光滑 並沒說AB間光滑啊
[/color]
Consider free body diagram for mass A,
Weight = 3 m g
Action force from mass B = m g sinθ
Normal reaction = 3 m g + m g cosθ cosθ
Net force = m g cosθ sinθ = 3 m a', => a' = 1/3 g sinθ cosθ

Consider mass B and apply equation of motion,
[color=red](a - b) / cosθ = (1/2) (g sinθ) t*2[/color], => t*2 = 2 (a - b) / (g sinθ cosθ)   [color=red](a - b) / cosθ係以A為參考系先係吧??  如果係要以A為參考系 咁對B受力分析唔係要考慮埋慣性力咩
[/color]Consider mass A and apply equation of motion,
s = (1/2) (1/3 g sinθ cosθ) 2 (a - b) / (g sinθ cosθ)
s = 1/3 (a - b)
[/quote]

呢條我諗左成日  發覺列晒D式出黎仲係有好多未知量:smile_39:

↗亢龍無悔↙ 2012-11-11 05:59 PM

做physics,除非題目講明,如果唔係都只會當自己係一個isolated observer
做呢條題目,我都自己當自己係一個isolated observer,我只不過係想提出mass B實際嘅locus
如果inclined plane有friction,即就會多左一個unknown,就要多一條equation先做到
咁我就唔識做喇

照你所講,我都可以話~條題目都無講friction有幾大,個friction都可以太大,令A唔會滑落
由於多左個friction, f cosθ向左, mg cosθ sinθ向右
咁我都可以話當f細時,net force會令B向右走;同時亦都可以話當f大時,net force會令B向左走

菊花仔 2012-11-11 06:27 PM

[quote]原帖由 [i]↗亢龍無悔↙[/i] 於 2012-11-11 05:59 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347180002&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
做physics,除非題目講明,如果唔係都只會當自己係一個isolated observer
做呢條題目,我都自己當自己係一個isolated observer,我只不過係想提出mass B實際嘅locus
如果inclined plane有friction,即就會多左一個 ... [/quote]
但係如果是以"自己"為參考系  這樣B的位移應該不只是(A-B)/COS 這樣簡單吧? 因為斜面自身都會郁  呢題真係好煩...

菊花仔 2012-11-11 11:31 PM

我嘗試過用伽利略變換  得到B相對於地面既位移   再分別計水平和豎直分量  計到最後仲有好多SIN 同COS 化簡唔到........   求高手指教

↗亢龍無悔↙ 2012-11-11 11:44 PM

[quote]原帖由 [i]菊花仔[/i] 於 11-11-2012 06:27 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347181587&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

但係如果是以"自己"為參考系  這樣B的位移應該不只是(A-B)/COS 這樣簡單吧? 因為斜面自身都會郁  呢題真係好煩... [/quote]
丫,sorry~一則更正,正如你所講,displacement (a - b) / cosθ 係當observer身到mass A
但係我想講,無論個observer係身在mass A,定係isolated object都好,計數的話個displacement仍然係,(a - b) / cosθ
咁當然,當你計哂所有acceleration, net force果啲出黎,你就可以用返佢地去搵返B嘅"actual displacement"

Assume all surfaces are smooth.
Consider the free body diagram for mass B [color=red]at any time on the inclined plane[/color]
weight同normal reaction永遠存在,weight就唔洗多加解釋
至於normal reaction,B永遠都會痴住A,咁個normal reaction永遠都係同一個value
因為normal reaction永遠balance唔到個weight,所以永遠都會有一個net force (m g sinθ)

Consider the free body diagram for mass A [color=red]at any time when B is sliding down[/color]
When B is sliding down, B exerts an action force to A (at any time when B is sliding down), this is the only force that make A to moves
其他forces就唔洗理
咁啲force永遠存在,點解唔可以用?

果條所謂嘅actual track其實先至係A嘅actual displacement
我幾肯定我所講嘅野,因為之前我去sit過一堂physics course,個professor就係拎左咁上下嘅example係到講,咁當然唔排除我聽少,聽錯啦

唔明我就用返我先前所講嘅
無論centripetal force定係centrifugal force都好,兩都都係= m v*2 / r

[color=red]有錯的話請其他師兄更正:smile_38: :smile_38: :smile_38: [/color]

↗亢龍無悔↙ 2012-11-11 11:49 PM

[quote]原帖由 [i]菊花仔[/i] 於 11-11-2012 11:31 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347206106&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
我嘗試過用伽利略變換  得到B相對於地面既位移   再分別計水平和豎直分量  計到最後仲有好多SIN 同COS 化簡唔到........   求高手指教 [/quote]
你咪玩啦,條題目都無collision,你點用momentum去計呀:smile_27:
我10000000% sure絕對唔係用energy或者momentum去計囉

菊花仔 2012-11-12 06:55 AM

[quote]原帖由 [i]↗亢龍無悔↙[/i] 於 2012-11-11 11:44 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347207590&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

丫,sorry~一則更正,正如你所講,displacement (a - b) / cosθ 係當observer身到mass A
但係我想講,[color=red]無論個observer係身在mass A,定係isolated object都好,計數的話個displacement仍然係,(a - b) / cosθ[/color]
咁當然, ... [/quote]
但係好似火車向左行 你向右行  如果車位移>你位移  地面睇落去人的位移就向左    參考系變左 個位移應該要修正吧

↗亢龍無悔↙ 2012-11-12 01:14 PM

你所提及嘅係relative motion
你都識講人係火車,因為火車行得比人快,所以望落個人會向左移
但係你講漏左,人向左移同火車向左移嘅速度唔同(relative to isolated observer)
由isolated observer黎睇,火車嘅displacement會比人大

或者再有少少更正
我係話如果你真係要搵mass B嘅actual displacement,你就係搵哂其他野,即係搵哂B嘅motion relative to A + A嘅motion relative to isolated observer
而且題目跟本都無叫你搵,加上呢隻題目已經係大學嘅level
題目要搵A嘅actual displacement,佢只係relative to isolated observer

不過有樣野想提嘅係,B嘅actual distance travelled = length of inclined plane
但由於A會向右走,如果用快拍去影住個motion,佢條track (actual displacement) 就會短左
呢兩行所講嘅,同你嘅火車同人嘅理道差唔多

但係呢個快拍,只不過係錯覺,B嘅motion係應該要諗B嘅free body diagram
net force, acceleration of B會唔會變(relative to A or relative to isolated observer)
係要留意我成日都提到嘅free body diagram "at any time"
首先,B落到底嘅時間,無論relative to A or relative to ioslated observer都係一樣
跟住,net force at any time都會一樣(先前解釋過)

如果你唔信我所講就等其他師兄幫你,或者去問呀sir

[[i] 本帖最後由 ↗亢龍無悔↙ 於 2012-11-12 01:38 PM 編輯 [/i]]

菊花仔 2012-11-12 06:36 PM

[quote]原帖由 [i]↗亢龍無悔↙[/i] 於 2012-11-12 01:14 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347241901&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
你所提及嘅係relative motion
你都識講人係火車,因為火車行得比人快,所以望落個人會向左移
但係你講漏左,人向左移同火車向左移嘅速度唔同(relative to isolated observer)
由isolated observer黎睇,火 ... [/quote]
我唔係唔信你  只係諗唔明想問下
如果B relative to A
因為A係非慣性系(A relative to floor 有加速度)
在非慣性系上考慮net force 唔係要計埋慣性力咩

同埋eg你向前面棵樹加速跑去 咁棵樹relative to you咪加速沖埋你到 但樹係豎直方向受重力同支持力 水平方向因為相對地面無滑動所以冇摩擦力
如果係咁講  佢受到合力係0 但又加速沖埋你到咪唔合理.  所以應該有個慣性力F=-m樹a人

[[i] 本帖最後由 菊花仔 於 2012-11-12 06:43 PM 編輯 [/i]]

↗亢龍無悔↙ 2012-11-12 06:43 PM

我睇唔明你問咩,可唔可以適當咁變成英文?
慣性英文係inertia喎,你係講緊weight?

菊花仔 2012-11-12 06:51 PM

[quote]原帖由 [i]↗亢龍無悔↙[/i] 於 2012-11-12 06:43 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347263303&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
我睇唔明你問咩,可唔可以適當咁變成英文?
慣性英文係inertia喎,你係講緊weight? [/quote]
非慣性系: non  inertial frame

↗亢龍無悔↙ 2012-11-12 11:14 PM

[quote]原帖由 [i]菊花仔[/i] 於 12-11-2012 06:51 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=347263743&ptid=21118133][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

非慣性系: non  inertial frame [/quote]
non inertial frame
omg呢個詞我係考完AL之後,再去睇啲更加advanced嘅physics書先見到嘅字
你究竟讀緊咩level嘅physics?:funk:

[[i] 本帖最後由 ↗亢龍無悔↙ 於 2012-11-12 11:16 PM 編輯 [/i]]

菊花仔 2012-11-13 01:08 PM

呢條係大學入學試 既試題....

↗亢龍無悔↙ 2012-11-13 01:21 PM

non inertial frame,我只係知大槪係咩東東,真正用係計數就未試過:smile_42:
其實你係咪想問,因為A會向右加速,所以會令B嘅net force同normal reaction有改變?

菊花仔 2012-11-13 06:52 PM

呢條老師解答了
x軸上系統不受外力係關鍵
由動量守恒得 mvbx = -3mva   負號即係va vb方向相反
即係 vbx = 3va  <<任何時刻都成立
將成個過程分成無限段
a的位移為s(a)=v1(bx)*t+v2(bx)*t+....+vn(bx)*t
b的水平位移s(bx)=3v1(bx)*t+3v2(bx)*t+....+3vn(bx)*t   
即係s(bx)=3s(a) -------(1)
又由位移同邊長關係得  s(bx)+s(a)=a-b------(2)
聯立(1)(2)  得s(a)=(a-b)/4

聽完我都o左 完全諗唔到會用到微元法

↗亢龍無悔↙ 2012-11-13 07:07 PM

即係我計錯哂:smile_38:
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