查看完整版本 : phy 求解答!!

kwan135 2012-10-24 06:03 PM

phy 求解答!!

1) A ship B is steaming on a straight course south-east at a uniform speed of 15km/h. Another ship A, is a distance of 10km due north of B and steams at a speed of 12km/h. Find the course that A must steer in order to get as close to B as possible, and their minimum distance apart.呢條我諗極都諗唔到:smile_39:



2)a light inextensible string fastened to a point in the ceiling passes under a smooth movable pulley of mass 5 kg, over a smooth fixed pulley and carries a 4 kg mass hanging freely. If the system is released from rest, find the rising speed acquired by the movable pulley in 2 seconds.
呢條我想問個兩個mass個acceleration係咪一樣??我問左miss佢又係到賣關子,叫我番屋企試下咁樣做係咪一樣=.=,,如果一樣我就計到個a係2.26..
[url=http://s1279.beta.photobucket.com/user/kwan135/media/Screenshot_2012-10-24-17-28-01_zps50be81af.png.html]http://s1279.beta.photobucket.com/user/kwan135/media/Screenshot_2012-10-24-17-28-01_zps50be81af.png.html[/url]  圖係呢度

求ching解答:smile_o03:

Chungaaabb 2012-10-24 08:41 PM

我覺得應該唔系一樣。當你用動滑輪提起一個野時,上升高度H,但系你你只手就要提高2H咁多。即系話,個SPEED V OF THE 5KG PULLEY應該系個MASS 的SPEED的一半。

有沒高手來解答??

kwan135 2012-10-26 06:48 PM

PUSH push !!

↗亢龍無悔↙ 2012-10-27 01:42 AM

2樓果位師兄....................@@ good try @@"

比一個hints你,左中右3條string嘅tension都會一樣
而且,中右兩條string可以將佢簡化做一個2.5kg嘅mass吊係另一邊
咁識做咩呀?

[[i] 本帖最後由 ↗亢龍無悔↙ 於 2012-10-27 01:53 AM 編輯 [/i]]

kwan135 2012-10-27 12:15 PM

回覆 4# 的帖子

我將 fix左係ceiling果條tension叫 T2  兩個pulley之間果條叫 T3
跟住我拆開兩個pulley用free body diagram睇 set兩條equations (take downward acceleration as +ve)
1. 4g-T3=4a
2. T2+T3-5g=-5A
咁岩唔岩=.=??  其實我知T2=T3(你話我馬後炮都好=3=),,BUT ching可唔可以解釋下點解T2+T3=2.5kg?
thx~

↗亢龍無悔↙ 2012-10-27 12:46 PM

我意思係,你所講嘅T2, T3,我將佢地simplify~~~
變成T2無左,得返T3,而T3係一個free end掛住一個2.5kg嘅mass

解親都係mechanics開始有啲悶:smile_45:

kwan135 2012-10-28 03:45 PM

咁第一條 ching點睇??

↗亢龍無悔↙ 2012-10-28 07:03 PM

無圖之餘,啲英文咁難,睇唔明,唔想諗:smile_45:

kwan135 2012-10-29 09:52 PM

第一條話  有隻船B向東南方行 15km/h (姐係45度)  另外一隻船A 在B的北方(姐係B上面),相距10km,行緊12km/hr. 問1) 船A應該向邊個方向行,姐係話向東南方o既邊個角度then A就可以最接近B 2)搵佢地之間相差的min distance  求ching指教:smile_13:

Zzlaz 2012-10-30 12:15 AM

[quote]原帖由 [i]kwan135[/i] 於 2012-10-29 09:52 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=346147422&ptid=21036634][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
第一條話  有隻船B向東南方行 15km/h (姐係45度)  另外一隻船A 在B的北方(姐係B上面),相距10km,行緊12km/hr. 問1) 船A應該向邊個方向行,姐係話向東南方o既邊個角度then A就可以最接近B 2)搵佢地之間相差的min distan ... [/quote]
佢地有無機會遇上先.

↗亢龍無悔↙ 2012-10-30 02:35 AM

咁我知個做法lu,不過啲數就唔想計
distance between ship A, B = z. Apply cosine law
z = root of [(10)*2 + (15t)*2 - (2) (10) (15t) (cos 135)] - 12t
differentiate both sides with respect to time
dz / dt = xxxxxxxxxxxxxxxxxxxxxxxxxx
因為ship B行得快過ship A,所以ship A永遠都追唔到ship B
即係話你set z = 0應該係搵唔到答案
咁你就要set dz / dt = 0,跟住用trial and error method,就會搵到一個value for t,
跟住sub返個t入條cosine law入面就計到哂,再用sine law計埋隻角

[[i] 本帖最後由 ↗亢龍無悔↙ 於 2012-10-30 12:39 PM 編輯 [/i]]

kwan135 2012-10-30 03:37 PM

回覆 10# 的帖子

A慢過b 當然遇唔到呢...

kwan135 2012-10-30 03:38 PM

回覆 11# 的帖子

thx ching 明白鳥 :)

↗亢龍無悔↙ 2012-10-30 11:22 PM

回覆 13# 的帖子

睇黎我低估左你嘅maths:smile_o05:

ivan711 2012-11-11 02:31 PM

借問一聲,第1題就一定唔係,咁第2題係唔係DSE syllabus內?
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查看完整版本: phy 求解答!!