mechanics

[S]Mangoray

q28點做??

看不清請到
http://i350.photobucket.com/albums/q437/mangorayyau/1.jpg

Zend

好似做過, 係take moment , about 個轉動點
唔多記得la

[S]Mangoray

a部答案是40N[[bao_29]]

piyopiyo

對Q29較有興趣[[tk_43]]
29a是否30N[[tk_30]]

[S]Mangoray

原帖由 piyopiyo 於 4-10-2008 15:11 發表
對Q29較有興趣[[tk_43]]
29a是否30N[[tk_30]]

係阿,點解係30N,而唔係70N[[bao_29]] ?

piyopiyo

let T1,T2,T3,,,T1係最低果個pulley ge tension against the 21kg mass,,,so on so forth

As they are in equilibrium,,,T1 = F
T2 = F+T1 = 2F
T3= T2 + 2F = 4F
T1+T2+T3 = 210
7F=210
F=30N[[tk_28]]

[S]Mangoray

原帖由 piyopiyo 於 4-10-2008 17:34 發表
let T1,T2,T3,,,T1係最低果個pulley ge tension against the 21kg mass,,,so on so forth

As they are in equilibrium,,,T1 = F
T2 = F+T1 = 2F
T3= T2 + 2F = 4F
T1+T2+T3 = 210
7F=210
F=30N[[tk_28]]

[[bao_29]]厲利厲利


至於q28

(a) Take moment at the point where the wheel touches the kerb,
50 x [(1+0.2-0.5] = W x [square-root(1^2-0.5^2)]
W=40N
where W is the weight of the wheel


可唔可以解一解個答案[[bao_29]] ??

Puppet

原帖由 [S]Mangoray 於 4-10-2008 17:48 發表


[]厲利厲利


至於q28

(a) Take moment at the point where the wheel touches the kerb,
50 x [(1+0.2-0.5] = W x [square-root(1^2-0.5^2)]
W=40N
where W is the weight of the wheel


可唔可以 ...

first, do u know what square-root(1^2-0.5^2) is?

[S]Mangoray

原帖由 Puppet 於 4-10-2008 17:58 發表

first, do u know what square-root(1^2-0.5^2) is?

我唔明點解 W 同 F 成直角都有MOMENT

Puppet

原帖由 [S]Mangoray 於 4-10-2008 18:06 發表


我唔明點解 W 同 F 成直角都有MOMENT

你解下點解W 同 F 成直角應該冇MOMENT先