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tim1993127 2016-3-6 07:26 PM

力學求教

希望各位大大可以教我因為做極都極唔到佢最尾個個答案...

可能初頭我畫右邊圖已經錯左...

[[i] 本帖最後由 tim1993127 於 2016-3-6 07:41 PM 編輯 [/i]]

hkpal 2016-3-7 06:15 PM

[font=新細明體][/font][font=Tahoma]Left diagram is correct with: [/font]
[font=新細明體][/font][font=Tahoma]W1=Mg, [/font]
[font=新細明體][/font][font=Tahoma]N=Mg cos [/font][font=Tahoma]α,[/font]
[font=新細明體][/font][font=Tahoma]F=μ Mg sin α;[/font]
[font=新細明體][/font][font=Tahoma]At equilibrium, along the slope direction, wehave:[/font]
[font=新細明體][/font][font=Tahoma]W1 sin α + F = T1[/font]
[font=新細明體][/font]

[font=新細明體][/font][font=Tahoma]Right diagram is wrong.
For mass m, there are only two forces, i.e.its downward weight W2 pulled by an upward rope tension T2 with:[/font]
[font=新細明體][/font][font=Tahoma]W2 = mg;[/font]
[font=新細明體][/font]

[font=新細明體][/font][font=Tahoma]For the relationship between T1 and T2, notethat for the pulley above mass m, it has one downward tension T2 balanced bytwo upward tensions both of T1, i.e. 2 T1 = T2[/font]
[font=新細明體][/font]

[font=新細明體][/font][font=Tahoma]Then eliminate T1 and T2, and work out theresult in Question (iv).[/font]
[font=新細明體][/font]

tim1993127 2016-3-7 08:47 PM

Thanks for you help!
Finally, I done it
For the pulley above mass m, it has one downward tension T2 balanced bytwo upward tensions
I assigned  T2 and T3 be that upward tension and then for model pulley use,
T1=T2=T3

Is it reasonable?:o  
I move the direction of T3 on right diagram to same as T2
Also can gain  same result....
Thanks!
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