推 ht15trep:1.阻力↑，重力↓，a=F/m，V^2=2as ...應該就降吧 05/02 15:18
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作者: ht15trep (柏林=Boring) 看板: TransPhys
標題: Re: [題目] 牛頓力學...
時間: Fri May 2 23:37:03 2008
※ 引述《alwyner (Time is money!!)》之銘言：
: ※ [本文轉錄自 Physics 看板]
: 作者: alwyner (Time is money!!) 看板: Physics
: 標題: [題目] 牛頓力學...
: 時間: Sun Apr 27 13:34:07 2008
: 1.阻力R = kv ，質量m的小鋼球從油的表面(y=0)靜止釋放，k為常數，導出在h深處，此
: 球所到達的速度v。
: Ans: h =?
: 2. ／
: m ／＼／ │
: ＼／ │
: 夾角 θ ／ M │
: ↘ ／ │
: ／ │
: │￣￣ │
: ￣￣￣￣￣￣￣￣￣
: 忽略所有摩擦，求a)M的加速度 b)m相對於M的加速度。
定義向右、向上為正
※(a)
M:
Fx = N1*sinθ = MA ---[1]
Fy = -N1*cosθ-Mg+N2 = 0 ---[2] (N2=Mg+mg)
m:
Fx = -N1*sinθ = m(A-a*cosθ) ---[3]
Fy = N1*cosθ-mg = -ma*sinθ ---[4]
要求出M的加速度，就必須想辦法把"a"消去
[1]+[3]
→ (M+m)A = ma*cosθ ---[5]
[3]*cosθ + [4]*sinθ
→ -mg*sinθ = mA*cosθ - ma
→ -mg*sinθ - mA*cosθ = -ma ---[6]
[5]+[6]*cosθ
→ -mg*sinθ*cosθ + MA + (1-(cosθ)^2)*mA = 0
→→→ A*( M + m*(sinθ)^2 ) = mg*sinθ*cosθ
mg*sinθ*cosθ
∴ A = ________________ ←M的加速度！！！
M + m*(sinθ)^2
※(b)
from [5]
(M+m)A = ma*cosθ
(M+m)
∴ a = __________* A
m*cosθ
(M+m)g*sinθ
= _____________ ←m的加速度！！！_
M+m*(sinθ)^2
m相對於M的加速度： a-A (懶的算了= =|||)
: 以上兩題希望大大幫個小忙^^" 謝謝!! 可以跟我說怎麼開始就好~ 力圖怎麼畫
: 謝謝唷!!
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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 124.8.24.152
推 howchiu:牛刀小用阿 嘖嘖... 05/12 00:44
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作者: Honor1984 (希望願望成真) 看板: TransPhys
標題: Re: [題目] 牛頓力學...
時間: Tue May 13 02:51:24 2008
※ 引述《alwyner (Time is money!!)》之銘言：
: ※ [本文轉錄自 Physics 看板]
: 作者: alwyner (Time is money!!) 看板: Physics
: 標題: [題目] 牛頓力學...
: 時間: Sun Apr 27 13:34:07 2008
: 1.阻力R = kv ，質量m的小鋼球從油的表面(y=0)靜止釋放，k為常數，導出在h深處，此
: 球所到達的速度v。
: Ans: h =?
y is negative for downward
v is negative for downward
v = y'
my" = -mg - ky'
set v = y'
=> v' = -g - (k/m)v
=> v = Aexp(-kt/m) - mg/k
v(t = 0) = 0 => A = mg/k
And y' = v = (mg/k)[exp(-kt/m) - 1]
2
=> y = -g(m/k) exp(-kt/m) - mgt/k + C
2
y(t = 0) = 0 => C = g(m/k)
Therefore,
2 2
y = -g(m/k) exp(-kt/m) - mgt/k + g(m/k)
v = (mg/k)[exp(-kt/m) - 1]
The question asks us to find the │v│ given h, but your answer is h = ??
I'm confused.
: 2. ／
: m ／＼／ │
: ＼／ │
: 夾角 θ ／ M │
: ↘ ／ │
: ／ │
: │￣￣ │
: ￣￣￣￣￣￣￣￣￣
: 忽略所有摩擦，求a)M的加速度 b)m相對於M的加速度。
Let A(->) be the acceleration of M, a the component of the acceleration of m
"relative" to M in the x-direction .
Let N' be the normal force between m and M, N between M and floor.
positive direction:upward and rightward
Assume A>0:(->), a>0:(<-)
Momentum conservation in the x-directon plus differentiation about time:
=> MA + m(-a + A) = 0 ---[1]
Force applied in the y-direction:
=> -ma*tanθ = -(m+M)g + N ---[2]
For big block in the y-direction:
=> -N'cosθ - Mg + N = 0 ---[3]
For big block in the x-direction:
=> N'sinθ = MA ---[4]
There is enough information for you to solve A.
----------------------------
From [3] and [4] => N = M(A*cotθ+ g ) ---[5]
2
Substitute [1] and [5] into [2] => A = mgsinθcosθ/(M + m*sinθ)
the acceleration of m relative to M is a*secθ
2
From [1] => a*secθ = (M + m)A*secθ/m = (M + m)g*sinθ/(M + m*sinθ)
Period.
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◆ From: 122.124.100.190
※ 編輯: Honor1984 來自: 122.124.100.190 (05/13 03:13)