已知有一輛巴士其懸吊系統的設計會使該輛巴士以a的加／減速率加／減速時，其車身的地面會與水平面傾斜的角度為ka，其中k是一個常數。
現在有一件質量m的重物放在該輛巴士的地面上，該重物與該輛巴士的地面構成的摩擦系數是μ。求該輛巴士在加／減速時不會令該重物滑動的最大加／減速率。請詳述其計算過程。
慣性力學計算題
2010-03-15 9:47 am
回答 (2)
2010-03-15 10:42 pm
✔ 最佳答案
巴士向前加速時, 車身會向後傾 (即乘客視會仰望天空). 而巴士向後加速 (即向前行走但減速) 時, 車身會向前傾 (即乘客視會俯瞰地面).所以兩個情況皆可以下圖描述 (設 a 為最大加/減速率):
圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Mar10/Doraemon1.jpg
將所有力拆成垂直及水平分量時:
圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Mar10/Doraemon2.jpg
參考: Myself
2010-03-15 5:26 pm
During acceleration, for equilibrium of an object rested on the floor of the bus, which inclined at an angle (ka) with the horizonta, the max frictional force must be equal to the weight component along the inclined floor. Hence,
mg.sin(ka) = μ[mg.cos(ka)
where m is the mass of the object, and g is the acceleration due to gravity
i.e. tan(ka) = μ
ka = arc-tan(μ)
a = [arc-tan(μ)]/k
mg.sin(ka) = μ[mg.cos(ka)
where m is the mass of the object, and g is the acceleration due to gravity
i.e. tan(ka) = μ
ka = arc-tan(μ)
a = [arc-tan(μ)]/k
收錄日期: 2021-04-29 17:33:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100315000051KK00180