天同師兄,有關Static Equilibrium 既題解

2011-04-01 8:59 pm

天同師兄,你係http://hk.knowledge.yahoo.com/question/question?qid=7011033100279 曾解答左小弟Static Equilibrium 既問題,但我有幾個地方想請教,希望你可以解一解答,唔該!

1."This force R is in a direction perpendicular to the tension T and is pointing towards the vertical that shown on the diagram." 有關呢句小弟有d唔明點解佢同時係R = mg.sin(theta) 而且又係圖中垂直向上既R.

2.大概小弟對向量問題唔純熟,係師兄係話resultant force R such that R = mg.sin(theta). 如果係咁咪姐係mg>T ?想請問點解會作出咁既假設 T不是有機會>mg的嗎?

3.小弟計算時,是將T cos(theta) balance左mg, ie.T cos(theta) =mg, resultant force,F=R'=Tsin(theta). BY solving,F=mg tan(theta) ,既然不知道(theta)的值,為何可以肯定 mg.sin(theta)<mg tan(theta)

回答 (1)

天同

2011-04-02 7:37 am

✔ 最佳答案

Be aware that the R in my answer is not the same as the "R" shown on the diagram of your question.

I have defined R as the resultant force of the tension T and weight of bob mg. Hence, R is given by mg.sin(theta).
In fact, it is not necessary to do any calculation in this problem. The simplest way is to solve it by the "triangle of force" using simple geometry.

The weight (of the bob) vector is vertically downward, whereas the tension T vector is making an angle (theta) with the weight vector. After sketching out these two vectors, the shortest length to complete the "force triangle" is to drop a perpendicular from the end of the weight vector to the tension vector, thus forming a right angle triangle. From the triangle, it is clear that the force F = mg.sin(theta). Since F is represented on the triangle by the perpendicular distance from the weight vector to the tension vector, it is thus the minimun force.

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