Two tugboats A and B are pulling a shop as shown.
a) Find the ratio of their pulling forces.
b) if the ship moves at a constant velocity and the water resistance on it is 10000N, what are the puuling forces?
(PLX BROWSE this website to see the figure...)
http://hk.geocities.com/yee_lamlam/fig2.bmp
PLS ANS STEP BY STEP ....THX A LOT!!!!
PHY !!!
2007-01-01 8:18 am
回答 (3)
2007-01-01 9:04 am
✔ 最佳答案
Sorry i want to use Chinese to answer your question.If you need, you may ask me to translate into english.
第一
因為力ａ和力ｂ是一個斜力，所以你要將佢地拆做打直同打橫的力。
如果隻船是沿著直線運動被拉動的話，
沒有向左或右被拉動。
因為水平的力數值一樣，可以約。
所以，（打橫力ａ）／力ａ＝ｓｉｎ４０
，（打橫力ｂ）／力ｂ＝ｓｉｎ３０
因為打橫力ａ＆ｂ一樣，
力ａ：力ｂ ＝ （１／ｓｉｎ４０）：（１／ｓｉｎ３０）
＝ ｓｉｎ３０ ： ｓｉｎ４０
第二
如果隻move with uniform velocity, acceleration = 0, 所以net force = 0 (Fnet=ma)
即是話ａ＆ｂ的向上力數值係１００００ｎ同水阻力一樣的
ａ＆ｂ的向上力又分別是
（打橫力ａ）／ｔａｎ４０ 及 （打橫力ｂ）／ｔａｎ３０
而且打橫力ａ＝ｂ
所以，橫力/tan40 + 橫力/tan30 = 10000
橫力=10000tan30tan40/(tan30+tan40)
= 3420N
係第一part我講左 ,所以
力a = 橫力/sin40 = 3420/sin40 =5320N
力b = 橫力/sin30 = 3420/sin30 =6840N
參考: 自己(F.5student)如果有另外的答案就用另一個，冇先用我果個= =學校冇教唔肯定（自己推的)
2007-01-01 8:49 am
a)
Using the fact that the ship is moving in the forward direction, we can deduce that the net force in the horizontal direction should be zero.
Let the tensions in the tugboats A and B are Ta and Tb resp.
Then Ta sin 40 = Tb sin 30 (to balance out the horizontal component)
Ta / Tb = sin 30 / sin 40 = 0.778
b) Subce the ship is moving forward with constant velocity, net force on it should be zero and hence for the forward component of the tensions:
Ta cos 40 + Tb cos 30 = 10000 (in order to balance the resistance of water)
Ta cos 40 + 0.778Ta cos 30 = 10000
So Ta = 6946 N
Hence Tb = Ta / 0.778 = 8930 N
2007-01-01 00:53:08 補充：
Sorry the last 3 steps should be amended as:Ta cos 40 (Ta/0.778) cos 30 = 10000So Ta = 5321 NHence Tb = Ta / 0.778 = 6840 N
Using the fact that the ship is moving in the forward direction, we can deduce that the net force in the horizontal direction should be zero.
Let the tensions in the tugboats A and B are Ta and Tb resp.
Then Ta sin 40 = Tb sin 30 (to balance out the horizontal component)
Ta / Tb = sin 30 / sin 40 = 0.778
b) Subce the ship is moving forward with constant velocity, net force on it should be zero and hence for the forward component of the tensions:
Ta cos 40 + Tb cos 30 = 10000 (in order to balance the resistance of water)
Ta cos 40 + 0.778Ta cos 30 = 10000
So Ta = 6946 N
Hence Tb = Ta / 0.778 = 8930 N
2007-01-01 00:53:08 補充：
Sorry the last 3 steps should be amended as:Ta cos 40 (Ta/0.778) cos 30 = 10000So Ta = 5321 NHence Tb = Ta / 0.778 = 6840 N
參考: My Physics knowledge
2007-01-01 8:38 am
a) Let TA and TB be the tensions by the tugboats A and B, respectively.
Since the ship is moving in the vertical direction, the resultant force of their horizontal components is zero.
TB sin 30º - TA sin 40º = 0
0.643TA = 0.5TB
TA:TB = 0.5/0.643
= 0.778:1
So the ratio of their pulling forces TA:TB is 0.778:1.
b) Since the ship is moving at a constant velocity, the acceleration of it in the vertical direction is also zero.
TA cos 40º + TB cos 30º - 10000 = 0
0.766TA + 0.866TB = 10000
0.766TA + 0.866(1/0.778)TA = 10000
1.879TA = 10000
TA = 5322 (N)
TB = TA / 0.778
= 5322 / 0.778
= 6841 (N)
Hence, the pulling forces by tugboats A and B are 5322 N and 6841 N, respectively.
Since the ship is moving in the vertical direction, the resultant force of their horizontal components is zero.
TB sin 30º - TA sin 40º = 0
0.643TA = 0.5TB
TA:TB = 0.5/0.643
= 0.778:1
So the ratio of their pulling forces TA:TB is 0.778:1.
b) Since the ship is moving at a constant velocity, the acceleration of it in the vertical direction is also zero.
TA cos 40º + TB cos 30º - 10000 = 0
0.766TA + 0.866TB = 10000
0.766TA + 0.866(1/0.778)TA = 10000
1.879TA = 10000
TA = 5322 (N)
TB = TA / 0.778
= 5322 / 0.778
= 6841 (N)
Hence, the pulling forces by tugboats A and B are 5322 N and 6841 N, respectively.
收錄日期: 2021-04-25 19:18:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070101000051KK00079