Resolve the forces into vertical and horizontal component
For the 8N one
vertical component=8cos45=4root2N(upward)
horizontal component=8cos45=4root2N(to the left)
For the 5N one
vertical component=5cos60=5/2N(upward)
horizontal component=5cos30=5root3/2(to the right)
resultant force of the y intercept=4root2+5/2=(8root2+5)/2N(upward)
resultant force of thex intercept=4root2-5root3/2=(8root2-5root3)/2N(to the left)
The resultant force=root(((8root2+5)/2)^2+((8root2-5root3)/2)^2)=8.26N
tan<==((8root2+5)/2)/((8root2-5root3)/2)
the angle=80.8
The resultant force is 8.26N which make a angle of 80.8 with the x-intercept
2.
Resolve the forces
net force=0
T1cos45=T2cos40---(1)
T1cos45=T2cos50+4000---(2)
sub(1) into (2)
T2cos40=T2cos50+4000
T2=4000/(cos40-cos50)
T2=32500N---(3)
sub(3)into(1)
T1=35200N 1.
Resolve the forces into vertical and horizontal component
For the 8N one
vertical component=8cos45=4root2N(upward)
horizontal component=8cos45=4root2N(to the left)
For the 5N one
vertical component ...
Soldier 發表於 17-4-2009 20:09 http://nakuz.com/bbs/images/common/back.gif
係wo...原來第一題咁計....我錯了/_\
我唔記得要畫番d力先可以加... 本帖最後由 ＊林、 於 17-4-2009 20:43 編輯
係wo...原來第一題咁計....我錯了/_\
我唔記得要畫番d力先可以加...
〝黑之神〞 發表於 17-4-2009 20:13 http://www.nakuz.com/bbs/images/common/back.gif
oh~睇完解釋清楚晒!
鬆左一口氣...
上面唔該晒! 1.
vertical:
8sin45+5sin30
horizontal:
8cos45-5cos30(一個左一個右)
之後用pyth thm就搵到resultant force
wilson1993912 發表於 17-4-2009 12:03 http://www.nakuz.com/bbs/images/common/back.gif
請問點解係咁計? re#14
一個一個向x軸負數走,一個正數,所以大個個減細
y軸個度因為同方向,所以加埋
再將2個square加埋佢,再開方就可以, 第10題係8.26==? 我計 8cos45+5cos60=8.16N
T2sin40+T1sin45=4000...I
T2cos40+T1cos45=0....II
I-II=T2sin40-T2cos40=4000
T2=-32452.6=-32500N
-32456.26177COS40+T1cos45=0
T1=-35161.5=35200N
〝黑之神〞 發表於 17-4-2009 18:01 http://www.nakuz.com/bbs/images/common/back.gif
我有D野唔明..
點解係加 唔係減-.-?
方向好似相反都係加? 16# screen
原本應該係-,但係佢寫做+,所以答案先至會計到負數=.=
第二條式都係一樣
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