Isibonelo Sombuzo Oxoxa Ngokungezwa Kwamavektha Amabili Kusetshenziswa Indlela Ye-Parallelogram
Ukwengezwa kwevektha kuwumqondo obalulekile ku-physics kanye nezibalo, ovame ukusetshenziselwa ukuchaza izenzakalo zemvelo nezinkinga zokuphila kwansuku zonke. Kunezindlela eziningana zokwengeza amavektha amabili, enye yazo indlela ye-parallelogram. Le ndlela ayigcini nje ngokunembile kodwa futhi inikeza umbono onamandla wokuthi amavektha amabili ahlangana kanjani ukwakha i-vektha ephumelayo. Kulesi sihloko, sizobheka izibonelo eziningana zokwengezwa kwevektha kusetshenziswa indlela ye-parallelogram, kanye nezixazululo zazo.
Kuyini i-Vector?
Ngaphambi kokuthi singene ezinkingeni zezibonelo, sidinga ukuqonda incazelo eyisisekelo yevektha. Ivektha iyinani elinobukhulu (ubude) kanye nesiqondiso. Izibonelo zakudala zamavektha zifaka phakathi ijubane, ukusheshisa, amandla, kanye nokufuduka. Ivektha ingamelwa njengezingxenye zayo (i, j, k) kuma-coordinates e-Cartesian noma njengobude bayo kanye nesiqondiso (i-angle).
Indlela ye-Paralelogram
Indlela ye-parallelogram iyindlela eyodwa yokwengeza amavekhtha amabili. Kule ndlela, simelela amavekhtha amabili njengezinhlangothi ezimbili ze-parallelogram. Ivekhtha ephumelayo iyi-diagonal ye-parallelogram eqala kusukela endaweni yokuqala yamavekhtha amabili. Ngokwezibalo, uma sinevekhtha ezimbili \(\vec{A}\) kanye \(\vec{B}\), umphumela ngu-\( \vec{R} = \vec{A} + \vec{B} \).
Indlela yesinyathelo ngesinyathelo yokusebenzisa indlela ye-parallelogram imi kanje:
1. Dweba i-vector \(\vec{A}\) kusukela ekuqaleni.
2. Kusukela ekugcineni kwevektha \(\vec{A}\), dweba ivektha \(\vec{B}\).
3. Dweba umugqa ohambisana nevektha \(\vec{B}\) kusukela endaweni yokuqala \(\vec{A}\).
4. Dweba umugqa ohambisana nevektha \(\vec{A}\) kusukela ekugcineni kwevektha \(\vec{B}\).
5. Dweba umugqa ovundlile kusukela endaweni yokuqala kuya ekhoneni eliphambene ukuze uthole i-vector ephumelayo \(\vec{R}\).
Imibuzo Eyisibonelo Nengxoxo
umbuzo 1
Ake sithi sinezivektha ezimbili \(\vec{A}\) kanye \(\vec{B}\):
– \(\vec{A}\) inobude (ubukhulu) bamayunithi ama-5 kanye nesiqondiso esingu-0° (noma eceleni kwe-x-axis enhle),
– \(\vec{B}\) inobude obungamayunithi ama-3 kanye nesiqondiso esingu-90° (noma eceleni kwe-y-axis enhle).
Iyini inani eliwumphumela lokwengeza la mavektha amabili usebenzisa indlela ye-parallelogram?
Ingxoxo:
1. Dweba i-vector \(\vec{A}\) eceleni kwe-x-axis elungile enobude obungamayunithi ama-5.
2. Kusukela ekugcineni kwevektha \(\vec{A}\), dweba ivektha \(\vec{B}\) eceleni kwe-y-axis elungile enobude obungamayunithi amathathu.
3. Kusukela ekuqaleni \(\vec{A}\), dweba umugqa ohambisana no \(\vec{B}\).
4. Kusukela ekugcineni kwe-\(\vec{B}\), dweba umugqa ohambisana ne-\(\vec{A}\).
5. Umphumela uba yi-parallelogram ene-diagonal eyi-vector ephumela \(\vec{R}\).
Njengoba i-\(\vec{A}\) kanye ne-\(\vec{B}\) ziqondene, singasebenzisa i-Pythagorean theorem ukubala ubude be-vector ephumela:
\[ R = \sqrt{A^2 + B^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \cishe 5.83 \]
Isiqondiso sevektha ephumayo singabalwa kusetshenziswa i-trigonometry. Uma i-\(\theta\) iyi-engeli ephakathi kwe-ephumayo kanye ne-\(\vec{A}\):
\[ \tan(\theta) = \frac{B}{A} = \frac{3}{5} \]
ngakho-ke:
\[ \theta = \tan^{-1}\left(\frac{3}{5}\right) \approx 30.96^\circ \]
Ngakho-ke, i-vector ephumelayo \(\vec{R}\) inobukhulu obungaba amayunithi angu-5.83 kanye nesiqondiso esingaba ngu-30.96° kusuka ku-\(\vec{A}\).
umbuzo 2
Amavekhtha amabili \(\vec{C}\) kanye \(\vec{D}\) anikezwe kanje:
– \(\vec{C}\) enobude obungamayunithi angu-4 kanye nesiqondiso esingu-45°.
– \(\vec{D}\) enobude obungamayunithi ayi-6 kanye nesiqondiso esingu-120°.
Nquma i-vector ephumela \(\vec{R}\) kusukela ekufakweni kwama-vector amabili.
Ingxoxo:
Ukuze wengeze amavekhtha amabili angaqondile komunye nomunye noma ngezimo ezihlukile, ungasebenzisa izingxenye zeCartesian.
1. Hlukanisa i-\(\vec{C}\) kanye ne-\(\vec{D}\) ibe izingxenye ze-x kanye ne-y.
Nge-\(\vec{C}\):
\[ C_x = C \cos(45^\circ) = 4 \cos(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \cishe 2.83 \]
\[ C_y = C \sin(45^\circ) = 4 \sin(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \cishe 2.83 \]
Ukuze \(\vec{D}\):
\[ D_x = D \cos(120^\circ) = 6 \cos(120^\circ) = 6 \cdot (-\frac{1}{2}) = -3 \]
\[ D_y = D \sin(120^\circ) = 6 \sin(120^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \cishe 5.20 \]
2. Engeza izingxenye zika-x no-y zazo zombili izivektha:
\[ R_x = C_x + D_x = 2.83 + (-3) = -0.17 \]
\[ R_y = C_y + D_y = 2.83 + 5.20 = 8.03 \]
3. Bala ubukhulu kanye nesiqondiso sevektha ephumela \(\vec{R}\):
\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-0.17)^2 + 8.03^2} = \sqrt{0.03 + 64.48} = \sqrt{64.51} \cishe 8.03 \]
\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{8.03}{-0.17}\right) \approx \tan^{-1}(-47.24) \]
Njengoba umphumela ungemuhle, sifaka u-180° ukuze sithole i-engeli ohlelweni olufanele lwe-quadrant:
\[ \theta \approx \tan^{-1}(47.24) + 180^\circ \approx 271.93^\circ \]
Ngakho-ke, i-vector ephumayo \(\vec{R}\) inobukhulu obungamayunithi angaba ngu-8.03 kanye nesiqondiso esingaba ngu-271.93°, noma singasho cishe u-91.93° kusukela ku-x-axis engemihle ku-quadrant yesine.
I-Penutup
Indlela ye-parallelogram iyindlela ephumelelayo nebonakalayo yokwengeza ama-vector amabili. Nakuba le ndlela ingase ibonakale ilula kuma-vector alula, kubalulekile ukuqonda ukuthi kuma-vector ayinkimbinkimbi kakhulu, sivame ukusebenzisa izingxenye ze-Cartesian kanye namasu e-algebraic athuthukile ukuze sithole imiphumela enembile. Ngethemba ukuthi izibonelo ezingenhla zinikeza isithombe esicacile sendlela le ndlela engasetshenziswa ngayo ezimweni ezahlukene.
