Page 6 - كتاب المعلم فيزياء اول ثانوي

‫اﻟﺘﻔﻜﻴﺮ اﻟﻨﺎﻗﺪ‬ ‫ال‪‬ضكل ‪  5-4‬‬ ‫ﹸﺗﺴـﻤﻰ ﻋﻤﻠﻴﺔ ﺗﺠﺰﺋﺔ ﺍﻟﻤﺘﺠﻪ ﺇﻟﻰ ﻣﺮﻛﺒﺎﺗﻪ ﺗﺤﻠﻴﻞ ﺍﻟﻤﺘﺠﻪ‪ .‬ﻻﺣﻆ ﺃﻥ ﺍﻟﻤﺘﺠﻪ ﺍﻷﺻﻠﻲ ﻳﻤﺜﻞ‬
‫العمل ‪‬يات ا‪�‬ضمو‪ ‬باإ‪‬رائها‬ ‫‪‬‬
‫ﺍﻟﻮﺗـﺮ ﻓـﻲ ﻣﺜﻠﺚ ﻗﺎﺋـﻢ ﺍﻟﺰﺍﻭﻳﺔ‪ ،‬ﻣﻤﺎ ﻳﻌﻨﻲ ﺃﻥ ﻣﻘـﺪﺍﺭ ﺍﻟﻤﺘﺠﻪ ﺍﻷﺻﻠﻲ ﻳﻜـﻮﻥ ﺩﺍﺋ ﹰﻤﺎ ﺃﻛﺒﺮ ﻣﻦ‬
‫ﺍﺳـﺄﻝ ﺍﻟﻄـﻼﺏ‪ :‬ﻣـﺎ ﺍﻟﻌﻤﻠ ﹼﻴـﺎﺕ ﺍﳊﺴـﺎﺑﻴﺔ ﺍﻟﺘـﻲ ﳚﻮﺯ‬ ‫‪‬‬ ‫ﻣﻘﺪﺍﺭ ﺃﻱ ﻣﺮ ﹼﻛﺒﺔ ﻣﻦ ﻣﺮﻛﺒﺘﻴﻪ‪.‬‬
‫ﺇﺟﺮﺍﺅﻫـﺎ ﺑﲔ ﺍﻟﻜﻤﻴـﺔ ﺍﳌ ﹼﺘﺠﻬﺔ ﻭﺑﲔ ﺍﻟﻜﻤﻴﺔ ﺍﻟﻘﻴﺎﺳـ ﹼﻴﺔ؟‬
‫ﻻ ﻳﻤﻜـﻦ ﺇﺟﺮﺍﺀ ﻋﻤﻠ ﹼﻴﺔ ﺍﳉﻤﻊ ﺃﻭ ﺍﻟ ﹼﻄﺮﺡ ﺑﲔ ﺍﳌ ﹼﺘﺠﻬﺎﺕ‬ ‫الربع ال‪‬ا‪‬‬ ‫الربع الأو‪‬‬ ‫ﻭﻫﻨﺎﻙ ﺳـﺒﺐ ﺁﺧﺮ ﻻﺧﺘﻴﺎﺭ ﺍﻟﻨﻈﺎﻡ ﺍﻹﺣﺪﺍﺛﻲ‪ ،‬ﻫﻮ ﺃﻥ ﺍﺗﺠﺎﻩ ﺃﻱ ﻣﺘﺠﻪ ﹸﻳﺤ ﱠﺪﺩ ﺑﺎﻟﻨﺴـﺒﺔ ﺇﻟﻰ ﻫﺬﻩ‬
‫ﻭﺍﻟﻜﻤ ﹼﻴـﺎﺕ ﺍﻟﻘﻴﺎﺳـ ﹼﻴﺔ‪ ،‬ﻭﻟﻜﻦ ﻳﻤﻜـﻦ ﴐﺏ ﺍﳌ ﹼﺘﺠﻪ ﰲ‬ ‫ﺍﻹﺣﺪﺍﺛﻴﺎﺕ‪ .‬ﻭﻳﻌ ﹼﺮﻑ ﺍﺗﺠﺎﻩ ﺍﻟﻤﺘﺠﻪ ﻋﻠﻰ ﺃﻧﻪ ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺘﻲ ﻳﺼﻨﻌﻬﺎ ﺍﻟﻤﺘﺠﻪ ﻣﻊ ﻣﺤﻮﺭ ‪ x‬ﻣﻘﻴﺴﺔ‬
‫ﻛﻤﻴﺔ ﻗﻴﺎﺳـ ﹼﻴﺔ‪ .‬ﹸﻳﻐ ﹼﲑ ﺍﻟﴬﺏ ﻃـﻮﻝ ﺍﳌ ﹼﺘﺠﻪ ﺩﻭﻥ ﺃﻥ ﻳﻐ ﹼﲑ‬ ‫ﻓـﻲ ﺍﺗﺠـﺎﻩ ﻋﻜﺲ ﻋﻘﺎﺭﺏ ﺍﻟﺴـﺎﻋﺔ‪ .‬ﻓﻔﻲ ﺍﻟﺸـﻜﻞ ‪ 5-3b‬ﺗﻤﺜـﻞ ﺍﻟﺰﺍﻭﻳﺔ ‪ θ‬ﺍﺗﺠـﺎﻩ ﺍﻟﻤﺘﺠﻪ ‪.A‬‬
‫ﺍﲡﺎﻫﻪ‪ ،‬ﺇﻻ ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻜﻤ ﹼﻴﺔ ﺍﻟﻘﻴﺎﺳـ ﹼﻴﺔ ﺳـﺎﻟﺒﺔ ﻓﻴﻨﻌﻜﺲ‬
‫ﺍﲡـﺎﻩ ﺍﳌﺘﺠﻪ‪ .‬ﻣﺜـ ﹰﻼ ﻋﻤﻠ ﹼﻴﺔ ﻃـﺮﺡ ﺍﳌ ﹼﺘﺠﻬـﺎﺕ ﻋﺒﺎﺭﺓ ﻋﻦ‬ ‫ﹸﻳﻤﻜـﻦ ﻗﻴﺎﺱ ﺃﻃﻮﺍﻝ ﻣﺮ ﹼﻛﺒﺎﺕ ﺍﻟﻤﺘﺠﻬﺎﺕ ﺑﻄﺮﻳﻘﺔ ﺍﻟﺮﺳـﻢ‪ ،‬ﻛﻤﺎ ﻳﻤﻜﻦ ﺃﻳ ﹰﻀﺎ ﺇﻳﺠﺎﺩ ﺍﻟﻤﺮﻛﺒﺎﺕ‬
‫ﴐﺏ ﺍﳌ ﹼﺘﺠـﻪ ﺍﻟﺜﺎﲏ ﰲ ﺍﻟﺮﻗﻢ ‪ ،-1‬ﺛـﻢ ﲨﻌﻪ ﻣﻊ ﺍﳌﺘﺠﻪ‬ ‫ﺑﺎﺳـﺘﻌﻤﺎﻝ ﻋﻠﻢ ﺍﻟﻤﺜﻠﺜﺎﺕ‪ .‬ﻓﺘﺤﺴﺐ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺑﺎﺳـﺘﻌﻤﺎﻝ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﺒﻴﻨﺔ ﺃﺩﻧﺎﻩ‪ ،‬ﻭﺗﻜﻮﻥ‬

‫ﺍﻷﻭﻝ‪1 .‬‬ ‫ﺍﻟﺰﺍﻭﻳﺔ ‪ θ‬ﻣﻘﻴﺴﺔ ﻓﻲ ﻋﻜﺲ ﺍﺗﺠﺎﻩ ﻋﻘﺎﺭﺏ ﺍﻟﺴﺎﻋﺔ ﻣﻦ ﻣﺤﻮﺭ ‪ x‬ﺍﻟﻤﻮﺟﺐ‪.‬‬

‫اﻟﺘﻘﻮﻳﺔ‬ ‫الربع ال‪‬ال‪‬‬ ‫الربع الرابع‬ ‫ﺍﻟﻀﻠﻊ ﺍﳌﺠﺎﻭﺭ‬ ‫=‬ ‫‪Ax‬‬ ‫‪⇒ Ax = Acos θ‬‬
‫‪‬م‪‬ع ا‪ ‬تجه‪‬ات ﻋﻨﺪﻣـﺎ ﺗﻄﺮﺡ ﻋـﲆ ﺍﻟﻄﻼﺏ ﻣﺴـﺄﻟﺔ‬ ‫ﺍﻟﻮﺗﺮ = ‪cos θ‬‬ ‫‪A‬‬

‫ﺷﻔﻬﻴﺔ ﺗﺸﺘﻤﻞ ﻋﲆ ﺍﳌ ﹼﺘﺠﻬﺎﺕ ﰲ ﻣﻀﻤﻮﳖﺎ‪ ،‬ﻗﺪ ﻳﻠﺠﺆﻭﻥ‬ ‫‪sin‬‬ ‫=‪θ‬‬ ‫ﺍﳌﻘﺎﺑﻞ‬ ‫=‬ ‫‪Ay‬‬ ‫‪⇒ Ay = Asin θ‬‬
‫ﺍﻟﻮﺗﺮ‬ ‫‪A‬‬
‫ﻋﻨـﺪ ﺣﻠﻬـﺎ ﺇﱃ ﻋﻤﻠ ﹼﻴﺔ ﺍﳉﻤـﻊ ﺍﳌﺄﻟﻮﻓـﺔ‪ .‬ﻭﻟﺘﺠﻨﺐ ﺫﻟﻚ‬ ‫ال‪‬ض‪‬ك‪‬ل‪  Rx 55 ‬‬
‫ﺍﻃﻠـﺐ ﺇﻟﻴﻬﻢ ﲣ ﹼﻴـﻞ ﺃﻥ ﻻﻋ ﹰﺒﺎ ﻳﺮﻛﺾ ‪ 50 m‬ﰲ ﻣﻠﻌﺐ‪،‬‬ ‫ﻋﻨﺪﻣﺎ ﺗﻜﻮﻥ ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺘﻲ ﻳﺼﻨﻌﻬﺎ ﺍﻟﻤﺘﺠﻪ ﻣﻊ ﻣﺤﻮﺭ ‪ x‬ﺍﻟﻤﻮﺟﺐ ﺃﻛﺒﺮ ﻣﻦ ‪ 90°‬ﻓﺈﻥ ﺇﺷـﺎﺭﺓ‬
‫ﺛـ ﹼﻢ ﻳﻨﻌﻄﻒ ﺑﺰﺍﻭﻳـﺔ ‪ 90°‬ﻭﻳﺮﻛـﺾ ‪ 50 m‬ﺇﺿﺎﻓ ﹼﻴﺔ‪ .‬ﺛﻢ‬ ‫‪ ‬‬ ‫ﺇﺣﺪ￯ ﺍﻟﻤﺮﻛﺒﺘﻴﻦ ﺃﻭ ﻛﻠﺘﻴﻬﻤﺎ ﺗﻜﻮﻥ ﺳﺎﻟﺒﺔ ﻛﻤﺎ ﻓﻲ ﺍﻟﺸﻜﻞ ‪.5-4‬‬
‫ﺍﺳـﺄﳍﻢ‪ :‬ﻛﻢ ﹸﻳﺒﻌﺪ ﺍﻟﻼﻋـﺐ ﻋﻦ ﻧﻘﻄـﺔ ﺍﻟﺒﺪﺍﻳﺔ؟ ‪71 m‬‬ ‫‪RyCBA‬‬
‫ﺇﺫﺍ ﻛﺎﻧﺖ ﺇﺟﺎﺑﺘﻬﻢ ‪ ،100 m‬ﻓﺎﺳﺄﳍﻢ‪ :‬ﳌﺎﺫﺍ ﻗﺎﻡ ﺍﻟﻼﻋﺐ‬ ‫‪ ‬‬ ‫ﺟﻤﻊ اﻟﻤﺘﺠﻬﺎت ﺟﺒﺮ ًّﻳﺎ ‪Algebraic Addition of Vectors‬‬

‫ﲠﺬﻩ ﺍﳉﻮﻟﺔ ﻭﱂ ﻳﺬﻫﺐ ﻣﻦ ﻧﻘﻄﺔ ﺍﻟﺒﺪﺍﻳﺔ ﺇﱃ ﻧﻘﻄﺔ ﺍﻟﻨﻬﺎﻳﺔ‬ ‫‪  ‬‬ ‫ﻟﻤﺎﺫﺍ ﹸﺗﺤ ﹼﻠﻞ ﺍﻟﻤﺘﺠﻬﺎﺕ ﺇﻟﻰ ﻣﺮ ﹼﻛﺒﺎﺗﻬﺎ؟ ﻷﻥ ﺫﻟﻚ ﹸﻳﺴﻬﻞ ﻋﻤﻠﻴﺔ ﺟﻤﻊ ﺍﻟﻤﺘﺠﻬﺎﺕ ﻣﻦ ﺍﻟﻨﺎﺣﻴﺔ‬
‫ﻣﺒﺎﴍﺓ ) ﻗﻄﺮ ﹰﹼﻳﺎ( ؟ ‪1‬‬ ‫‪  C  B  A‬‬
‫‪RyRx ‬‬ ‫ﺍﻟﺤﺴﺎﺑﻴﺔ‪ .‬ﻓﻴﻤﻜﻦ ﺟﻤﻊ ﻣﺘﺠﻬﻴﻦ ﺃﻭ ﺃﻛﺜﺮ ﻣﺜﻞ‪ …. ،C ، B ، A‬ﺇﻟﺦ‪ ،‬ﻭﺫﻟﻚ ﺑﺘﺤﻠﻴﻞ ﻛﻞ ﻣﺘﺠﻪ‬
‫‪       ‬‬ ‫ﺇﻟـﻰ ﻣﺮﻛﺒﺘﻴـﻪ ‪ x‬ﻭ‪ y‬ﺃﻭ ﹰﻻ‪ ،‬ﺛـﻢ ﺗﺠﻤﻊ ﺍﻟﻤﺮﻛﺒـﺎﺕ ﺍﻷﻓﻘﻴﺔ )ﻣﺮﻛﺒﺎﺕ ﺍﻟﻤﺤـﻮﺭ ‪ ( x‬ﻟﻠﻤﺘﺠﻬﺎﺕ‬

‫‪.CBA‬‬ ‫ﻟﺘﻜ ﹼﻮﻥ ﺍﻟﻤﺮﻛﺒﺔ ﺍﻷﻓﻘﻴﺔ ﻟﻠﻤﺤﺼﻠﺔ‪. Rx = Ax + Bx + Cy :‬‬

‫ﻭﺑﺎﻟﻤﺜـﻞ ﺗﺠﻤـﻊ ﺍﻟﻤﺮﻛﺒﺎﺕ ﺍﻟﺮﺃﺳـﻴﺔ )ﻣﺮﻛﺒـﺎﺕ ﺍﻟﻤﺤﻮﺭ‬

‫‪ (y‬ﻟﻠﻤﺘﺠﻬـﺎﺕ ﻟﺘﻜـﻮﻥ ﺍﻟﻤﺮﻛﺒـﺔ ﺍﻟﺮﺃﺳـﻴﺔ ﻟﻠﻤﺤﺼﻠـﺔ‪:‬‬

‫‪ Ry = Ay + By + Cy‬ﻭﻫـﺬﻩ ﺍﻟﻌﻤﻠﻴـﺔ ﻣﻮﺿﺤـﺔ ﺑﻴﺎﻧ ﹰﹼﻴـﺎ‬
‫ﻓﻲ ﺍﻟﺸـﻜﻞ‪ .5-5‬ﻭﻷﻥ ‪ Rx‬ﻭ ‪ Ry‬ﻣﺘﻌﺎﻣﺪﺗﺎﻥ؛ ﻟﺬﺍ ﻳﻤﻜﻦ‬

‫ﺣﺴـﺎﺏ ﻣﻘـﺪﺍﺭ ﺍﻟﻤﺘﺠـﻪ ﺍﻟﻤﺤﺼـﻞ ﺑﺎﺳـﺘﻌﻤﺎﻝ ﻧﻈﺮﻳﺔ‬

‫ﻓﻴﺜﺎﻏﻮﺭﺱ ‪R2 = Rx2 + Ry2‬‬

‫ﻭﻹﻳﺠـﺎﺩ ﺍﻟﺰﺍﻭﻳـﺔ ﺃﻭ ﺍﺗﺠـﺎﻩ ﺍﻟﻤﺤﺼﻠـﺔ ﺗﺬﻛـﺮ ﺃﻥ ﻇـﻞ‬

‫ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺬﻱ ﻳﺼﻨﻌﻪ ﺍﻟﻤﺘﺠﻪ ﺍﻟﻤﺤﺼﻞ ﻣﻊ ﻣﺤﻮﺭ ‪ x‬ﹸﻳﻌ ﹼﺒﺮ ﻋﻨﻪ ﺑﺎﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬

‫ﻟﻠﻤﺘﺠﻪ‬ ‫‪x‬‬ ‫ﺍﳌﺮﻛﺒﺔ‬ ‫ﻋﲆ‬ ‫ﻣﻘﺴﻮﻣﺔ‬ ‫‪y‬‬ ‫ﺍﳌﺮﻛﺒﺔ‬ ‫ﻗﺴﻤﺔ‬ ‫ﳋﺎﺭﺝ‬ ‫=‪θ‬‬ ‫"ﺯﺍﺯﺍﻭﻳﻭﺔﻳ ﺍﺔﳌﺍﺘﳌﺘﺠﻪﺠ ﺍﻪﳌﺍﳌﺤﺤﺼﺼﻞﻞ)ﺗ‪yx‬ﺴ‪R‬ﺎ‪R‬ﻭ(ﻱ‪ 1‬ﺍ‪-‬ﻟ‪n‬ﻈ‪a‬ﻞ‪t‬‬

‫ﺍﻟﻌﻜﴘ‬

‫ﺍﳌﺤﺼﻞ"‪.‬‬

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