حلول الممارسات – 1 – الإستمرارية

14. 2.4

في هذه الحالة يؤثر المركِّبان الاثنان «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨22px¨»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»F«/mi»«mrow mathcolor=¨#0000FF¨»«mi mathvariant=¨bold¨»X«/mi»«mo mathvariant=¨bold¨»§#160;«/mo»«/mrow»«/msub»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»-«/mo»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#1608;«/mi»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#160;«/mo»«msub»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»W«/mi»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»X«/mi»«/msub»«/mstyle»«/math» في اتجاه أسفل المستوى المائل، ولذلك يمكن تحديد أن قوة الاحتكاك الساكن تؤثِّر إلى أعلى المستوى المائل.

سنُجري تحليلًا متعامدًا لكلٍّ من قوة الضغط العمودية (قوة النورمال) وقوة الجاذبية:

الجسم في حالة سكون، ووفقًا للقانون الأوّل لنيوتن فإن محصّلة القوى المؤثِّرة على الجسم تساوي صفرًا.

نكتب معادلات الحركة، ونُرمز لزاوية ميل المستوى عندما يكون الجسم على عتبة الحركة بـ - «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨22px¨»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#945;«/mi»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»`«/mo»«/mstyle»«/math» :

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نُعوِّض قوة الضغط العمودية (قوة النورمال) من معادلة الحركة على المحور Y في معادلة الحركة على المحور X.

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لإيجاد «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨22px¨»«mi mathvariant=¨bold¨ mathcolor=¨#0000FF¨»§#945;«/mi»«mo mathvariant=¨bold¨ mathcolor=¨#0000FF¨»`«/mo»«/mstyle»«/math» نُعوِّض القيم المعطاة:

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خلاصةً: في هذه الحالة، لكي يبدأ الجسم بالانزلاق إلى أسفل المستوى المائل، يجب أن تكون زاوية ميل المستوى أكبر من 7.23 درجات.