פתרונות פרקטיקות- 1- התמדה

14. 2.4

במקרה זה שני הרכיבים «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¨»§#1493;«/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» פועלים בכיוון מורד המישור, לכן ניתן לקבוע שכוח החיכוך הסטטי פועל במעלה המישור.

נבצע הפרדה ישרת זווית לכוח הנורמל ולכוח הכובד:

הגוף מתמיד בתנועתו , מהחוק הראשון של ניוטון שקול הכוחות הפועלים על הגוף שווה לאפס.

נכתוב את משוואות התנועה נסמן את זווית נטיית המישור כאשר הגוף נמצא בסף תנועה ב- «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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נציב את כוח הנורמל ממשוואת התנועה בציר Y במשוואת התנועה בציר X.

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כדי למצוא את «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» נציב את הערכים הנתונים:

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לסיכום: במקרה זה כדי שהגוף יחליק במורד המישור זווית נטיית המישור צריכה להיות גדולה מ- 7.23 מעלות.