Lecture 1 Statics (Parallelogram Law) General 2025 PDF
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2025
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This lecture covers the parallelogram law for vector addition of forces, and solving problems in statics. It demonstrates vector analysis techniques and provides worked examples of calculating forces and resultant forces, also includes calculations.
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Introduction Parallelogram Law 𝜽𝑹 𝑭𝑹 = 𝟐𝟐 + 𝟔𝟐 − 𝟐 × 𝟐 × 𝟔𝒄𝒐𝒔𝟏𝟎𝟓 = 𝟔. 𝟖 𝑵 𝑭𝑹 𝟔 𝜷 = 𝒔𝒊𝒏𝟏𝟎𝟓 𝒔𝒊𝒏𝜷 𝜷 = 𝟓𝟖. 𝟓° 𝜽𝑹 = 𝟒𝟓 + 𝟓𝟖. 𝟓 = 𝟏𝟎𝟑. 𝟓° 𝟏𝟎𝟓°...
Introduction Parallelogram Law 𝜽𝑹 𝑭𝑹 = 𝟐𝟐 + 𝟔𝟐 − 𝟐 × 𝟐 × 𝟔𝒄𝒐𝒔𝟏𝟎𝟓 = 𝟔. 𝟖 𝑵 𝑭𝑹 𝟔 𝜷 = 𝒔𝒊𝒏𝟏𝟎𝟓 𝒔𝒊𝒏𝜷 𝜷 = 𝟓𝟖. 𝟓° 𝜽𝑹 = 𝟒𝟓 + 𝟓𝟖. 𝟓 = 𝟏𝟎𝟑. 𝟓° 𝟏𝟎𝟓° 𝟒𝟓𝟎 𝑭𝑨𝑩 𝑭𝑨𝑪 = = 𝒔𝒊𝒏𝟑𝟎 𝒔𝒊𝒏𝟏𝟎𝟓 𝒔𝒊𝒏𝟒𝟓 𝟒𝟓𝟎𝒔𝒊𝒏𝟏𝟎𝟓 𝑭𝑨𝑩 = = 𝟖𝟔𝟗. 𝟑 𝑵 𝒔𝒊𝒏𝟑𝟎 𝟒𝟓𝟎𝒔𝒊𝒏𝟒𝟓 𝑭𝑨𝑪 = = 𝟔𝟑𝟔. 𝟒 𝑵 𝒔𝒊𝒏𝟑𝟎 𝑭𝑨𝑪 𝑭𝑩 𝑭𝑹 ° 𝑩 𝑹 𝑩 𝑨 𝑹 𝑩 𝑭𝑹 𝟓𝟎𝟎 𝑵 𝜽 𝜑 𝑭 = 𝟔𝟎𝟎 𝑵 𝑭 𝟓𝟎𝟎 𝑭𝑹 = = 𝒔𝒊𝒏𝜽 𝒔𝒊𝒏𝟑𝟎 𝒔𝒊𝒏𝝋 𝟔𝟎𝟎 𝟓𝟎𝟎 𝑰𝒇 𝑭 = 𝟔𝟎𝟎𝑵, = 𝒔𝒊𝒏𝜽 𝒔𝒊𝒏 𝟑𝟎 𝟔𝟎𝟎𝒔𝒊𝒏𝟑𝟎 𝒔𝒊𝒏𝜽 = = 𝟑𝟔. 𝟗° 𝟓𝟎𝟎 𝝋 = 𝟏𝟖𝟎 − (𝟑𝟎 + 𝜽) = 𝟏𝟖𝟎 − (𝟑𝟎 + 𝟑𝟔. 𝟗) = 𝟏𝟏𝟑. 𝟏° 𝟓𝟎𝟎𝒔𝒊𝒏𝟏𝟏𝟑. 𝟏 𝑭𝑹 = = 𝟗𝟏𝟗. 𝟔 𝑵 𝒔𝒊𝒏𝟑𝟎
