In what direction is the ring getting pulled?

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[Class 12] In what direction is the ring getting pulled? - Teachoo - CBSE Class 12 Sample Paper for 2024 Boards

part 2 - Question 37 (iii) (Choice 2) - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 37 (iii) (Choice 2) - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Transcript

To find direction, we need to find Resultant force Now, Resultant force exerted by teams (š‘­ āƒ—) = š‘Ž āƒ—+š‘ āƒ—+š‘ āƒ— = āˆ’š’Š ˆ + š’‹ ˆ This is in direction of š’ƒ āƒ— . Finding angle of vector š’ƒ āƒ— with respect to x-axis Now, Resultant force (š‘­ āƒ—) = āˆ’š’Š ˆ + š’‹ ˆ And, Vector of x-axis = š‘„ āƒ— = š’Š ˆ Now, š¹ āƒ—.š‘„ āƒ—=|š¹ āƒ— ||š‘„ āƒ— | cosā”ć€–šœƒ 怗 (āˆ’š’Š ˆ + š’‹ ˆ) . š’Š ˆ = āˆššŸ Ɨ 1 Ɨ cos Īø āˆ’1 = √2 Ɨ cos Īø (āˆ’1)/√2 = cos Īø cos Īø = (āˆ’šŸ)/āˆššŸ Since cos is negative ∓ Angle is in second quadrant And, cos š…/šŸ’ = šŸ/āˆššŸ Therefore, Required angle = šœ‹ āˆ’ šœ‹/4 = šŸ‘š…/šŸ’

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