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In what direction is the ring getting pulled?

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To find direction, we need to find Resultant force Now, Resultant force exerted by teams (๐‘ญ โƒ—) = ๐‘Ž โƒ—+๐‘ โƒ—+๐‘ โƒ— = โˆ’๐’Š ห† + ๐’‹ ห† This is in direction of ๐’ƒ โƒ— . Finding angle of vector ๐’ƒ โƒ— with respect to x-axis Now, Resultant force (๐‘ญ โƒ—) = โˆ’๐’Š ห† + ๐’‹ ห† And, Vector of x-axis = ๐‘ฅ โƒ— = ๐’Š ห† Now, ๐น โƒ—.๐‘ฅ โƒ—=|๐น โƒ— ||๐‘ฅ โƒ— | cosโกใ€–๐œƒ ใ€— (โˆ’๐’Š ห† + ๐’‹ ห†) . ๐’Š ห† = โˆš๐Ÿ ร— 1 ร— cos ฮธ โˆ’1 = โˆš2 ร— cos ฮธ (โˆ’1)/โˆš2 = cos ฮธ cos ฮธ = (โˆ’๐Ÿ)/โˆš๐Ÿ Since cos is negative โˆด Angle is in second quadrant And, cos ๐…/๐Ÿ’ = ๐Ÿ/โˆš๐Ÿ Therefore, Required angle = ๐œ‹ โˆ’ ๐œ‹/4 = ๐Ÿ‘๐…/๐Ÿ’

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.