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Methods of Finding Vector Sum

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1 Methods of Finding Vector Sum
Phys 13 General Physics 1 Methods of Finding Vector Sum MARLON FLORES SACEDON

2 Vector analysis Graphical Method Analytical Method
Methods of finding Vector sum or resultant of forces Parallelogram Method Polygon Method Graphical Method Cosine law Method Component Method Analytical Method

3 Vector analysis Two forces Graphical Method Parallelogram Method ๐‘…
๐œƒ ๐›ฝ ๐›ฝ ๐น 2 ๐นโ€ฒ 2 ๐œƒ ๐น 1 ๐นโ€ฒ 1 ๐œƒ ๐›ฝ Two forces (Drawn to scale)

4 Vector analysis Two forces Graphical Method Polygon Method ๐น 2 ๐น 1
๐œƒ ๐›ฝ Two forces (Drawn to scale)

5 Vector analysis Two forces Graphical Method Polygon Method ๐‘… ๐น 2 ๐น 2
๐›ฝ ๐œƒ ๐น 1 ๐น 1 ๐œƒ ๐›ฝ ๐œƒ Two forces (Drawn to scale) This is Polygon method

6 Vector analysis Two forces Analytical Method Cosine law Method ๐‘… ๐น 2
๐œƒ ๐›ฝ ๐น 2 ๐น 1 Equivalent polygon of vectors (Drawn not to scale)) ๐น 1 ๐›ฝ ๐œƒ ๐น 2 Two forces Drawn not to scale Apply law sines to solve for ๐›ผ: ๐‘ ๐‘–๐‘›๐›ผ ๐น 2 = ๐‘ ๐‘–๐‘› ๐œƒ+๐›ฝ ๐‘… ๐›ผ ๐›พ Magnitude of the Resultant R (apply law of cosines) ๐‘… or ๐‘…= ๐น ๐น 2 2 โˆ’2 ๐น 1 ๐น 2 cosโก(๐œƒ+๐›ฝ) Direction of the Resultant R ๐›พ= 180 ๐‘œ โˆ’๐œƒโˆ’๐›ผ

7 Vector analysis Analytical Method Component Method
Vector can be resolve in x & y components + ๐ด ๐‘ฆ ๐ด ๐œƒ + ๐ด ๐‘ฅ

8 Vector analysis Analytical Method Component Method
Vector can be resolve in x & y components Components and its vector can be formed into a right triangle. + ๐ด ๐‘ฆ ๐œƒ ๐ด + ๐ด ๐‘ฆ ๐ด ๐‘จ = ๐‘จ ๐’™ + ๐‘จ ๐’š Resolution of vectors Composition of vectors ๐œƒ + ๐ด ๐‘ฅ + ๐ด ๐‘ฅ From right triangle (left side figure) ๐ด= ๐ด ๐‘ฅ ๐ด ๐‘ฆ 2 ๐‘๐‘œ๐‘ ๐œƒ= ๐ด ๐‘ฅ ๐ด ๐ด ๐‘ฅ =๐ด๐‘๐‘œ๐‘ ๐œƒ ๐œƒ= ๐‘‡๐‘Ž๐‘› โˆ’1 ๐ด ๐‘ฆ ๐ด ๐‘ฅ ๐‘ ๐‘–๐‘›๐œƒ= ๐ด ๐‘ฆ ๐ด ๐ด ๐‘ฆ =๐ด๐‘ ๐‘–๐‘›๐œƒ

9 Vector analysis Two forces or more Analytical Method Component Method
Steps in Component Method Step 2: Sum up all components along x-axis and all components along y-axis algebraically. Step 1: Resolve the vectors into each components and solve their values using ๐ด ๐‘ฅ =๐ด๐‘๐‘œ๐‘ ๐œƒ & ๐ด ๐‘ฆ =๐ด๐‘ ๐‘–๐‘›๐œƒ. ๐น ๐‘–๐‘ฅ = ๐น ๐‘– ๐‘๐‘œ๐‘ ๐œƒ From the Figure (left) we have, ๐›ฝ ๐น 2 + ๐น 2๐‘ฆ + ๐น 1๐‘ฆ ๐‘… ๐‘ฅ =โˆ’ ๐น 1๐‘ฅ + ๐น 2๐‘ฅ +โ€ฆ ๐น 1 ๐œƒ ๐น ๐‘–๐‘ฆ = ๐น ๐‘– ๐‘ ๐‘–๐‘›๐œƒ Components of Resultant ๐‘… along x & y axis ๐‘… ๐‘ฆ = ๐น 1๐‘ฆ + ๐น 2๐‘ฆ +โ€ฆ Step 3: Calculate the magnitude and direction of the resultant using ๐‘… ๐‘ฅ and ๐‘… ๐‘ฆ . โˆ’ ๐น 1๐‘ฅ ๐น 2๐‘ฅ ๐‘…= ฮฃ๐‘… ๐‘ฅ ฮฃ๐‘… ๐‘ฆ 2 Magnitude of resultant Two forces or more Drawn not to scale ๐›พ= ๐‘‡๐‘Ž๐‘› โˆ’1 ๐‘… ๐‘ฆ ๐‘… ๐‘ฅ direction of resultant

10 Vector analysis Steps in Component Method + ๐น 1๐‘ฆ ๐‘… ๐‘ฅ =โˆ’ ๐น 1๐‘ฅ + ๐น 2๐‘ฅ +โ€ฆ
Step 1: Resolve the vectors into each components and solve their values using ๐ด ๐‘ฅ =๐ด๐‘๐‘œ๐‘ ๐œƒ & ๐ด ๐‘ฆ =๐ด๐‘ ๐‘–๐‘›๐œƒ. Step 2: Sum up all components along x-axis and all components along y-axis algebraically. From the Figure (left) we have, ๐›ฝ ๐น 2 + ๐น 1๐‘ฆ ๐‘น ๐‘… ๐‘ฅ =โˆ’ ๐น 1๐‘ฅ + ๐น 2๐‘ฅ +โ€ฆ ๐น 1 ๐œƒ ๐‘… ๐‘ฆ = ๐น 1๐‘ฆ + ๐น 2๐‘ฆ +โ€ฆ Step 3: Calculate the magnitude and direction of the resultant using ๐‘… ๐‘ฅ and ๐‘… ๐‘ฆ . โˆ’ ๐น 1๐‘ฅ + ๐น 2๐‘ฆ ๐‘…= ฮฃ๐‘… ๐‘ฅ ฮฃ๐‘… ๐‘ฅ 2 Magnitude of resultant ๐›พ= ๐‘‡๐‘Ž๐‘› โˆ’1 ๐‘… ๐‘ฆ ๐‘… ๐‘ฅ direction of resultant ๐›พ ๐น 2๐‘ฅ

11 Vector analysis Problem Find the magnitude and direction of resultant
using polygon method, cosine law and component method. 200๐‘”๐‘“ 30 ๐‘œ 65 ๐‘œ 500๐‘”๐‘“ 35 ๐‘œ 400๐‘”๐‘“

12 Vector analysis ๐œธ Application Problem
end Displacement (D) ๐ท ๐‘ฆ 8 km Application Problem A cross-country skier skis 5.00 km in the direction 500 east of south, then 3.00 km in the direction N 600 E, and finally 8.00 km with bearing angle of Find the displacement of the skier. N E S W ๐œธ Solving for the x component of displacement =2.82 ๐‘˜๐‘š ๐ท ๐‘ฅ start ๐ท ๐‘ฅ = +5 cos 50 ๐‘œ 50o + 3 cos 30 ๐‘œ โˆ’8 cos 68 ๐‘œ 5 km Solving for the y component of displacement N E S W ๐ท ๐‘ฆ = โˆ’ 5 sin 50 ๐‘œ + 3 sin 30 ๐‘œ + 8 sin 68 ๐‘œ =5.09 ๐‘˜๐‘š 68o 338o Solving for the magnitude and direction of displacement 3 km N E S W 60o ๐ท= =5.82 ๐‘˜๐‘š 30o ๐›พ= ๐‘‡๐‘Ž๐‘› โˆ’ = 61 ๐‘œ East of North

13 eNd


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