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![CROSS PRODUCT
STEP 1: WRITE THE CROSS PRODUCT IN MATRIX FORMAT
STEP 2: WRITE THE COMPONENT USING THE DETERMINANT FROM THE STEP 1.
STEP 3: SIMPLIFY THE DETERMINANT
STEP 4: SIMPLIFY
𝐴⃑ × 𝐵⃑ =
𝑖 𝑗 𝑘
𝐴𝑥 𝐴𝑦 𝐴𝑧
𝐵𝑥 𝐵𝑦 𝐵𝑧
𝐴⃑ × 𝐵⃑ =
𝐴𝑦 𝐴𝑧
𝐵𝑦 𝐵𝑧
𝑖 −
𝐴𝑥 𝐴𝑧
𝐵𝑥 𝐵𝑧
j+
𝐴𝑥 𝐴𝑦
𝐵𝑥 𝐵𝑦
k
𝐴⃑ × 𝐵⃑ = [(𝐴𝑦𝐵𝑧 − 𝐴𝑧𝐵𝑦)𝑖̂ - (𝐴x𝐵z − 𝐴z𝐵x)𝑗̂ + (𝐴𝑥𝐵𝑦 − 𝐴𝑦𝐵𝑥)𝑘̂]](https://image.slidesharecdn.com/work-220420150043/85/WORK-pdf-6-320.jpg)
![EXAMPLE
𝐴⃑ = -2𝑖̂+ 4𝑗̂-7 𝑘̂
𝐵⃑ = -4𝑖̂+ 10𝑗̂+ 5𝑘̂
𝐴⃑ × 𝐵⃑ =
𝑖 𝑗 𝑘
−2 4 −7
−4 10 5
𝐴⃑ × 𝐵⃑ =
4 −7
10 5
𝑖 −
−2 −7
−4 5
j+
−2 4
−4 10
k
𝐴⃑ × 𝐵⃑ = [(4·5 − (-7)·10)𝑖̂ - (-2·5 − (-7)·(-4))𝑗̂ + (-2·10 − 4·(-4))𝑘̂]
𝐴⃑ × 𝐵⃑ = [(20+70)𝑖̂ - (-10-28)𝑗̂ + (-20+16)𝑘̂] = [90𝑖̂ ,38 𝑗̂ ,-4 𝑘̂]
𝐴⃑ × 𝐵⃑ =
𝑖 𝑗 𝑘
−2 4 −7
−4 10 5
𝐴⃑ × 𝐵⃑ =
𝑖 𝑗 𝑘
−2 4 −7
−4 10 5](https://image.slidesharecdn.com/work-220420150043/85/WORK-pdf-7-320.jpg)
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WORK.pdf
- 1. WORK DONE TEACHER: Mr.TRISTAN RYAN TINGSON
- 2. OBJECTIVES At the endof this lesson, you will be able to: 1. calculate the dot or scalar product of vectors; 2. determine the work done by a force acting on a system; 3. define work as a scalar or dot product of force and displacement; and 4. interpret the work done by a force in one- dimension as an area under a Force vs. Position curve. 2
- 3. MULTIPLICATION OF VECTORS DOTPRODUCT The dot product of two vectors A and B is called scalar product. CROSS PRODUCT Also is called as vector product 3
- 4. DOT PRODUCT 4 4 If, 𝐴⃑ =𝐴𝑥𝑖̂+ 𝐴𝑦𝑗̂+ 𝐴𝑧𝑘̂ 𝐵⃑ = 𝐵𝑥𝑖̂+ 𝐵𝑦𝑗̂+ 𝐵𝑧𝑘̂ then the dot product of 𝐴⃑ and 𝐵⃑⃑ is, 𝐴⃑ ∙ 𝐵⃑ = 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦 + 𝐴𝑧𝐵z Angle between two vectors Magnitude
- 5. 5 EXAMPLE MAGNITUDE OF INDIVIDUALVECTORS MAGNITUDE SOLVE FOR THE COS θ 𝐴⃑ = (3i,6j,7k) , 𝐵⃑ = (8i,4j,9k) 𝐴⃑· 𝐵⃑ = (3)(8)+(6)(4)+(7)(9) 𝐴⃑· 𝐵⃑ = 111 𝐴 = 32 + 62 + 72 𝐵 = 82 + 42 + 92 𝐴 = 94 𝐵 = 161 θ = 𝑐𝑜𝑠−1 111 94· 161 θ = 25.54° 𝐴⃑ ∙ 𝐵⃑ = 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦 + 𝐴𝑧𝐵z
- 6. CROSS PRODUCT STEP 1:WRITE THE CROSS PRODUCT IN MATRIX FORMAT STEP 2: WRITE THE COMPONENT USING THE DETERMINANT FROM THE STEP 1. STEP 3: SIMPLIFY THE DETERMINANT STEP 4: SIMPLIFY 𝐴⃑ × 𝐵⃑ = 𝑖 𝑗 𝑘 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 𝐴⃑ × 𝐵⃑ = 𝐴𝑦 𝐴𝑧 𝐵𝑦 𝐵𝑧 𝑖 − 𝐴𝑥 𝐴𝑧 𝐵𝑥 𝐵𝑧 j+ 𝐴𝑥 𝐴𝑦 𝐵𝑥 𝐵𝑦 k 𝐴⃑ × 𝐵⃑ = [(𝐴𝑦𝐵𝑧 − 𝐴𝑧𝐵𝑦)𝑖̂ - (𝐴x𝐵z − 𝐴z𝐵x)𝑗̂ + (𝐴𝑥𝐵𝑦 − 𝐴𝑦𝐵𝑥)𝑘̂]
- 7. EXAMPLE 𝐴⃑ = -2𝑖̂+4𝑗̂-7 𝑘̂ 𝐵⃑ = -4𝑖̂+ 10𝑗̂+ 5𝑘̂ 𝐴⃑ × 𝐵⃑ = 𝑖 𝑗 𝑘 −2 4 −7 −4 10 5 𝐴⃑ × 𝐵⃑ = 4 −7 10 5 𝑖 − −2 −7 −4 5 j+ −2 4 −4 10 k 𝐴⃑ × 𝐵⃑ = [(4·5 − (-7)·10)𝑖̂ - (-2·5 − (-7)·(-4))𝑗̂ + (-2·10 − 4·(-4))𝑘̂] 𝐴⃑ × 𝐵⃑ = [(20+70)𝑖̂ - (-10-28)𝑗̂ + (-20+16)𝑘̂] = [90𝑖̂ ,38 𝑗̂ ,-4 𝑘̂] 𝐴⃑ × 𝐵⃑ = 𝑖 𝑗 𝑘 −2 4 −7 −4 10 5 𝐴⃑ × 𝐵⃑ = 𝑖 𝑗 𝑘 −2 4 −7 −4 10 5
- 8. WORK 8 work done onan object by an applied force is defined as the product of the magnitude of the displacement multiplied by the component of the force parallel to the displacement.
- 9. 9 WORK FORMULA SI UNITTYPE OF QUANTITY 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 𝑊 = 𝐹𝑑 Where: F = force W = work d = displacement Newton-meter (Nm) or Joule (J) Scalar
- 10. 10 WORK 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 𝑊= 𝐹𝑐𝑜𝑠0𝑑 𝑊 = 𝐹(1)𝑑 𝑊 = 𝐹𝑑 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 𝑊 = 𝐹𝑐𝑜𝑠90𝑑 𝑊 = 𝐹(0)𝑑 𝑊 = 0 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 𝑊 = 𝐹𝑐𝑜𝑠(180)𝑑 𝑊 = 𝐹(−1)𝑑 𝑊 = −𝐹𝑑 Force Displacement
- 11. 11 EXAMPLE 1 You mustexert a force of 4.5 N on a book to slide it across a table. If you do 2.7 J of work in the process, how far have you moved the book? Given: F= 4.5 N W = 2.7 J 𝜃 = 0 d = ? Solution: 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 Formula:
- 12. 12 EXAMPLE 2 A childpulls a sled up a snow-covered hill. The child does 405 J of work on the sled. If the child walks 15 m up the hill, how large of a force must the child exert? Given: F= ? W = 405 J 𝜃 = 0 d = 15m Solution: 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 Formula:
- 13. 13 EXAMPLE 3 A porterpulls a 10-kg luggage along a level road for 5 m by exerting a force of 20 N at an angle of 30o with the horizontal shoulder through a vertical distance of 1.5 m and carries it for another 5 m. a. How much work he do in pulling b. How much work he do in lifting c. How much work carrying the luggage on his shoulder?
- 14. 14 EXAMPLE 3 A porterpulls a 10-kg luggage along a level road for 5 m by exerting a force of 20 N at an angle of 30o with the horizontal shoulder through a vertical distance of 1.5 m and carries it for another 5 m. a. How much work he do in pulling Given: F = 20N ; θ = 30o ; d = 5m W = ? Formula: Solution:
- 15. 15 EXAMPLE 3 A porterpulls a 10-kg luggage along a level road for 5 m by exerting a force of 20 N at an angle of 30o with the horizontal shoulder through a vertical distance of 1.5 m and carries it for another 5 m. b. How much work he do in lifting Given: m = 10kg ; θ = 30o ; d = 1.5m W = ? Formula: Solution:
- 16. 16 EXAMPLE 3 A porterpulls a 10-kg luggage along a level road for 5 m by exerting a force of 20 N at an angle of 30o with the horizontal shoulder through a vertical distance of 1.5 m and carries it for another 5 m. c. How much work carrying the luggage on his shoulder? Given: F = 98N ; θ = 90o ; d = 5m W = ? Formula: Solution: 𝑊 = 98𝑐𝑜𝑠 90 5 𝑊 = 98𝑐𝑜𝑠 0 5 𝑊 = 0J
- 17. AREA UNDER THE LINEON THE GRAPH
- 18.
- 19.
- 20. 20 EXAMPLE 1 A constantforce of 4 N acts on an object in the direction of its motion. If the object moves 2 meters, how much work is done? Given: F= 4 N W = ? 𝜃 = 0 d = 2m Solution: 𝑊 = 𝐹𝑐𝑜𝑠𝜃𝑑 Formula: 𝑊 = 4 cos 0 2 𝑊 = 8J 𝑊 = 𝐿𝑊
- 21. 21 EXAMPLE 2 The nethorizontal force on a box F as a function of the horizontal position is shown below. What is the work done on the box from x = 0 m to x = 2 m?
- 22. 22 EXAMPLE 3 The nethorizontal force on a box as a function of the horizontal position is shown below. What is the work done on the box from x = 0 m to x = 6 m?