Work MCV4U: Calculus & Vectors Work is when a force is applied - PDF document

a l g e b r a i c v e c t o r s a l g e b r a i c v e c t o r s Work MCV4U: Calculus & Vectors Work is when a force is applied to an object, causing it to move. It is defined as the the product of the object’s displacement and the component of the force applied along the line of displacement. Applications of the Dot and Cross Products Work can be calculated using the dot product, using either Part 2: Work and Torque geometric or algebraic vectors. J. Garvin J. Garvin — Applications of the Dot and Cross Products Slide 1/13 Slide 2/13 a l g e b r a i c v e c t o r s a l g e b r a i c v e c t o r s Work Work Work Example The work, W , done by a force is given by Calculate the work done if a sled is pulled forward 50 m W = � F · � d = | � F | · | � d | cos θ , where | � F | is the magnitude of the along a frictionless surface by a force of 250 N at an angle of 35 ◦ to the horizontal. applied force (in Newtons), | � d | is the magnitude of the object’s displacement (in metres), and θ is the angle (in degrees) between the force and the displacement vectors. W = (250)(50) cos 35 ◦ The standard unit of work is the Joule (J), or N · m. ≈ 10 239 J J. Garvin — Applications of the Dot and Cross Products J. Garvin — Applications of the Dot and Cross Products Slide 3/13 Slide 4/13 a l g e b r a i c v e c t o r s a l g e b r a i c v e c t o r s Work Work Example Example A 30 kg box is placed 10 m up a ramp that is inclined at 23 ◦ It takes 12 000 J of work to pull a sled 200 m with a 150 N force. Determine the angle of the rope with the horizontal. to the horizontal. Calculate the work done by the force of gravity as the box slides down to the bottom of the ramp. The force of gravity acting downward on the box is 12 000 = (150)(200) cos θ 30 × 9 . 8 = 294 N. � 12 000 � The angle between the displacement down the ramp and the θ = cos − 1 (150)(200) force of gravity is 90 ◦ − 23 ◦ = 67 ◦ . ≈ 66 ◦ The work done by gravity, then, is W = (294)(10) cos 67 ◦ ≈ 1149 J J. Garvin — Applications of the Dot and Cross Products J. Garvin — Applications of the Dot and Cross Products Slide 5/13 Slide 6/13

a l g e b r a i c v e c t o r s a l g e b r a i c v e c t o r s Work Work The displacement vector, � Example d , can be calculated via subtraction. A force of 20 N, applied in the direction of � m = (1 , 2), moves an object from P (3 , 2) to Q (6 , 15). Determine the work � d = (6 − 3 , 15 − 2) done. = (3 , 13) The vector representing the applied force, � f , must have the Now use the dot product for algebraic vectors to calculate same direction as � m , and a magnitude of 20. the work done. Begin by representing � f as a scalar multiple of � m . √ √ � 1 2 + 2 2 W = (4 5 , 8 5) · (3 , 13) 20 = k √ √ k = 20 = 12 5 + 104 5 √ √ 5 = 116 5 � � � 20 5 , 40 f = √ √ 5 ≈ 259 . 4 J √ √ = (4 5 , 8 5) J. Garvin — Applications of the Dot and Cross Products J. Garvin — Applications of the Dot and Cross Products Slide 7/13 Slide 8/13 a l g e b r a i c v e c t o r s a l g e b r a i c v e c t o r s Work Torque The last example illustrates an alternate method of The turning effect of a force around a fixed point is known as calculating work, given algebraic vectors. torque . Any time a force is applied to an object that results in some Work rotation about a point (pedalling a bicycle, opening a door, The work, W , required to move an object from P ( x p , y p , z p ) turning a wrench), a force perpendicular to the plane to Q ( x q , y q , z q ) using some force, � f , in the direction of containing the applied force is produced. m | | � � m = ( x m , y m , z m ) is � m f | · ( x q − x p , y q − y p , z q − z p ). | � If � r is the vector representing the radius of the rotation, and f is the vector representing the applied force, then the torque r × � is � τ = � f . r × � The magnitude, | � τ | = | � f | is a scalar measure of the overall twisting effect. Like work, torque is measured in Joules. J. Garvin — Applications of the Dot and Cross Products J. Garvin — Applications of the Dot and Cross Products Slide 9/13 Slide 10/13 a l g e b r a i c v e c t o r s a l g e b r a i c v e c t o r s Torque Torque Example Your Turn A bolt is tightened using a 20 N force, applied at an angle of A bicycle pedal, 20 cm in length, has a 50 N force applied to 60 ◦ to the end of a wrench that is 30 cm long. Calculate the it at an angle of 45 ◦ . Determine the magnitude of the torque. magnitude of the torque about its point of rotation. Since Joules are N · m, convert cm to m first. τ | = (0 . 2)(50) sin 45 ◦ | � τ | = (0 . 3)(20) sin 60 ◦ | � = 10 √ ≈ 5 . 2 J 2 √ = 5 2 ≈ 7 . 1 J J. Garvin — Applications of the Dot and Cross Products J. Garvin — Applications of the Dot and Cross Products Slide 11/13 Slide 12/13

a l g e b r a i c v e c t o r s Questions? J. Garvin — Applications of the Dot and Cross Products Slide 13/13