Force and Newton’s Laws Insert Gif of box being pushed in different dimensions 1.1 Different types of forces: Every student attempting the Senior Certificate Exam for Physical Sciences should learn all the definitions as given in the Exam guidelines of 2017. It is important to know that a force can be a push, a pull or it can be a force in a rope (tension). Because most examples in this section is linked to a sketch, it is also important to know all the SI-units for each force. • • Normal force (N) is the force or component of the exerted force by the surface an object rests on. The force is exerted perpendicular to the surface. Friction (f) is the force that resists the movement of an object and is always parallel to the surface. It is important to differentiate with symbols between static- and kinetic friction. o Static friction (fs) is the force resisting the tendency of an object at rest to start moving relative to the surface. It is much more difficult to get an object moving and breaking loose from the surface it is resting on than to keep a moving object in motion. For this reason the maximum static friction on an object is always bigger than the kinetic friction of the two surfaces on each other. (This is true at the point where an object breaks loose from the surface in order to start moving.) o o o o o Kinetic friction (fk) is the force resisting the movement of a moving object relative to the surface it moves on. Friction forces is always: ▪ proportional to the normal force ▪ independent of the contact surface (This implies that a brick on a concrete floor will experience the same friction irrespective of the way it is positioned... ▪ independent of the velocity of the moving object. If a force, F, exerted parallel to the surface, does not cause the object to move, it will mean that the magnitude of that force is equal in magnitude to the static friction force. The static friction force is at a maximum (fsmax) at the moment the object starts to break loose from the surface to move. If the applied force is bigger than fsmax, the resultant/ net force will cause the object to accelerate. 1.2 The following formulas can be used for questions on forces and Newton’s Laws on the formula page: Fnet = ma fsmaks = µs N F=G 1.3 fk = µkN m1 m2 g=G d2 M d2 Force diagrams and free body diagrams It is recommended that you ALWAYS draw a small force diagram for physics questions involving forces, even if it is not requested. A force diagram uses a box to represent a body’s centre of mass. Vector arrows are then drawn from the box with relative magnitudes and directions that are isolated in the area of the object. A free-body diagram only uses a dot in the centre of the represented object for the centre of mass. Because all equations are only applicable in ONE dimension, it is sometimes necessary to break two-dimensional forces (for example a force forming an angle with the horizontal or an object pussing down on an inclined plane) into it’s two components. It is important though to remember that a sketch drawn for marks in the paper must ONLY consist of EITHER the main force or the components of the force, not both on the same force diagram. By making the mistake of drawing both, your vectors represent a vector of double the magnitude. The force diagrams enables you, with the necessary background, to calculate the net/ resultant forces. 1.4 Example 1. Force diagrams: 1.4.1 Exemplar 2014 Question 2 Draw a labeled free-body diagram of all the forces working in on the 6kg-block as it moves up the inclined plane. Example 1.4.2: November 2015 Question 2 Draw a labeled force diagram for both blok M and the 2,5 kg- block. Example 1.4.3 DBE Feb/Mrt 2018 Question 2 Draw a fully labeled free body diagram for Block Q. Solution 1.4.1 DBE Exemplar 2014 Question 2 Solution 1.4.2 DBE November 2015 Question 2 Adapted Solution 1.4.3 DBE Feb/Mrt 2018 Question 2 1.5 Example 1.5.1 DBE Feb 2017 Question 2 Solution 1.5.1 DBE Feb 2017 Question 2 2.1 2.2 2.3.1 2.3.2 Zero See to it that every force has a label with the name written down below, next to the symbol. Unnecessary marks will be forfeited if the forces are not labeled. 2.3.3 This particular part of the question is for Grade 12 learners only, but all the previous questions is on Grade 11 level. This chapter forms part of the final Grade 12 exams. 2. Scale drawings (Only applicable to grade 11, not for grade 12) Only in the grade 11 curriculum does scale drawings form part of the required set of skills. The reason behind this chapter in the work is that some concepts on vectors are understood more easily if you draw a sketch thereof. The sketching brings about inside knowledge of certain techniques, that would otherwise be neglected. Many people believe that technology already replaced hand-drawn sketches, but many technical problems in programming cannot be corrected without a proper knowledge of the background of the problem. In this section we look at drawings that will be done with a protractor, pencil and a ruler: • Draw a sketch of the vertical vectors (y-axis) and horizontal vectors (x-axis) on a Cartesian plane. This involves one dimensional vectors that will fit exactly on the vertical or horizontal axes of the Cartesian plane. For example: 8 6 F1:6N North 4 2 10 5 5 10 15 20 F2: 5N East 2 4 6 • Add co-linear vertical vectors and co-linear horizontal vectors together to find the net vertical vector (Ry) and the net horizontal vector (Rx). (Co-linear vectors refers to vectors in a straight line). In this example it is still about vectors in the same dimension. For example: 8 6 4 2 10 5 5 10 15 20 2 4 6 • Solution: Horizontal – Right/ East as positive Rx = 6 +7 -8 = 5 Units East/right Vertical – Upwards/North as positive Ry = -6 + (-3) = -9 of 9 Units downward/ South Sketch the resultant (R) by using the head-to-tail or tail-to-tail method. 2.1 Head-to-tail method This method draws the vectors as vector arrows following one another. The following vector will be drawn where the previous one ends. The magnitude (and thus the length of the arrow) is of great importance and therefor it is also important to write down the scale used for the drawing. For example 1cm:10m or what ever will be convenient for the magnitude of force vectors you are working with. (Remember that it is not only forces that can be drawn according to these methods, but any other vector as well.) The past few years a scale was provided according to which the drawings had to be done. The head-to-tail method can be used for an unlimited number of vectors, where the resultant will be drawn connecting the start of the FIRST vector, in a straight line with the end of the LAST vector. Measure the angle of the resultant to determine the direction of the resultant vector. The length of the resultant vector should firstly be measured and then converted by means of the scale that was used to see exactly how big the vector really is. 2.2 Tail-by-tail method: This method can only be used for TWO vectors to determine the resultant. That tail-bytail method forms a parallelogram with the two vectors as the two sets of parallel sides of the parallelogram from the same corner. The resultant will be the diagonal starting from the same corner as the other two vectors (all tails of the vector angles together) to the opposite corner of the parallelogram. The two vectors of which the resultant needs to be calculated is drawn in the same Cartesian plane. Measure carefully to draw the remaining two sides of a parallelogram with each opposite side equal. The length and direction of the diagonal line that is ALSO drawn from the same starting point in the same Cartesian plane will be the values for the Resultant. The length again as in the previous method needs to be converted by the scale that was used to draw the sketch. Please make sure whether the question says that you have to determine a resultant vector (may include up to four vectors in one or in two dimensions) GRAPHICALLY or if it should be CALCULATED. Through calculation it will be necessary to use the Pythagoras theory, the Sinus-rule in mathematics or easy trigonometry functions. A closed vector diagram shows that the forces acting on an object are in equilibrium. This means the net resultant on the object is equal to zero as in Newton’s first law. The object on which the forces in equilibrium is exerted will either stay at rest OR continue movement at a constant velocity. If there are three forces involved, the forces can be rearranged to form a closed triangle where other mathematical principles can be applied. This type of questions can include calculating the breaking tension of a rope or calculating which rope carries the most tension. Look at the following examples from GDE question papers: 2.3 Example June 2014 1.1 Three objects A, B and C is hanging from a rope that has a knot at O, as indicated in the diagram below: O F3 F2 F1 Which vector diagram best represents the relation between the forces exerted through point O? F3 A. F3 B. F1 F1 Answer: 1.2 F3 D. F1 F1 F2 F2 F3 C. F2 F2 D The figure below represents a book on a table. Two forces are exerted on the book. According to Newton’s third law, the reaction force of F will be... F A. B. C. D. the gravitational force of the earth exerted on the book. the force of the table on the earth. the force of the table on the book. the force of the book on the table. Answer: 1.3 C (F is the force of the book on the table) A vehicle moves horizontally at a constant velocity of 60 km∙h-1. Which ONE of the force diagrams below correctly represents the forces acting in on the car? A. B. Answer: C. D. D Question 2 2. An 8 kg mass is hanged from the roof by a rope. A second rope pulls the mass in a horizontal direction as shown in the diagram below. Calculate the tension in each of the ropes. (g=9,8 m∙s-2) T1 30° Ring in equilibrium 8kg T2 W= mg = 78,4 N 2.1 Draw a vector diagram to show that forces T1, T2 and W is in equilibrium. (3) 2.2 Determine the value of the forces T1 and T2. (6) Answer: 2.1 30° T1 W=78,4 N 60° T2 2.2 Fg = ma = 8x9,8 =78,4 N W cos 30° =T 1 T1 = 90,52 N tan 30° = T2 W T2 = 45,26 N November 2014 3. The diagram below shows a rope-and-pulley system used lift an 800 N -object from the ground. Accept that the ropes are almost weightless and unelastic and that the pulleys are light and frictionless. T2 Pulley P T1 140° 120° 800 N Calculate the: 3.1 Magnitude of the tension in T1 and T2 graphically. (7) 3.2 Magnitude and direction of the reaction force at pulley P. (4) Answer: 3.1 3.2 3 Newton’s first, second and third laws: Please follow the link below for an interesting phet simulation done by the University of Colorado... A force that is exerted horizontally, can be much smaller than the weight of an object, depending on the magnitude of the friction force: https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motionbasics_en.html NEWTON I: A body will remain in its state of motion (at rest or moving at constant velocity) until a net force acts on it. NEWTON II: When a net force acts on an object, the object will accelerate in the direction of the net force and the acceleration is directly proportional to the force and inversely proportional to the mass of the object. NEWTON III: When one body exerts a force on a second body, the second body exerts a force of equal magnitude in the opposite direction on the first body. NEWTONS UNIVERSAL GRAVITATIONAL LAW: Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. 3.1 Application: Each of the above laws can be applied in real-life situations. What is important in the answering of the questions is to choose ONE direction as positive to calculate the other forces. Vertical and diagonal forces cannot be calculated together in the same equation. The force making an angle with the horizontal should rather be broken down to its’ components and then only take the direction you are working with into account. Calculations with Newton’s Laws can be applied to objects in rest as well as moving objects. The symbols used all refers back to: Fnet = ma The net force of the system can be zero, when the forces in the system is in equilibrium as stated in Newton 1(Equilibrium problems). Equilibrium allows an object to have a certain velocity, as Newton 1 explains that if the net force working in on the object is zero, the velocity will remain constant. In such cases you can also substitute a zero into the acceleration placeholder, a. The other scenario is that there is a non-zero net force as is the case with the other laws (Nonequilibrium problems). This means there will be an acceleration. It is here where the Grade 10 horizontal laws of motion also comes in handy and must sometimes be applied. Remember that the Senior Certificate Exam includes this Grade 10 chapter. The Newton Laws can be applied in the following situations: • A single object: o Moving on a horizontal plane with or without friction o Moving on an inclined plane with and without friction o Moving in the vertical plane (lifts, rockets, etc.) • Two-body systems (joined by a light inextensible string): o Both on a flat horizontal plane with and without friction o One on a horizontal plane with and without friction, and a second hanging vertically from a string over a frictionless pulley o Both on an inclined plane with or without friction o Both hanging vertically from a string over a frictionless pulley Look at the following example that is addapted for Grade 11 learners: 3.2 Example 3.3: DBE November 2015 Question 2 Solution 3.3: DBE November 2015 Question 2 4 Breakdown of diagonal forces: It is important to know how to handle a two-body system as in the example above. The two blocks M and the 2,5 kg block moves together as a system, even though the 2,5 kg block moves vertical and block M moves horizontally. It is therefore also important to choose one direction as positive before any substitutions is made into some formulas. This kind of problems are solved by simultaneous equations. If you choose the DOWNWARD movement of the 2,5 kg block as positive, then the RIGHT direction of block M should also be positive to keep the systems vectors positive in the same direction. It may happen that the friction of block M is bigger than the tension in the rope between the blocks, which will result in a zero acceleration or a negative answer for some of the vectors. That information is important at the end of solving the equations in order to apply the right directions for the system... Continue the rest of the sub-questions by still applying the positive directions for certain directions... As a result, it is also important to break down forces working on an incline into parallel and perpendicular components in order that there is still only one direction as positive when substituting into the equations. Start with a force/free body diagram. 4.1 Step 1: Draw all the forces involved on your sketch and also indicate the degree of the inclined plane. FN f Fg 30° Take care that you always draw the free-body diagram like the diagram above, if it is needed as part of a sub-question, but ALSO a key below it with the key as well as the name of each force, for example: FN: Normal force Fg: Gravitational force f: Friction force BUT on your own answer sheet you ALWAYS need to draw a free-body diagram for each object and that would be the drawing where you can “color” the picture as you need to help you with calculations. Write as much information on it as you possibly can. This will help you also with directions for vectors and applying your negative answers. 4.2 Step 2: Extend the normal force to cut through the inclined plane. This line is supposed to be perpendicular to the incline of the plane!! FN Fg 30° 30° Now the red newly added line will represent Fg┴. The angle between the extension of the normal force and Fg will be the same as the angle of the incline of the plane. Write it on your sketch!!! This will enable you to determine the trigonometry relation of the red side of the triangle that is formed. See the last step to ensure that the triangle is drawn correctly... 4.3 Step 3: Complete the last side of the right angled triangle by drawing one last dashed line, forming a right angle (90 degrees) with the red line. This blue line represents the parallel component of weight Fgll. Substituting the different forces into an equation is not as much of a problem as learners struggle to apply their trigonometry knowledge. Assure that you break down the forces of each Newton question into its’ components and only substitute forces in ONE dimension into each equation. 5. Newton’s Universal Gravitational Law Newtons’ Universal Law: Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Weight is the gravitational force exerted by the earth on any object on or near it’s surface. m1 m2 r2 F=G