Newton's Second Law. net force= F 1 F 2 F 3... F n

Transcription

1 Newton's second law is the relationship between an object's rate of velocity change, the forces acting on that object, and the object's resistance to changing its velocity. The rate of velocity change is called the object's acceleration. Acceleration has units of m/s 2. Newton's second law is a mathematical relationship between the sum of the external vector forces acting on an object, the object's mass, and the object's vector acceleration. A force F is an external influence that can change an object's velocity. Forces are vector quantities because they have both a magnitude and a direction. The total influence of all of the external forces is called the net force and is the vector sum all of the forces acting on an object. The net force is a vector quantity. Force has units of newtons, N. net force= F 1 F 2 F 3... F n An object's resistance to changing its velocity is called the object's inertia. The amount of inertia is called the object's mass, m. Mass has units of kilograms, kg. An object's acceleration is directly proportional to the net force acting on that object and inversely proportional to the object's mass. The acceleration direction and the net force direction are the same direction. The acceleration direction is the same as the velocity change direction v. a= net force m Newton's second law Using the rules of algebra, this relationship can be rearranged to produce the following relationships. net force=m a m= net force a With a constant net force, the product of the mass and acceleration is constant which demonstrates that the acceleration is inversely proportional to the mass. With a constant mass, the ratio of the net force and acceleration is a constant ratio which demonstrates that the acceleration is directly proportional to the net force. Since the net force is the vector sum of the external forces acting on an object, and mathematically the symbol Σ is used for a sum, the net force is often written as F net force= F 1 F 2 F 3... F n = F Using the symbol Σ for the net force, Newton's second law can then be written in as F =m a. 1 of 6

2 F =m a is an extremely compact mathematical expression that can be reliably applied to a wide variety of physical situations. However, correctly applying Newton's second law requires the organization of a lot of detailed information about the acceleration and about all of the forces. Polar and Rectangular Vector Forms In two dimensions a vector can be represented in either polar form or rectangular form. Polar form is convenient when the direction angles of the vector quantities are known. Rectangular form is required to add vectors mathematically. Polar form examples: Rectangular form examples: F =F a=a F =F x i F y j a=a x i a y j Applying Newton's Second Law Newton's second law relates the forces, mass, and acceleration for a single object. Steps for applying Newton's Second Law F =m a Newton's Second Law (short form) F 1 F 2 F 3... F n =m a Newton's Second Law 1) Select one object. (2 nd law only applies to one object) Example: Select the block 2) List the forces acting on that object. (Need to know which forces to add to get the net force.) Example: gravity force, support force, pull force 3) Write the 2 nd law vector equation explicitly showing the forces in your list. Write it in vector notation form and in polar form. The polar form shows the angles that need to be determined. F Example: g F N F p =m a vector notation form F g g F N N F p p =m a a polar form 4) Determine the direction angles. Draw a force diagram showing a vector arrow for each force to visually show the vector directions. For more information read the Guidelines for Drawing Force 2 of 6

3 Diagrams. Example: Select x and y coordinate axes. The positive x axis direction is the 0 direction. Vector equations require accurate vector directions. For more information read the Guidelines for Coordinate Axes. Example: acceleration 0, gravity force -90, support force +90, and pull force 0 5) Rewrite the 2 nd law vector equation in polar form showing the known direction angles. Example: F g 90 F N 90 F p 0 =m a 0 6) Determine as much as you can about the vector magnitudes using available information and the force laws such as F g =m g and F k = k F N. Example: mass = 2 kg, and pulling force = 10 N F g =m g= 2 kg 9.8 N kg =19.6 N 19.6 N 90 F N N 0 = 2kg a 0 7) Convert the polar form equation to rectangular form so that the vectors can be added. Derive the x component equation by equating the sum of the x components of the forces to the x component of the mass times the acceleration. Derive the y component equation by equating the sum of the y components of the forces to the y component of the mass times the acceleration. You can use the unit vector notation style or organize the x and y components in a table. Examples are show on page 4. 8) Determine whether enough information is available for a mathematical solution. Rule: A mathematical solution exists only if the number of independent scalar equations is equal to the number of unknowns. Since a single vector equation has both x components and y components, a single vector equation contains two independent scalar equations. You can usually solve one vector equation for two unknown scalar quantities such as two unknown magnitudes. If you have more than two unknowns, you need to find more independent equations using other available information such information about friction. Example: A vector equation can be solved for the two unknown magnitudes F N and a because there are only two unknowns in this vector equation. 3 of 6

4 9) If you have the same number of independent equations as the number of unknowns, solve for the unknown quantities. Solution using unit-vector-notation 19.6 N 90 F N N 0 = 2kg a 0 [ 0 i 19.6 N j] [ 0 i F N j] [ 10 N i 0 j]=[ 2kg a i 0 j] x component scalar equation : N = 2kg a ycomponent scalar equation : 19.6 N F N 0=0 Solution :a= 10 N 2kg =5 m s 2 and F N=19.6 N Solution using a vector-component table Force Magnitude Direction Angle F x = F cos θ ma x = ma cos θ F y = F sin θ ma y = ma sin θ F g 19.6 N N F N F N 90 0 F N F p 10 N 0 10 N 0 F =m a (2 kg) a 0 (2 kg) a 0 x component scalar equation: N = (2 kg) a y component scalar equation: N + F N + 0 = 0 Solution: a= 10 N 2 kg =5 m s 2 and F N=19.6 N 10) If you have two objects, repeat steps 1-7 for each object to derive your independent equations. 11) If you have two objects that are connected together, you need to identify quantities that have the same magnitude for both of the objects. For example, if two objects are connected by something that has a fixed length, such as a string, those two objects always have the same acceleration magnitude and at any specific time will always have the same velocity magnitude. The tension force law also requires the string force magnitude has the same magnitude at each end of the string. If the acceleration magnitudes of the two objects are the same, then use a single letter a for that magnitude instead of magnitude a 1 for object 1 and a 2 for object 2. Example: a 1 =a 0 a 2 =a 90 4 of 6

5 If force magnitudes are the same for the two objects, such as a string force that is the same magnitude at each end of the string, use a single variable name for that magnitude. Example: F string 1 = F string 0 F string 2 =F string 90 Example for simplifying the vector equations for two objects connected by a string object 1 F g1 F N1 F string1 =m 1 a 1 object 2 F g2 F N2 F string2 =m 2 a 2 object 1 F g1 g1 F N1 N1 F string1 string1 =m 1 a 1 a1 object 2 F g2 g2 F N2 N2 F string2 string2 =m 2 a 2 a2 The last 2 equations have a combined total of 18 unknowns. The number of unknowns can be reduced to 10 by determining all the direction angles (which is the task you accomplish by drawing a good force diagram). If m 1, m 2, F g1, and F g2 are also known, you are still left with 6 unknowns. F N1, F N2, F string1, F string2, a 1, and a 2. The vector equation for object 1 can generate 2 independent equations. The x components on the left are equal to the x components on the right and the y components on the left are equal to the y components on the right. Similarly, the vector equation for object 2 can generate independent x and y component equations. A mathematical solution is not possible with only 4 independent equations and 6 unknowns. However, more independent equations can be derived because the two objects are connected to each other. If F string1 =F string2 = F string and a 1 =a 2 =a then the vector equations for objects 1 and 2 can be simplified. object 1 F g1 g1 F N1 N1 F string string1 =m 1 a a1 object 2 F g2 g2 F N2 N2 F string string2 =m 2 a a2 These 2 equations now have only 4 unknowns are F N1, F N2, F string, and a. Write each polar equation in rectangular form to produce 4 scalar equations and solve for the 4 unknown quantities. 5 of 6

6 Name Assignment Date Checklist for One Object Applying Newton's Second Law Completed? Step Brief Description 1 Select one object. Which object did you select? 2 List the forces acting on the object 3 Write the 2 nd law vector equation explicitly showing the forces in your list. a) Vector notation form equation and b) polar form equation 4 Draw a force diagram. Select x-y axes. Determine vector directions. 5 Write the 2 nd law equation in polar form with correct direction angles. 6 Determine as much as you can about the vector magnitudes. Write the 2 nd law equation in polar form with updated magnitudes. 7 Use vector components to derive the scalar equations. 8 Check that you have the same number of independent equations as unknowns. 9 Solve for the unknown values. Checklist for Two Objects Applying Newton's Second Law Completed? Step Brief Description 1 Identify the magnitudes that are common to the 2 objects such as acceleration and string force. (Use single variable for magnitudes that are the same.) 2 Do the checklist for one object steps 1-7 for object 1. 3 Do the checklist for one object steps 1-7 for object 2. 4 Check that you have the same number of independent equations as unknowns. 5 Solve for the unknown values. 6 of 6