PHYS 210 Chapter 5, Spring 2012

Chapter 5 Chapter 5 Force and Motion I Force and Motion I Classical Mechanics Classical Mechanics  Describes the relationship between Describes the relationship between the motion of objects in our everyday the motion of objects in our everyday world and the forces acting on them world and the forces acting on them  Conditions when Classical Mechanics Conditions when Classical Mechanics does not apply does not apply  very tiny objects (< atomic sizes) very tiny objects (< atomic sizes)  objects moving near the speed of light objects moving near the speed of light Forces Forces  Usually think of a force as a push or Usually think of a force as a push or pull pull  Vector quantity Vector quantity  May be contact or field force May be contact or field force Contact and Field Forces Contact and Field Forces Fundamental Forces Fundamental Forces  Types Types  Strong nuclear force Strong nuclear force  Electromagnetic force Electromagnetic force  Weak nuclear force Weak nuclear force  Gravity Gravity  Characteristics Characteristics  All field forces All field forces  Listed in order of decreasing strength Listed in order of decreasing strength  Only gravity and electromagnetic in mechanics Only gravity and electromagnetic in mechanics Measuring the Strength of a Measuring the Strength of a Force Force  Use the deformation of a Use the deformation of a spring to determine the spring to determine the strength of F. strength of F.  The spring elongates The spring elongates under the action of F under the action of F11 and and the pointer on the scale the pointer on the scale indicates the magnitude indicates the magnitude of F of F11. .  Calibration is necessary Calibration is necessary Newton’s First Law Newton’s First Law  If no forces act on an object, it If no forces act on an object, it continues in its original state of continues in its original state of motion; that is, unless something motion; that is, unless something exerts an external force on it, an exerts an external force on it, an object at rest remains at rest and an object at rest remains at rest and an object moving with some velocity object moving with some velocity continues with that same velocity. continues with that same velocity. Newton’s First Law, cont. Newton’s First Law, cont.  External force External force  any force that results from the interaction any force that results from the interaction between the object and its environment between the object and its environment  Alternative statement of Newton’s First Alternative statement of Newton’s First Law Law  When there are no external forces acting When there are no external forces acting on an object, the acceleration of the on an object, the acceleration of the object is zero. object is zero. Inertia Inertia  Is the tendency of an object to Is the tendency of an object to continue in its original motion continue in its original motion Mass Mass  A measure of the resistance of an A measure of the resistance of an object to changes in its motion due to object to changes in its motion due to a force a force  Scalar quantity Scalar quantity  SI units are kg SI units are kg Newton’s Second Law Newton’s Second Law  The acceleration of an object is directly The acceleration of an object is directly proportional to the net force acting on it proportional to the net force acting on it and inversely proportional to its mass. and inversely proportional to its mass.  FF and and aa are both vectors are both vectors  Can also be applied three-dimensionally Can also be applied three-dimensionally Units of Force Units of Force  SI unit of force is a Newton (N) SI unit of force is a Newton (N)  US Customary unit of force is a pound US Customary unit of force is a pound (lb) N. B. (1 N = 0.225 lb) (lb) N. B. (1 N = 0.225 lb) 2 s 1kg m 1N  A sphere and a block are connected by a string placed over a pulley. The two objects are at rest at the same level, as shown in the Figure. A teacher performs the following demonstration: The block is now pulled down and held while in its new position, as shown in Figure 2. The teacher asks the class to predict what will happen in this real situation when the block is released, and to give a reason supporting the prediction. A teacher performs the following demonstration: 1. The block will move down because the gravitational force on the block will be greater at this lower level. 2. The block will remain stationary because there will be no resultant force on the block. 3. The block will move up and return to its original position because this will conserve potential energy. 4. The block will move up and return to its original position because that was its equilibrium position. More Question: The adjacent figure shows overhead views of four situations in which forces act on a block that lies on a frictionless floor. If the force magnitudes are chosen properly, in which situations is it possible that the block is (a) stationary and (b) moving with a constant velocity? Gravitational Force Gravitational Force  Mutual force of attraction between Mutual force of attraction between any two objects any two objects  Expressed by Newton’s Law of Expressed by Newton’s Law of Universal Gravitation: Universal Gravitation: 2 2 1 g r F G m m Weight Weight  The magnitude of the gravitational The magnitude of the gravitational force acting on an object of mass force acting on an object of mass m m near the Earth’s surface is called the near the Earth’s surface is called the weight weight ww of the object of the object  w = m g w = m g is a special case of Newton’s is a special case of Newton’s Second Law Second Law  g g can also be found from the Law of can also be found from the Law of Universal Gravitation Universal Gravitation More about weight More about weight  Weight is Weight is not not an inherent property of an inherent property of an object an object  mass mass is is an inherent property an inherent property  Weight depends upon location Weight depends upon location Newton’s Third Law Newton’s Third Law  If two objects interact, the force F If two objects interact, the force F12 12 exerted by object 1 on object 2 is exerted by object 1 on object 2 is equal in magnitude but opposite in equal in magnitude but opposite in direction to the force F direction to the force F21 21 exerted by exerted by object 2 on object 1. object 2 on object 1.  Equivalent to saying a single isolated Equivalent to saying a single isolated force cannot exist force cannot exist Newton’s Third Law cont. Newton’s Third Law cont.  FF12 12 may be called may be called the the action action force force and F and F21 21 the the reaction reaction force force  Actually, either Actually, either force can be the force can be the action or the action or the reaction force reaction force  The action and The action and reaction forces act reaction forces act on on different different objects objects Some Action-Reaction Pairs Some Action-Reaction Pairs  n and n’ n and n’  n is the n is the normal normal force, the force the force, the force the table exerts on the table exerts on the TV TV  n is always n is always perpendicular to perpendicular to the surface the surface  n’ is the reaction – n’ is the reaction – the TV on the table the TV on the table  n = - n’ n = - n’ More Action-Reaction pairs More Action-Reaction pairs  FFgg and F and Fgg’’  FFgg is the force the is the force the Earth exerts on the Earth exerts on the object object  FFgg’ is the force the ’ is the force the object exerts on the object exerts on the earth earth  FFgg = -F = -Fgg’’ Forces Acting on an Object Forces Acting on an Object  Newton’s Law uses Newton’s Law uses the forces acting the forces acting on on an object an object  n and F n and Fgg are acting are acting on the object on the object  n’ and F n’ and Fgg’ are acting ’ are acting on other objects on other objects The adjacent Figure shows four choices for the direction of a force of magnitude F to be applied to a block on an inclined plane. The directions are either horizontal or vertical. (For choices a and b, the force is not enough to lift the block off the plane.) Rank the choices according to the magnitude of the normal force on the block from the plane, greatest first. Forces Acting on an Object Ex.: Forces Acting on an Object Ex.: Applying Newton’s Laws Applying Newton’s Laws  Assumptions Assumptions  Objects behave as particles Objects behave as particles  can ignore rotational motion (for now) can ignore rotational motion (for now)  Masses of strings or ropes are negligible Masses of strings or ropes are negligible  Interested only in the forces acting on Interested only in the forces acting on the object the object Free Body Diagram Free Body Diagram  Must identify all the forces acting on Must identify all the forces acting on the object of interest the object of interest  Choose an appropriate coordinate Choose an appropriate coordinate system system  If the free body diagram is incorrect, If the free body diagram is incorrect, the solution will likely be incorrect the solution will likely be incorrect Equilibrium Equilibrium  An object either at rest or moving An object either at rest or moving with a constant velocity is said to be with a constant velocity is said to be in in equilibrium equilibrium  The net force acting on the object is The net force acting on the object is zero zero  F 0 Equilibrium cont. Equilibrium cont.  Easier to work with the equation in Easier to work with the equation in terms of its components: terms of its components:  Fx 0  Fy 0 Solving Equilibrium Solving Equilibrium Problems Problems  Make a sketch of the situation described in Make a sketch of the situation described in the problem the problem  Draw a free body diagram for the isolated Draw a free body diagram for the isolated object under consideration and label all the object under consideration and label all the forces acting on it forces acting on it  Resolve the forces into x- and y- Resolve the forces into x- and y- components, using a convenient components, using a convenient coordinate system coordinate system  Apply equations, keeping track of signs Apply equations, keeping track of signs  Solve the resulting equations Solve the resulting equations Equilibrium Example – Free Equilibrium Example – Free Body Diagrams Body Diagrams Newton’s Second Law Newton’s Second Law Problems Problems  Similar to equilibrium except Similar to equilibrium except  Use components Use components  aaxx or a or ayy may be zero may be zero  ma F x x ma F  y y ma F  Solving Newton s Second Solving Newton s Second Law Law Problems Problems  Make a sketch of the situation described in Make a sketch of the situation described in the problem the problem  Draw a free body diagram for the isolated Draw a free body diagram for the isolated object under consideration and label all the object under consideration and label all the forces acting on it forces acting on it  If more than one object is present, draw free body If more than one object is present, draw free body diagram for each object diagram for each object  Resolve the forces into x- and y- Resolve the forces into x- and y- components, using a convenient components, using a convenient coordinate system coordinate system  Apply equations, keeping track of signs Apply equations, keeping track of signs  Solve the resulting equations Solve the resulting equations Inclined Planes Inclined Planes  Choose the Choose the coordinate system coordinate system with x along the with x along the incline and y incline and y perpendicular to perpendicular to the incline the incline  Replace the force Replace the force of gravity with its of gravity with its components components Example: Example:  A car of mass m is on A car of mass m is on a icy driveway inclined a icy driveway inclined at angle at angle  as shown in as shown in the adjacent Figure. the adjacent Figure. Find the acceleration Find the acceleration of the car, assuming of the car, assuming No Friction No Friction.. Solution: Solution:  Chose of x- and y-axes. Chose of x- and y-axes.  Free Body diagram. Free Body diagram.  Newton’s 2 Newton’s 2nd nd Law: Law: m mgg + + nn = m = maa Projection along the x-axis: Projection along the x-axis: mgsin( mgsin() = m ) = maaxx Projection along the y-axis: Projection along the y-axis: mgcos( mgcos() + n = 0 ) + n = 0  ax = gsin() More Examples: More Examples:  Two Blocks of masses m Two Blocks of masses m11 and m and m22, with m , with m11 > > m m22, are placed in contact with each other on a , are placed in contact with each other on a frictionless horizontal surface. A constant frictionless horizontal surface. A constant force force FF is applied to m is applied to m11 as shown. as shown. a) Find the magnitude of the acceleration of a) Find the magnitude of the acceleration of the system? the system? b) Determine the magnitude of the contact b) Determine the magnitude of the contact force between the two blocks? force between the two blocks? Solution: Solution:  Both Blocks must Both Blocks must experience the experience the same acceleration same acceleration since they are in since they are in contact. contact. FF is the only force is the only force acting horizontally: acting horizontally: F = (m F = (m11+m +m22))aax x 2 1 m m F  ax  Solution Cont’d: Solution Cont’d:  Free body diagram of Free body diagram of m m11.. Along the x-axis, we Along the x-axis, we have: have: F – P F – P21 21 = m = m11aaxx x y 2 1 m m F  x  a 2 1 1 21 m m F m F P     m F m m P 2 1 2 21   Solution: Solution:  Note that one can get Note that one can get PP12 12 from m from m22:: Along the x-axis: Along the x-axis: PP12 12 = m = m22aaxx 2 1 m m F  x  a m F m m P 2 1 2 12   We see that | We see that |PP12 12|= | |= |PP21 21|| Example, The Atwood Machine: Example, The Atwood Machine:  Two object of unequal masses Two object of unequal masses (m (m22>m >m11) are hung vertically ) are hung vertically over a frictionless pulley of over a frictionless pulley of negligible mass. negligible mass. Determine the magnitude of Determine the magnitude of the acceleration of the two the acceleration of the two objects and the tension in the objects and the tension in the lightweight cord. lightweight cord. Solution Atwood Machine: Solution Atwood Machine:  m m22 will move down and m will move down and m11 will move up with the same will move up with the same aa..  N.B. The tensions in the cord N.B. The tensions in the cord from both side are equal from both side are equal since mass of pulley is since mass of pulley is negligible and there is no negligible and there is no Friction. Friction. a a Atwood Cont’d: Atwood Cont’d: a  Drawing the free body diagram and Drawing the free body diagram and applying Newton’s second Law. applying Newton’s second Law. Projection along the direction of motion (upwards for m1) -m1g + T = m1a  T = m1g + m1a a   1 1 m T m g   a   2 2 m T m g   Projection along the direction of motion (downwards for m2) m2g - T = m2a  T = m2g – m2a Atwood Cont’d: Atwood Cont’d:  T = T  m1g + m1a = m2g – m2a m g m m m 1 2 1 2    a (m2 – m1)g = Net Force (m2 + m1) = Total mass N.B.