PHY183-Lecture12 - W e ig h t a n d M a s s ! Magnitude of...

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Unformatted text preview: W e ig h t a n d M a s s ! Magnitude of gravitational force on an object = weight ! Gravitational force on an object, Fg, is always proportional to its mass ! Near surface of Earth (altitude of 10 km or less): Weight is constant and is the product of object’s mass and the Earth’s gravitational acceleration: Physics for Scientists & Engineers 1 Fall Semester 2006 Lecture 12 Fg = mg g = 9.81 m/s 2 September 14, 2006 Physics for Scientists&Engineers 1 1 September 14, 2006 Physics for Scientists&Engineers 1 2 W e ig h t a n d M a s s ! Example: • Object with mass m = 5.00 kg • Gravitational force Fg = mg = (5.00 kg)(9.81 m/s2) = 49.05 kg m/s2 N e t F o rc e ! Definition: Net force = vector sum of all forces acting on a given object n! ! !! ! Fnet = ! Fi = F1 + F2 + ... + Fn i =1 ! Force Unit 1 kg m/s2 = 1 N ! Cartesian components of net force are given by: n i =1 n • Named after British physicist Sir Isaac Newton (1642-1727), father of modern mechanics and perhaps the most influential scientist who ever lived Fnet , x = ! Fi , x = F1, x + F2 , x + ... + Fn , x Fnet , y = ! Fi , y = F1, y + F2 , y + ... + Fn , y i =1 n ! Summary: mass, m, is measured in units of kg, whereas weight (a force!), mg, is measured in units of N September 14, 2006 Physics for Scientists&Engineers 1 3 September 14, 2006 Fnet , z = ! Fi , z = F1, z + F2 , z + ... + Fn , z i =1 Physics for Scientists&Engineers 1 4 N o t e b o o k C o m p u t e r , a g a in ! Hand exerts a force on computer: N (normal force) • This force points in opposite direction to weight, and has same magnitude F r e e -B o d y D ia g r a m ! First observation: We do not really need the hand in the picture. It’s entire effect is represented by the normal force arrow. ! ! N = ! Fg ! Calculate net force: n! ! Fnet = ! Fi i =1 ! ! = Fg + N !! = "N + N =0 Physics for Scientists&Engineers 1 5 ! Second observation: We don’t really need an exact representation of the notebook computer either, a sketch is sufficient. ! Free-body diagram! September 14, 2006 Physics for Scientists&Engineers 1 6 September 14, 2006 E q u ilib r iu m C o n d it io n ! Static Equilibrium Condition 1: An object can only stay at rest, if the net force acting on it is exactly 0. E q u ilib r iu m C o n d it io n ! In Cartesian coordinates, this vector equation is actually three independent equations: ! Fnet = 0 Fnet , x = ! Fi , x = F1, x + F2 , x + ... + Fn , x = 0 i =1 n n ! We can use this 0 net force equilibrium condition to solve for unknown forces. • Example: If object 1 rests on object 2, then the normal force, ! N, keeps object 1 at rest and therefore the net force on that ! object 0. If N would be larger, object 1 would have to lift off. ! If N were smaller, it would sink into object 2. September 14, 2006 Physics for Scientists&Engineers 1 7 September 14, 2006 Fnet , y = ! Fi , y = F1, y + F2 , y + ... + Fn , y = 0 i =1 n Fnet , z = ! Fi , z = F1, z + F2 , z + ... + Fn , z = 0 i =1 Physics for Scientists&Engineers 1 8 Two Object’s Forces on Each Other ! Consider two arbitrary objects at rest and in isolation, ! with no net external forces acting on them, Fext = 0 ! ! Add two intern! l forces F1, 2 (force of object 1 acting on a object 2) and F2 ,1 (force of object 2 acting on object 1) to the sum of the ext! rnal !orces to obtain the total net e f ! ! force: Fnet = Fext + F1, 2 + F2 ,1 ! Since the objects are at rest, the total net force vanishes. Since in addition the external force is 0, we get: ! ! ! ! G e n e r a l R u le ! If an object 1 exerts a force F1, 2 on object 2, then ! the force F that object 2 exerts on object 1 is 2 ,1 exactly equal in magnitude an opposite in direction: ! ! ! ! ! F1, 2 + F2 ,1 = 0 ! F1, 2 = " F2 ,1 • “Newton’s Third Law” F1, 2 + F2 ,1 = 0 ! F1, 2 = " F2 ,1 ! In static equilibrium the two objects forces on each other are exactly equal in magnitude and opposite. September 14, 2006 Physics for Scientists&Engineers 1 9 September 14, 2006 Physics for Scientists&Engineers 1 10 Example: Two Books ! What is the magnitude of the force that the table exerts on lower book? ! Answer: • First draw free-body diagram for upper book: F2 ,1 = N1 = F = m1g 1 • Newton’s Third Law: F , 2 = F2 ,1 1 • Then draw free-body diagram for ! ! ! lower book: F + N + F = 0 ! R o p e s a n d P u lle y s ! The force is transmitted exactly in the direction of the rope ! Force inside the rope is called “string tension” 2! ! ! 1, 2 ! 2 ! N 2 = "( F1, 2 + F2 ) = "( F1 + F2 ) N 2 = g( m1 + m2 ) • Normal force = sum of both weights September 14, 2006 Physics for Scientists&Engineers 1 11 September 14, 2006 Physics for Scientists&Engineers 1 12 R o p e s a n d P u lle y s ! String tension is the same everywhere inside the rope ! Ropes can be redirected via pulleys => direction of force changes, but not absolute magnitude N o n -t r iv ia l e x a m p le : 3 w e ig h t s ( 1 ) ! Three strings are tied together. They are on top of a circular table and meet exactly in the center of the table, as shown in the figure (top view). Each of the strings hangs over the edge of the table, and a weight is supported from it as shown. The masses m1 = 3.9 kg and m2 = 5.2 kg are known. The angle ! = 74.2 degrees between strings 1 and 2 is also known. ! Question: What is the mass m3 that is needed so that the system is in equilibrium? Example September 14, 2006 Physics for Scientists&Engineers 1 13 September 14, 2006 Physics for Scientists&Engineers 1 14 N o n -t r iv ia l e x a m p le : 3 w e ig h t s ( 2 ) ! Answer: ! Step 1: pick a suitable coordinate system • P ic k x - a x is a lo n g th e d ir e c tio n o f s tr in g 1 N o n -t r iv ia l e x a m p le : 3 w e ig h t s ( 3 ) ! Step 4: Solve for the components of F3: F3, x = ! m1g ! m2 g cos " F3, y = ! m2 g sin " ! Step 5: Don’t put in the numbers yet; instead calculate absolute value of F3: ! Step 2: calculate components of forces due to m1 and m2: Example Example F1, x = m1g; F1, y = 0 F2 , x = m2 g cos ! ; F2 , y = m2 g sin ! ! Step 3: Use static equilibrium condition: F3 = F3, x 2 + F3, y 2 = g ( m1 + m2 cos ! )2 + ( m2 sin ! )2 = g( 7.29 kg ) ! This is the answer: F1, x + F2 , x + F3, x = 0 F1, y + F2 , y + F3, y = 0 September 14, 2006 Physics for Scientists&Engineers 1 15 September 14, 2006 m3 = 7.29 kg Physics for Scientists&Engineers 1 16 ...
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This note was uploaded on 03/01/2010 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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