P161Sp08Chapter_07

# P161Sp08Chapter_07 - Chapter 7 Newton's Third Law The...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 7 Newton's Third Law The Interaction Law n n So far we have consider an object being acted upon by an external force and applied Newton's 2nd law to determine the motion. We may also have the situation where two objects interact with each other. n This is called an interaction. n Newton's Third Law addresses this situation Newton's Third Law n If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to a force F21 exerted by object 2 on object 1 n F12 = - F21 n Note on notation: FAB is the force exerted by A on B n Note: read FAB as Force of A on B Newton's Third Law, Additional Statements n n n Forces always occur in pairs A single isolated force cannot exist. Forces act on something The action force is equal in magnitude to the reaction force and opposite in direction n n n One of the forces is the action force, the other is the reaction force It doesn't matter which is considered the action and which the reaction The action and reaction forces must act on different objects and be of the same type Hammer & Nail Force Pair Newton's 3rd Law Action-Reaction Example n n The force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to F21 exerted by object 2 on object 1 F12 = - F21 Action-Reaction Examples F = mg F = - M ea If a 1.00kg ball drops 1.00 m (Dy=-1.00) how far does the earth move in that time ? Earth: The force of the earth's gravity on the ball is Feb = -mg. The acceleration of the ball is Feb/m = -g. By action and reaction the force of the ball on the earth is Fbe = -Feb mg = -Meae Me = 5.98 1024 kg Action-Reaction Pair How long does it take the ball to fall 1.00m from rest? y = yi +vi t +(1/2) a t2 yi-y = Dy vi =0 a = -g=-9.80m/s2 Ball: Dy = -(1/2) g t2 t = (2.00/g) = 0.452 s Action-Reaction Pair During the 0.452s the ball takes to fall to the earth how far is the earth pulled to the ball? Dye = (1/2) aet2 ae =? Feb=-Fbe mbab= -Meae ae = -ab(mb/Me) ae = 9.80m/s2(1kg/5.981024 kg) ae = 1.6410-24m/s2 Dye = (1/2) ae t2 Dye = 1.6710-25m Fan/Cart Demo n Fan on Cart, w &w/o sail n n n Fan on cart - Will the cart move? If so which way? Why? Fan off cart w/sail. Will the cart move? Which way? Why? Fan on cart w/sail. Will the cart move? Which way? Why? Start 3/5 Action-Reaction Example n The normal force (table on monitor) is the reaction of the force the monitor exerts on the table n Normal means perpendicular, in this case n The action (Fg, Earth on monitor) force is equal in magnitude and opposite in direction to the reaction force, the force the monitor exerts on the Earth Free Body Diagram n n In a free body diagram, you want the forces acting on a particular object The normal force and the force of gravity are the forces that act on the monitor Demo: Interacting Masses Spring released between 1 & 2 F12 a2 = m2 v 2 = a2 Dt F21 a1 = m1 v1 = a1Dt v 2 F12 m1 F12 v 2 m2 = fi = v1 m2 F21 F21 v1 m1 Interacting Masses A 10.0 N force pushes on a 4.00 kg box (m1) which is touching a 2 kg box( m2). What force does the 2.00 kg box experience and what is its acceleration? Apply Newton's 2nd law to each box noting that there are reaction forces between the boxes. Interacting Boxes F - F12 = m1 a F21 = m 2 a The a's are the same because the boxes are in contact. Add the equations and make use of the action-reaction force concept: F12 = -F21. 10.0N F = (m1 + m 2 )a fi a = = 1.67m / s 2 6.00kg F21 = m 2 a = (2.00kg)1.67m / s = 3.33N 2 Strings Does (a) or (b) produce a larger tension in the rope? Student - Wall Pull Imagine cutting the rope in two pieces. There are two forces on the left end of the rope. -100N + TRonL= 0 TRonL= 100N The left end and right end are an action reaction pair. TRonL= - TLonR = 100N The right end of the rope is not accelerating: -TLonR+ F wall= 0 Fwall = 100N Student - Student Pull The left half of the rope has the same forces on it as in the wall case. TRonL = 100N. There are two forces on the right half of the rope,and the rope doesn't accelerate: TLonR = 100N. As a result the ends of the rope have 100N forces acting in opposite directions just as in the wall case. Accelerating String In the last example the string was not accelerating and the tension was constant throughout the rope. Is the tension the same at both ends if the string is accelerating? Newton's 2nd Law for the string: F = TBonS-TAonS =msta So if the string is accelerating the tensions on the ends of a rope w/mass cannot be the same TBonS = TAonS + msta. If the rope has zero mass the tensions will be the same at each end. Often we make this approximation. Pulling Two Boxes Connected by a Rope Take T1 = 10N, mB = 4kg and m1 = 2 kg. What is the acceleration and the forces on the blocks. Assume massless ropes. Also assume a non-stretch rope. Both blocks will then have the same acceleration. Pulling Two Boxes Connected by a Rope - 2 Massles rope fi TAonB = TBonA = TR m B : TR = m b a Magnitudes m A : -TR +T1 = m A a Adding the equations we see : T1 = (m b + m A )a a= 10.0N = 1.67m / s 2 6.00kg TR = m b a = (4kg)(1.67m / s 2 ) = 6.67N T1 -TR = 10.N - 6.67N = 3.33N Pulleys & Strings Assume a massless and non-stretch string. Pulley & String TSonA = TAonS TAonS = TBonS TBonS = TSonB Let all string tensions = T Newton's x Eq. for A: T = mAa Newton's y Eq. for B: T - mBg = -mBa a = g(mB/(mB+mA) T = mAmBg/(mB+mA) ...
View Full Document

## This note was uploaded on 05/19/2008 for the course PHYS 161 taught by Professor Hammer during the Spring '07 term at Maryland.

Ask a homework question - tutors are online