Main points of today s lecture: Normal force Newton s 3 d Law Frictional forces: kinetic friction: static friction Examples. Physic 231 Lecture 9

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1 Main points of today s lecture: Normal force Newton s 3 d Law Frictional forces: kinetic friction: static friction Examples. Physic 3 Lecture 9 f N k = µ k f N s < µ s

2 Atwood s machine Consider the Atwood machine to the right. The massless string passes over a massless and frictionless pulley. It is under tension T, which we define to be the magnitude of the tension force. By this definition it is a positive number. Choosing up to be positive, what is the net force on mass? a) T-m g b) T+m g c) m g-t d) none of the above m m

3 Choosing up to be positive, what is the net force on mass? a) m g-t b) T+m g c) T-m g d) none of the above Atwood s machine m m

4 From Newton s nd law: T m g = T = m a m a + m g Atwood s machine Also T m g = T = m a m a + m g m If m exceeds m, m goes down and m goes up. If m exceeds m, m goes down and m goes up. In either case Δy = Δy v a = v = a m Putting it together: m a + m g = m a + m g m a + m g = m a + m g m a + m a = m g m g ( a = m m )gçnet gravitational m + m force pulling m up total mass è and m down.

5 Static Equilibrium All objects are at rest and remain so. The net force on any object must vanish. I.e. on an object:! F i = 0 i Example: Three ropes are arranged so as to support a 4 kg mass as shown below. Determine the tension in each rope. T! T! 3 A 4 kg T! 60 o Solve by considering forces on junction at point A i i i,x ( 0 ) F = 0= T + T cos 60 all T s here are positive 0= T + T /; T = T / ( 0 ) F = 0 = T sin 60 T i,y 3 ( 0) T = T /sin 60 =.6T 3 3 T 3 = mg = 39.N;T = 45.3N T =.6N

6 Normal Force N W=mg The Normal force acts perpendicular to the surface of the block on which the mass sits. It assumes whatever value is needed to prevent the block from penetrating the surface. In other words N=mg regardless of the mass of the block.

7 Example Two forces, F! and F!, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F =60 N and F = 5 N. What is the horizontal acceleration (magnitude and direction) of the block? 60 0 F!! Normal! W Normal force cancels the y components of F! F,y + Normaly W = 0 ( o ) F = 60cos 60 N= 30N,x! F and! W Fnet = F,x F = 30N 5N = 5N F F m θ 60 N 5 N 5 kg 60 o a = F/m= 5N/5kg= m/s Block accelerates to the right.

8 conceptual question Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is a) larger than b) identical to c) smaller than the downward weight W of the person.

5 s and comes to rest. What does the scale register (a) before the elevator starts to move? (b) during the first 0.8 s? (c) while the elevator is traveling at constant speed?

9 Example: with balance scale marked in Newtons, not the usual scale, but a quite reasonable one. A 00 kg man stands on a bathroom scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of. m/s in 0.80 s. It travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for.5 s and comes to rest. What does the scale register (a) before the elevator starts to move? (b) during the first 0.8 s? (c) while the elevator is traveling at constant speed? (d) during the negative acceleration? N? Scale reading =Normal force a): N = mg: scale = N = 980Newtons m 00 kg v f Δt. m/s 0.8 s b) : N mg = ma : scale = Newtons a =Δv / Δ t = (.m / s) / 0.8s =.5m / s N= m(g+ a);scale= m(g+ a) = 30Newtons c) : N = mg : scale = N = 980Newtons v i Δt. m/s.5 s d) : a = (v v ) / Δ t =.m / s / (.5s) = 0.8m / s 0 N= m(g+ a);scale= m(g+ a) = 900Newtons

10 Reading Quiz 4. An action/reaction pair of forces A. point in the same direction. B. act on the same object. C. are always long-range forces. D. act on two different objects. Slide 4-3

11 Newton s Third Law When a body exerts a force on another, the second body exerts an equal oppositely directed force on the first body. Note: the two forces act on different bodies Force on body due to body : F Force on body due to body : body body F 3 d law: F =- F

12 ! F = 60 N Example N! N! m =0 kg m = 5 kg A block with mass 5 kg and a second block with mass 0 kg are supported by a frictionless surface. A force of 60 N is applied to the 0 kg mass. What is the force of the 5 kg block on the 0 kg block? x components: body : N,x = m ; N,x = N,x from Newtons 3 d law body : F x + N,x = m ; F x = N,x + m F x = N,x + m = m + m = ( m + m ) = F x / ( m + m ) N,x = N,x = m = m F x / ( m + m ) N,x = 5kg i 60N / (5kg) = 0N N,x = 0N to the left If objects move together, the acceleration is governed by the total mass

13 Example Two skaters, an 8 kg man and a 48 kg woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes the man with a force of 45 N due east. Determine the accelerations (magnitude and direction) of the man and the woman. East x components (West is positive) a a man woman 45N = = 0.55m/s east 8kg 45N = = 0.94m / s west 48kg

Example. F and W. Normal. F = 60cos 60 N = 30N. Block accelerates to the right. θ 1 F 1 F 2

Example. F and W. Normal. F = 60cos 60 N = 30N. Block accelerates to the right. θ 1 F 1 F 2 Physic 3 Lecture 7 Newton s 3 d Law: When a body exerts a force on another, the second body exerts an equal oppositely directed force on the first body. Frictional forces: kinetic friction: fk = μk N static