Applications of Newton’s Laws Lecture 7 Applications of Newton’s Laws (Chapter 6)

Reading and Review

Going Up I v m a) N > mg b) N = mg c) N < mg (but not zero) A block of mass m rests on the floor of an elevator that is moving upward at constant speed. What is the relationship between the force due to gravity and the normal force on the block? a) N > mg b) N = mg c) N < mg (but not zero) d) N = 0 e) depends on the size of the elevator m v

Going Up I v m a) N > mg b) N = mg c) N < mg (but not zero) A block of mass m rests on the floor of an elevator that is moving upward at constant speed. What is the relationship between the force due to gravity and the normal force on the block? a) N > mg b) N = mg c) N < mg (but not zero) d) N = 0 e) depends on the size of the elevator m v The block is moving at constant speed, so it must have no net force on it. The forces on it are N (up) and mg (down), so N = mg, just like the block at rest on a table.

Going Up II A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? a) N > mg b) N = mg c) N < mg (but not zero) d) N = 0 e) depends on the size of the elevator m a

Going Up II N m a > 0 mg F = N – mg = ma > 0 A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? a) N > mg b) N = mg c) N < mg (but not zero) d) N = 0 e) depends on the size of the elevator m a > 0 mg N The block is accelerating upward, so it must have a net upward force. The forces on it are N (up) and mg (down), so N must be greater than mg in order to give the net upward force! F = N – mg = ma > 0 →N = mg + ma > mg Follow-up: What is the normal force if the elevator is in free fall downward?

Frictional Forces Friction has its basis in surfaces that are not completely smooth:

Kinetic friction Kinetic friction: the friction experienced by surfaces sliding against one another. This frictional force is proportional to the contact force between the two surfaces (normal force): The constant is called the coefficient of kinetic friction. fk always points in the direction opposing motion of two surfaces

Frictional Forces fk Naturally, for any frictional force on a body, there is an opposing reaction force on the other body

fk fs Kinetic Friction ≤ Static Friction Frictional Forces when relative motion stops, surfaces settle into one another static friction when moving, one bumps “skip” over each other kinetic friction Kinetic Friction ≤ Static Friction

Static Friction The static frictional force tries to keep an object from starting to move when other forces are applied. The static frictional force has a maximum value, but may take on any value from zero to the maximum... depending on what is needed to keep the sum of forces to zero. The maximum static frictional force is also proportional to the contact force

Similarities between →normal forces and →static friction Variable; as strong as necessary to prevent relative motion either - perpendicular to surface (normal) - parallel to surface (friction)

Characteristics of Frictional Forces Frictional forces always oppose relative motion Static and kinetic frictional forces are independent of the area of contact between objects Kinetic frictional force is also independent of the relative speed of the surfaces. Coefficients of friction are independent of the mass of objects, but in (most) cases forces are not: (twice the mass → twice the weight → twice the normal force → twice the frictional force)

Coefficients of Friction Q: what units?

Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? a) pushing her from behind b) pulling her from the front c) both are equivalent d) it is impossible to move the sled e) tell her to get out and walk 1 2

Going Sledding Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this? a) pushing her from behind b) pulling her from the front c) both are equivalent d) it is impossible to move the sled e) tell her to get out and walk 1 2 In case 1, the force F is pushing down (in addition to mg), so the normal force is larger. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. Recall that the frictional force is proportional to the normal force.

  Measuring static coefficient of friction If the block doesn’t move, a=0. x y N W fs Wx Wy at the critical point  

  Acceleration of a block on an incline y x N W v If the object is sliding down - x y v N W fk Wx Wy  

  Acceleration of a block on an incline y x N W If the object is sliding up - x y v N Wx Wy fk  W What will happen when it stops? 

A mass m, initially moving with a speed of 5 A mass m, initially moving with a speed of 5.0 m/s, slides up a 30o ramp. If the coefficient of kinetic friction is 0.4, how far up the ramp will the mass slide? If the coefficient of static friction is 0.6, will the mass eventually slide down the ramp?

A mass m, initially moving with a speed of 5 A mass m, initially moving with a speed of 5.0 m/s, slides up a 30o ramp. If the coefficient of kinetic friction is 0.4, how far up the ramp will the mass slide? If the coefficient of static friction is 0.6, will the mass eventually slide down the ramp?

Sliding Down II A mass m is placed on an inclined plane ( > 0) and slides down the plane with constant speed. If a similar block (same ) of mass 2m were placed on the same incline, it would: a) not move at all b) slide a bit, slow down, then stop c) accelerate down the incline d) slide down at constant speed e) slide up at constant speed Answer: d m

Sliding Down II A mass m is placed on an inclined plane ( > 0) and slides down the plane with constant speed. If a similar block (same ) of mass 2m were placed on the same incline, it would: a) not move at all b) slide a bit, slow down, then stop c) accelerate down the incline d) slide down at constant speed e) slide up at constant speed  W N f Wx Wy The component of gravity acting down the plane is double for 2m. However, the normal force (and hence the friction force) is also double (the same factor!). This means the two forces still cancel to give a net force of zero.

Translational Equilibrium “translational equilibrium” = fancy term for not accelerating = the net force on an object is zero example: book on a table example: book on a table in an elevator at constant velocity

Tension When you pull on a string or rope, it becomes taut. We say that there is tension in the string. Note: strings are “floppy”, so force from a string is along the string!

Tension in a chain Tup = Tdown when W = 0 W Tup Tdown

Tension is the same everywhere in a massless rope! The tension in a real rope will vary along its length, due to the weight of the rope. T3 = mg + Wr In this class: we will assume that all ropes, strings, wires, etc. are massless unless otherwise stated. T2 = mg + Wr/2 T1 = mg Tension is the same everywhere in a massless rope! m

Idealization: The Pulley An ideal pulley is one that only changes the direction of the tension along with a rope: useful class of problems of combined motion distance box moves = distance hands move speed of box = speed of hands acceleration of box = acceleration of hands

Tension in the rope? Translational equilibrium?

Translational equilibrium? 2.00 kg Translational equilibrium? W T Tension in the rope?

y : m1 : x : m2 : y :

Three Blocks Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? a) T1 > T2 > T3 b) T1 < T2 < T3 c) T1 = T2 = T3 d) all tensions are zero e) tensions are random a T3 T2 T1 3m 2m m

Three Blocks Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? a) T1 > T2 > T3 b) T1 < T2 < T3 c) T1 = T2 = T3 d) all tensions are zero e) tensions are random T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass. a T3 T2 T1 3m 2m m Follow-up: What is T1 in terms of m and a?

Over the Edge m m a a F = 98 N Case (1) Case (2) In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes. a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) m a F = 98 N m a 10 kg Case (1) Case (2)

Over the Edge m m a a F = 98 N Case (1) Case (2) In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes. a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) m a F = 98 N In case (2) the tension is 98 N due to the hand. In case (1) the tension is less than 98 N because the block is accelerating down. Only if the block were at rest would the tension be equal to 98 N. m a 10 kg Case (1) Case (2)

Elevate Me You are holding your 2.0 kg physics text book while standing on an elevator. Strangely, the book feels as if it weighs exactly 2.5 kg. From this, you conclude that the elevator is: a) in freefall b) moving upwards with a constant velocity of 4.9 m/s c) moving down with a constant velocity of 4.9 m/s d) experiencing a constant acceleration of about 2.5 m/s2 upward e) experiencing a constant acceleration of about 2.5 m/s2 downward Use Newton’s 2nd law! the apparent weight: and the sum of forces: give a positive acceleration ay