Lecture 7 Newton ’ s Laws of Motion

Lecture 7 Newton’s Laws of Motion

Midterm Test #1 - Thursday! • 21 multiple-choice problems • A calculator will be needed. • CHECK YOUR BATTERIES! • NO equations or information may be stored in your calculator. This is part of your pledge on the exam. • Scratch paper will be provided, to be turned in at the end of the exam. • A sign-in sheets will be used, photographs of the class will be taken, all test papers will be collected

A weight on a string... if I pull the bottom string down, which string will break first? Ftop a) top string b) bottom string c) there is not enough information to answer this question W Fbot How quickly is the string pulled? A sudden, strong tug is resisted by the inertia of the mass, protecting the top string. A gradual pull forces the top string to keep the system in equilibrium.

Translation 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

Before practicing his routine on the rings, a 67-kg gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal. (a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet? (b) If the angle his arms make with the horizontal is greater that 24°, and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the value found in part (a)? Explain.

Before practicing his routine on the rings, a 67-kg gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both arms slope upward at an angle of 24° above the horizontal. (a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet? (b) If the angle his arms make with the horizontal is greater that 24°, and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the value found in part (a)? Explain. N a) b) if the angle is larger and everything else remains the same, the applied forces are more vertical. With more upward force from the arms, LESS normal force is required for zero acceleration

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

Going Up II a) N > mg b) N = mg c) N < mg (but not zero) d) N = 0 e) depends on the size of the elevator 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? m a

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 mgin order to give the net upward force! m mg a > 0 Going Up II a) N > mg b) N = mg c) N < mg (but not zero) d) N = 0 e) depends on the size of the elevator 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? F = N – mg = ma > 0 Thus, N =mg+ma > mg Follow-up: What is the normal force if the elevator is in free fall downward?

Elevate Me 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 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: 10

Elevate Me 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 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: Use Newton’s 2nd law! the apparent weight: and the sum of forces: give a positive acceleration ay

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

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

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

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

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 tozero. The maximum staticfrictional force is also proportional to the contact force

Static Friction A block sits on a flat table. What is the force of static friction? a) zero b) infinite c) you need to tell me stuff, like the mass of the block, μs, and what planet this is happening on

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?

Wx at the critical point Measuring static coefficient of friction N y fs Wy x W If the block doesn’t move, a=0. Given the “critical angle” at which the block starts to slip, what is μs?  

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

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

m Sliding Down II 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 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:

f N Wy  W Wx  Sliding Down II 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 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: 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.

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!

Tup = Tdown when W = 0 Tup Tdown W Tension in a chain In this class: we will assume that all ropes, strings, wires, etc. are massless unless otherwise stated. Tension is the same everywhere in a massless rope!

Massive vs. 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

Three Blocks a) T1 > T2 > T3 b) T1 < T2< T3 c) T1= T2= T3 d) all tensions are zero e) tensions are random 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 T3 T2 T1 3m 2m m

Three Blocks a) T1 > T2 > T3 b) T1 < T2< T3 c) T1= T2= T3 d) all tensions are zero e) tensions are random 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? 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 T1in terms of m and a?

Ty T T T T W Tension Force is always along a rope

Idealization: The Pulley An ideal pulley is one that simply changes the direction of the tension distance box moves = distance hands move speed of box = speed of hands acceleration of box = acceleration of hands

Tension in the rope?

T T W 2.00 kg W Tension in the rope?

m1 : x : m2 : y : μk μk y : μk fk μk μk

m a F = 98 N Over the Edge a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) 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 rope and frictionless table. m a 10 kg Case (1) Case (2) 35

m a F = 98 N Over the Edge 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 rope and frictionless table. a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) 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) 36

Springs Hooke’s law for springs states that the force increases with the amount the spring is stretched or compressed: The constant k is called the spring constant.

Springs Note: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite.

A mass M hangs on spring 1, stretching it length L1 Springs and Tension Mass M hangs on spring 2, stretching it length L2 S1 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S2 a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2

A mass M hangs on spring 1, stretching it length L1 Fs=T Springs and Tension W Mass M hangs on spring 2, stretching it length L2 S1 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S2 a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2

A mass M hangs on spring 1, stretching it length L1 Fs=T Springs and Tension W Mass M hangs on spring 2, stretching it length L2 S1 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S2 a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2 Spring 1 supports the weight. Spring 2 supports the weight. Both feel the same force, and stretch the same distance as before.

Lecture 7 Newton ’ s Laws of Motion