Lecture 5 Newton ’ s Laws of Motion

Lecture 5 Newton’s Laws of Motion (sections 5.5-5.7)

Exam #1 - next 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 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push. (a) Is the force experienced by the child more than, less than, or the same as the force experienced by the parent? (b) Is the acceleration of the child more than, less than, or the same as the acceleration of the parent? Explain. (c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the magnitude of the parent’s acceleration? 3

On vacation, your 1300-kg car pulls a 540-kg trailer away from a stoplight with an acceleration of 1.9 m/s2 (a) What is the net force exerted by the car on the trailer? (b) What force does the trailer exert on the car? (c) What is the net force acting on the car? 5

Collision Course I a) the car b) the truck c) both the same d) it depends on the velocity of each e) it depends on the mass of each A small car collides with a large truck. Which experiences the greater impact force?

Collision Course I a) the car b) the truck c) both the same d) it depends on the velocity of each e) it depends on the mass of each A small car collides with a large truck. Which experiences the greater impact force? According to Newton’s Third Law, both vehicles experience the same magnitude of force.

a2 = 2a1 m1 m1 m2 a1 F m2 a3 F F F = m1 a1 Force and Two Masses a) ¾a1 b) 3/2a1 c) ½a1 d) 4/3a1 e) 2/3a1 A force Facts on mass m1giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2= 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? F = m2 a2 a3 = ?

a2 = 2a1 m1 m1 F m2 a1 m2 a3 F F F = m1 a1 F = (3/2)m1 a3 => a3 = (2/3) a1 Force and Two Masses a) ¾a1 b) 3/2a1 c) ½a1 d) 4/3a1 e) 2/3a1 A force Facts on mass m1giving acceleration a1. The same force acts on a different mass m2 giving acceleration a2= 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? Mass m2 must be ( m1)because its acceleration was 2a1 with the same force. Adding the two masses together gives ( )m1, leading to an acceleration of ( )a1 for the same applied force. F = m2 a2 = (1/2 m1 )(2a1 )

In order to change the velocity of an object – magnitude or direction – a net force is required. Newton’s third law: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1. (I) (II) (III) Newton’s Laws

An archer shoots a 0.022-kg arrow at a target with a speed of 57 m/s. When it hits the target, it penetrates to a depth of 0.085 m. (a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain. 12

When you lift a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.

b) a) When you lift a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.

F m 2m Contact Force a) 2F b) F c) ½ F d) 1/3 F e) ¼ F Two blocks of masses 2mand mare in contact on a horizontal frictionless surface. If a force F0 is applied to mass 2m, what is the force on mass m?

F m 2m Contact Force a) 2F b) F c) ½ F d) 1/3 F e) ¼ F Two blocks of masses 2mand mare in contact on a horizontal frictionless surface. If a force F0 is applied to mass 2m, what is the force on mass m? The force F0 leads to a specific acceleration of the entire system. In order for mass mto accelerate at the same rate, the force on it must be smaller! For the two blocks together, F0 = (3m)a. Since a is the same for both blocks, Fm = ma

(a) (b) m The lighter box will require less net force than the heavier box did (for the same acceleration). If the external force is instead applied to the heavier box, the contact force must be less than it was before: mM Mm Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass of 5.2 kg, the heavier box has a mass of 7.4 kg. (a) Find the contact force between these boxes when a horizontal force of 5.0 N is applied to the light box. (b) If the 5.0-N force is applied to the heavy box instead, is the contact force between the boxes the same as, greater than, or less than the contact force in part (a)? Explain. (c) Verify your answer to part (b) by calculating the contact force in this case. The boxes will remain in contact, so must have the same acceleration. Use Newton’s second law to determine the magnitude of the contact forces

Forces in Two Dimensions The easiest way to handle forces in two dimensions is to treat each dimension separately, as we did for kinematics.

Normal Forces The normal force is the force exerted by a surface on an object, to keep an object above the surface. The normal force is always perpendicular to the surface.

The normal force may be equal to, greater than, or less than the weight.

No vertical motion, so

Weight Apparent weight: Your perception of your weight is based on the contact forces between your body and your surroundings. If your surroundings are accelerating, your apparent weight may be more or less than your actual weight. In this case the “apparent weight” is the normal force, and is equal to the sum of the gravitational attaction (actual weight) and the force required to accelerate the body, as specified

Case 1 Case 2 Normal Force a) case 1 b) case 2 c) it’s the same for both d) depends on the magnitude of the force F e) depends on the ice surface Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle . In which case is the normal force greater?

Case 1 Case 2 Normal Force a) case 1 b) case 2 c) it’s the same for both d) depends on the magnitude of the force F 5) depends on the ice surface Below you see two cases: a physics student pulling or pushing a sled with a force F that is applied at an angle . In which case is the normal force greater? In case 1, the force F is pushing down (in addition to mg), so the normal force needs to be larger. In case 2, the force F is pulling up, against gravity, so the normal force is lessened.

A B On an Incline a) case A b) case B c) both the same (N = mg) d) both the same (0 < N < mg) e) both the same (N = 0) Consider two identical blocks, one resting on a flat surface and the other resting on an incline. For which case is the normal force greater?

A B On an Incline a) case A b) case B c) both the same (N = mg) d) both the same (0 < N < mg) e) both the same (N = 0) Consider two identical blocks, one resting on a flat surface and the other resting on an incline. For which case is the normal force greater? In case A, we know that N = W. In case B, due to the angle of the incline, N < W. In fact, we can see that N = W cos(). y x N f θ Wy W 

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

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.

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)

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? 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 m a

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? 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 Σ F = N – mg = ma > 0 ∴N > mg Follow-up: What is the normal force if the elevator is in free fall 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: 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 35

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: 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 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 

Lecture 5 Newton ’ s Laws of Motion