Lecture 8 Applications of Newton’s Laws

Lecture 8 Applications of Newton’s Laws (Chapter 6)

Announcements Assignment #4: due tomorrow night, 11:59pm Midterm Exam #1 scores are up: Class Average 75.0%

Reading and Review

Static friction(s= 0.4) T m Will It Budge? a) moves to the left, because the force of static friction is larger than the applied force b) moves to the right, because the applied force is larger than the static friction force c) the box does not move, because the static friction force is larger than the applied force d) the box does not move, because the static friction force is exactly equal the applied force e) The answer depends on the value for μk. A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?

Static friction(s= 0.4) T m Will It Budge? a) moves to the left, because the force of static friction is larger than the applied force b) moves to the right, because the applied force is larger than the static friction force c) the box does not move, because the static friction force is larger than the applied force d) the box does not move, because the static friction force is exactly equal the applied force e) The answer depends on the value for μk. A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? The static friction force has a maximum of sN= 40 N. The tension in the rope is only30 N. So the pulling force is not big enough to overcome friction. Follow-up: What happens if the tension is 35 N? What about 45 N?

T T Fs W W Tension

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

m1 : x : m2 : y : y : fk

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 ropes. m a 10 kg Case (1) Case (2)

m T a W 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 ropes. 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) a>0 downward implies T<W

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

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.

Instantaneous acceleration Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing: From Lecture 3

The magnitude of this centripetal force is given by: If the speed is constant, the direction of the force and the acceleration is towards the center of the circle. a a For circular motion problems, it is often convenient to choose coordinate axes with one pointing along the direction of this centripetal force Circular Motion An object moving in a circle must have a force acting on it; otherwise it would move in a straight line.

Circular Motion This force may be provided by the tension in a string, the normal force, or friction, among others.

Examples of centripetal force no friction is needed to hold the track! when

Circular Motion An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well:

A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

To support mass M, the necessary tension is: necessary centripetal force: Only force on puck is tension in the string! A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?

Circular motion and apparent weight This normal force is the apparent, or perceived, weight

Key points re: circular motion 1) If object moving in a circular path Then 2) is NOT a separate force; it represents the sum of the physical forces acting on m

Going in Circles I a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?

Going in Circles I a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion? You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg. Follow-up: Where is N larger than mg?

Vertical circular motion http://www.youtube.com/watch?v=BHu8LAWSKxU

So, as long as: vertical (down) at C: Centripetal acceleration must be C at the top, then N>0 and pointing down. (now apparent weight is in the opposite direction to true weight!) B horizontal vertical (up) A Vertical circular motion Condition for falling: N=0

The Centrifuge Other common examples: • spin cycle on washing machine • salad spinner • artificial gravity on giant space station in show on the SciFi channel

c e b a d Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

c e b a d Barrel of Fun A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically. Follow-up: What happens if the rotation of the ride slows down?

Lecture 8 Applications of Newton’s Laws