Dynamics I Motion Along a Line

Example: consider the situation below where two ropes hold up a weight:  left =30 o  right = 55 o T left T right W = 100 Nt Static Equilibrium

Dynamic Equilibrium An elevator that is moving upward at 2 m/s. Find the tension T in the cable, if elevator has mass 400 kg

Ramps Why is it easier to push something up a ramp than it is to lift it? P

Ramps  F // = P - mg sin(  ) = mg //  F  = F N - mg cos(  ) = mg  = 0 W=mg P  FF  F // FNFN P balance = mg sin(  )

Ramps  F // = P – F // - F f = ma // W=mg FNFN P F f =  F c F //

Ramps W=mg FNFN P F f =  F N F N = mg cos(  ) P = mg sin(  ) +  mg cos(  ) = mg[sin(  ) +  cos(  )] sin(  ) +  cos(  ) < 1 → P < mg

Pulleys

P W = mg P = Tension = W

Pulleys P W T T P=T, 2T = W; so P=W/2

Pulleys P

P W

Newton’s law of gravitation G = 6.67 x Nm 2 /kg 2

Problem 1 Three masses are each at a vertex of an isosceles right triangle as shown. Write an expression for the force on mass three due to the other two. r r m3m3 m1m1 m2m2

Gravity at earth’s surface

Gravity: What is the force of gravity exerted by the earth on a typical physics student? –Typical student mass m = 55kg –g = 9.81 m/s 2. –F g = mg = (55 kg)x(9.81 m/s 2 ) –F g = 540 N = WEIGHT F= g F g = mg

Test your Understanding An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force. His foot hurts... (a) more (b) less (c) the same Ouch!

However the weights of the bowling ball and the astronaut are less: Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. W = mg Moon g Moon < g Earth Test your Understanding

Friction The forces shown are an action-reaction pair. Acme Hand Grenades f (force on table due to box) f v (force on box due to table)

Friction... Friction is caused by the “microscopic” interactions between the two surfaces:

Two Kinds of Friction Static friction –Must be overcome in order to budge an object –Present only when there is no relative motion between the bodies, e.g., the box & table top Kinetic friction –Weaker than static friction –Present only when objects are moving with respect to each other (skidding) FAFA fkfk F net is to the right. a is to the right. v is left or right. FAFA fsfs Objects are still or moving together. F net = 0.

Friction Facts Friction is due to electrostatic attraction between the atoms of the objects in contact. It can speed you up, slow you down, or make you turn. It allows you to walk, turn a corner on your bike, warm your hands in the winter, and see a meteor shower. Friction often creates waste heat. It makes you push harder / longer to attain a given acceleration. Like any force, it always has an action-reaction pair.

Friction Strength The magnitude of the friction force is proportional to: how hard the two bodies are pressed together (the normal force, N ). the materials from which the bodies are made (the coefficient of friction,  ). Attributes that have little or no effect: sliding speed contact area

Coefficients of Friction Static coefficient …  s. Kinetic coefficient …  k. Both depend on the materials in contact. –Small for steel on ice or scrambled egg on Teflon frying pan –Large for rubber on concrete or cardboard box on carpeting The bigger the coefficient of friction, the bigger the frictional force.

Friction... lForce of friction acts to oppose motion: çParallel to surface. N çPerpendicular to Normal force. amaama F ffFffF gmggmg N i j

Model for Sliding Friction The “heavier” something is, the greater the friction will be...makes sense!

Dynamics i :F  K N = ma j :N = mg soF  K mg = ma amaama F gmggmg N i j  K mg

Test your Understanding A box of mass m 1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (  k = 0.51) on top of a second box having mass m 2 = 3 kg, which in turn slides on a frictionless floor. –What is the acceleration of the second box ? (a) a = 0 m/s 2 (b) a = 2.5 m/s 2 (c) a = 3.0 m/s 2 m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.51  ) slides without friction a = ?

Solution First draw FBD of the top box: m1m1 N1N1 m1gm1g T f =  K N 1 =  K m 1 g

Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. m1m1 f f 1,2 m2m2 f f 2,1 =  K m 1 g Solution

Now consider the FBD of box 2: m2m2 f f 2,1 =  k m 1 g m2gm2g N2N2 m1gm1g Solution

Finally, solve F = ma in the horizontal direction: m2m2 f f 2,1 =  K m 1 g  K m 1 g = m 2 a a = 2.5 m/s 2 Solution

Inclined Plane with Friction: Draw free-body diagram:  i j mg N KNKN ma 

Inclined plane... ij lConsider i and j components of F NET = ma : i i mg sin  K N = ma j j N = mg cos   i j mg N  KNKN ma mg sin  mg cos  mg sin  K mg cos  = ma a / g = sin  K cos 

Static Friction... F gmggmg N i j fFfF So far we have considered friction acting when something moves. –We also know that it acts in un-moving “static” systems: In these cases, the force provided by friction will depend on the forces applied on the system.

Static Friction... Just like in the sliding case except a = 0. i :F  f F = 0 j :N = mg F gmggmg N i j fFfF While the block is static: f F  F

Static Friction... F gmggmg N i j fFfF

F  S is discovered by increasing F until the block starts to slide: i : F MAX  S N = 0 j : N = mg  S  F MAX / mg F F MAX gmggmg N i j  S mg

A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is  s = 0.4. A rope is attached to the box and pulled at an angle of  = 30 o above horizontal with tension T = 40 N. –Does the box move? (a) yes (b) no (c) too close to call T m  static friction (  s  = 0.4  )  Test Your Understanding

Pick axes & draw FBD of box: Apply F NET = ma y: N + T sin  - mg = ma Y = 0 N = mg - T sin q = 80 N x: T cos  - f FR = ma X The box will move if T cos  - f FR > 0 y x T m  N mg f FR Solution

x: T cos  - f FR = ma X y: N = 80 N The box will move if T cos  - f FR > 0 T cos  = 34.6 N f MAX =  s N = (.4)(80N) = 32 N So T cos  > f MAX and the box does move T m f MAX =  s N N mg y x  Solution

Static Friction: We can also consider  S on an inclined plane. In this case, the force provided by friction will depend on the angle  of the plane. 

Static Friction... ma = 0 (block is not moving) The force provided by friction, f F, depends on . mg sin  f f  (Newton’s 2nd Law along x-axis)  mg N  fFfFi j

Static Friction... We can find  s by increasing the ramp angle until the block slides: In this case: mg sin  M  S mg cos  M   S  tan  M   M mg N SNSN  i j mg sin  f f   f f  S N  S mg cos  M

Friction Facts Since f F =  N, the force of friction does not depend on the area of the surfaces in contact. By definition, it must be true that  S >  K for any system (think about it...).

Friction vs Applied force fFfF FAFA f F = F A f F =  K N f F =  S N

Problem: Box on Truck A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is  S. a çWhat is the maximum acceleration a that the truck can have without the box slipping? m SS a

Problem: Box on Truck Draw Free Body Diagram for box: –Consider case where f F is max... (i.e. if the acceleration were any larger, the box would slip). N f F =  S N mg i j

Problem: Box on Truck ij Use F NET = ma for both i and j components  I  I :  S N = ma MAX çJ çJ: N = mg a MAX =  S g N f F =  S N mg a MAX i j

Forces and Motion An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. What is the direction of the static frictional force? (a) (b) (c) FfFf FfFf FfFf SS a

Solution First consider the case where the inclined plane is not accelerating. mg FfFf N l All the forces add up to zero! mg N FfFf

If the inclined plane is accelerating, the normal force decreases and the frictional force increases, but the frictional force still points along the plane: mg N FfFf a All the forces add up to ma! çF = ma çThe answer is (a) mg FfFf N ma Solution

Putting on the brakes Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since  S >  K. The driver of a car moving with speed v o slams on the brakes. The coefficient of static friction between the wheels and the road is  S. What is the stopping distance D? abab vovo v = 0 D

Putting on the brakes ij Use F NET = ma for both i and j components  i :  S N = ma çj: N = mg a =  S g N f F =  S N mg a i j

Putting on the brakes As in the last example, find a b =  S g. l Using the kinematic equation: v 2 - v 0 2 = 2a( x -x 0 ) In our problem:0 - v 0 2 =  2a b ( D ) abab vovo v = 0 D

Putting on the brakes In our problem: 0 - v 0 2 =  2a b ( D ) l Solving for D: Putting in a b =  S g abab vovo v = 0 D

Dynamics Problem xA box of mass m = 2 kg slides on a horizontal frictionless floor. A force F x = 10 N pushes on it in the x direction. What is the acceleration of the box? y Fi F = F x i a a = ? m x

Solution Draw a picture showing all of the forces F F F B,F F F F,B F F B,E F F E,B yx

Solution Draw a picture showing all of the forces. Isolate the forces acting on the block. F F F B,F F F F,B Fg F B,E = mg F F E,B y x

Solution Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. F F F B,F gmggmg y x

Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. Solve Newton’s equations for each component. – F X = ma X – F B,F - mg = ma Y F F F B,F gmggmg yx Solution

F X = ma X – So a X = F X / m = (10 N)/(2 kg) = 5 m/s 2. F B,F - mg = ma Y – But a Y = 0 – So F B,F = mg. Normal ForceThe vertical component of the force of the floor on the object (F B,F ) is called the Normal Force (N). Since a Y = 0, N = mg in this case. FXFX N mgmg y x Solution

Recap FXFX N = mg mg a X = F X / m y x

Normal Force 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 > mg (b) N = mg (c) N < mg a

Solution All forces are acting in the y direction, so use: F total = ma N - mg = ma N = ma + mg therefore N > mg m N mg a

Tools: Ropes & Strings Can be used to pull from a distance. TensionTension (T) at a certain position in a rope is the magnitude of the force acting across a cross- section of the rope at that position. –The force you would feel if you cut the rope and grabbed the ends. –An action-reaction pair. cut T T T

Tools: Ropes & Strings... Consider a horizontal segment of rope having mass m: –Draw a free-body diagram (ignore gravity). xUsing Newton’s 2nd law (in x direction): F NET = T 2 - T 1 = ma So if m = 0 (i.e. the rope is light) then T 1 =  T 2 T1T1 T2T2 m ax

Tools: Ropes & Strings... An ideal (massless) rope has constant tension along the rope. If a rope has mass, the tension can vary along the rope – For example, a heavy rope hanging from the ceiling... We will deal mostly with ideal massless ropes. T = T g T = 0 TT

Tools: Ropes & Strings... The direction of the force provided by a rope is along the direction of the rope: mg T m Since a y = 0 (box not moving), T = mg

Force and acceleration A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s 2. What is the mass of the fish? (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg m = ? a = 12.2 m/s 2 snap !

Solution: Draw a Free Body Diagram!! T mg m = ? a = 12.2 m/s 2 l Use Newton’s 2nd law in the upward direction: F TOT = ma T - mg = ma T = ma + mg = m(g+a)

Tools: Pegs & Pulleys Used to change the direction of forces –An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: FF1FF1 ideal peg or pulley FF2FF2 FF | F 1 | = | F 2 |

Tools: Pegs & Pulleys Used to change the direction of forces –An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: mg T m T = mg F W,S = mg

Springs Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. –F X = -k x –Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = 0 x

Springs... Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. –F X = -k x –Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = -kx > 0 x x  0

Springs... Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. –F X = -k x –Where x is the displacement from the relaxed position and k is the constant of proportionality. F X = - kx < 0 x x > 0 relaxed position

Scales: Springs can be calibrated to tell us the applied force. – We can calibrate scales to read Newtons, or... –Fishing scales usually read weight in kg or lbs lb = 4.45 N

(a) (b) (c) (a) 0 lbs. (b) 4 lbs. (c) 8 lbs. mmm (1) (2) ? Force and acceleration A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs?

Solution: Draw a Free Body Diagram of one of the blocks!! l Use Newton’s 2nd Law in the y direction: F TOT = 0 T - mg = 0 T = mg = 4 lbs. mg T m T = mg a = 0 since the blocks are stationary

Solution: The scale reads the tension in the rope, which is T = 4 lbs in both cases! mmm T T T T T T T

Car going around a turn What are the forces that cause this circular acceleration? Gravity (weight) acts down Contact Force acts up (perpendicular to the surface) Friction - which way does it act (left or right)? a=v 2 /r W=mg FcFc Car is heading into the screen.

Car going around a turn Without friction the car will NOT make the turn, it will continue straight - into the left ditch! Therefore, friction by the road must push on the car to the right. (Friction by the car on the road will be opposite - to the left.) The fastest the car can go around the turn without sliding is when the friction is maximum: F f =  F c. a=v 2 /r W=mg FcFc F f  F c

Car going around a turn We now apply Newton’s Second Law - in rectangular components:  F x = F f = ma = mv 2 /r To make the car go around the turn fastest, we need the maximum force of friction: F f =  F c  F y = F c - W = 0 which says F c = mg, so  mg = mv 2 /r, or v max =  [  gr] a=v 2 /r W=mg FcFc F f  F c

Car going around a turn v max =  [  gr] Note that the maximum speed (without slipping) around a turn depends on the coefficient of friction, the amount of gravity (not usually under our control), and the sharpness of the turn (radius). If we go at a slower speed around the turn, friction will be less than the maximum: F f <  F c. There is one other thing we can do to go faster around the turn - bank the road! How does this work?

Banked turn By banking the road, we have not added any forces, but we have changed the directions of both the contact force and the friction force! Have we changed the direction of the acceleration? No - the car is still travelling in a horizontal circle. W=mg FcFc  a=v 2 /r FfFf

Banked turn Since the acceleration is still in the x direction, we will again use x and y components (rather than // and   F x = F c sin(  ) + F f cos(  ) = mv 2 /r  F y = F c cos(  ) - F f sin(  ) - mg = 0 If we are looking for the max speed, we will need the max friction: F f =  F c. This gives 3 equations for 3 unknowns: F f, F c, and v. FcFc W=mg  a=v 2 /r FfFf

Banked turn  F x = F c sin(  ) + F f cos(  ) = mv 2 /r  F y = F c cos(  ) - F f sin(  ) - mg = 0 F f =  F c. Using the third equation, we can eliminate F f in the first two: F c sin(  ) +  F c cos(  ) = mv 2 /r F c cos(  ) -  F c sin(  ) - mg = 0 We can now use the second equation to find F c : F c = mg / [cos(  ) -  sin(  )], and use this in the first equation to get: v = [gr {sin(  )+  cos(  )} / {cos(  ) -  sin(  )} ] 1/2

Banked Turn v = [g r {sin(  )+  cos(  )} / {cos(  ) -  sin(  )} ] 1/2 Notice that the mass cancels out. This means that the mass of the car does not matter! (Big heavy trucks slip on slippery streets just like small cars. When going fast, big heavy trucks flip over rather than slide off the road; little cars don’t flip over like big trucks. But flipping over is not the same as slipping! We’ll look at flipping in Part 4 of the course.) Note also that when  0, the above expression reduces to the one we had for a flat road: v max = [  gr] 1/2.

Banked Turn v = [g r {sin(  )+  cos(  )} / {cos(  ) -  sin(  )} ] 1/2 Note that as  sin(  ) approaches cos(  ), the denominator approaches zero, so the maximum speed approaches infinity! What force really supports such large speeds (and so large accelerations)? As the angle increases, the contact force begins to act more and more to cause the acceleration. And as the contact force increases, so does friction. Actually, there is a limit on the maximum speed because there is a limit to the contact force.

Banked turn - Minimum Speed Is there a minimum speed for going around a banked turn? Consider the case where the coefficient of friction is small and the angle of bank is large. In that case the car, if going too slow, will tend to slide down (to the right) so friction should act to the left. Can you get an equation for the minimum speed necessary? What changes in what we did for max speed? FcFc W=mg  a=v 2 /r FfFf