Lab 4 Newtons Laws PDF

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Physics 110 Online Lab 4 Newtons Laws...

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Introductory Physics

Hunter College

Newton’s Laws and Forces in 1 Dimension Abstract: Newtons 3 Laws of motion summarize the nature of movements of objects through effects of external forces. The first law indicates an object at rest stays at rest and an object in motion stays in motion unless acted upon an external force. The second law indicates that multiple forces acted upon an object can be summed to yield a Net force, also represented by mass times acceleration. Newton’s 3rd law of motion states that for every action there is an equal and opposite reaction, also seen as any force between two objects existing in equal magnitudes but opposite directions. Frictional forces also exist in opposite directions of external applied forces upon objects on a surface. Static friction is a set force that must be overcome to set an object in motion and kinetic friction acts on an object that is already in motion. Kinetic friction will always be less than static friction because it always requires more force to set object in motion than to keep it moving. Through stimulations and algebraic calculations Newtons 3 laws and the effects of static and kinetic friction on various objects are examined.

Introduction Newton’s 3 Laws of Motion marks the foundation for nearly all of Classical Mechanics. They describe the nature of how objects move, and are affected by external forces. To Review:

1. Newton’s 1st Law, also known as the law of Inertia, states that an object (in a constant frame of reference) at rest will stay at rest, whereas an object moving at a constant velocity (in a constant frame of reference) will keep moving at that velocity, unless acted upon by an external force. a. This is the reason that objects in deep space, relatively unaffected by the gravity of stars, can float off in one direction forever. 2. Newton’s 2nd Law, arguably the most famous, states that when multiple forces act on an object, the net force, equal to the vector sum total, is equal to the object’s mass times its net acceleration: ❑

∑ ❑ F i= F net=mobject ∗a net i

a. NOTE: Forces are vectors, and as such they add like vectors. In the case of 1-dimensional motion, this is easy to compute, but becomes non-trivial in 2 and 3 dimensions. b. Example: Gravity. The Earth constantly exerts a force on all objects around it, equal to the objects mass * the acceleration due to gravity, g:

F g=m object∗g On the Earth’s surface, this value g is approximately equal to 9.8 m/s^2. 3. Newton’s 3rd Law, or the principle of action and reaction, states that any force between two objects exists in equal magnitude and opposite in direction. That is to say, if an object A exerts a

2 force F A on object B, then object B will exert a force states that these two forces are equal:

F B on object A. Newton’s 3rd Law

F A =−F B a. Example: Normal Forces. Normal Forces are the reason that objects that collide with one another, or are in contact, don’t go through one another. As an example, we stand on the ground because the force of gravity pulls us down. However, because we’re not just constantly sinking into the ground, there must exist a force exerted by the ground back at us, which leads to 0 net acceleration:

F g−F N =0

This normal force is a consequence of Newton’s 3rd Law. It is called normal, because the force is always perpendicular, or ‘normal’, to the point of contact.

As a special case of normal forces, let’s look at friction. Friction arises when two objects move against one another, due to molecular interactions at the surface of two objects, and can differ depending on the types of objects involved. However, the friction force is always opposite the direction of the applied force/direction of motion. There are two types of friction: 1. Static Friction: Static Friction usually comes from the interlocking of edges or irregularities of two surfaces and can increase to prevent any relative motion between two objects, up to a point. The static friction force will increase to counteract the applied force, until it eventually fails, and the object begins to move. This force is proportional to the contact force between the two objects:

F static ≤ μ s∗F N where μs

is the coefficient of static friction.

3 2. Kinetic Friction: Once two objects are in motion against one another, friction still exists between them (e.g. rubbing your hands together produces heat from friction), although this force is constant and usually less than the force of static friction:

F kinetic=μk F N where μk

is the coefficient of kinetic friction.

Pre-Lab Questions (show your work): 1. Imagine you are in a car without a seat belt, travelling at a constant speed. Suddenly, you traffic light turns red, and you are forced to brake, and you find yourself lurching forward. Why is that so? 2. A force F acts on a mass M, for some time interval T, giving it an initial speed v. If the same force acted on another mass 3M, what would be the initial speed of the new mass (in terms of v)?

Experiment: 1. Open the Simulation using this link. You should come across this window:

This window contains a Force vs. time graph and a choice of objects to push. When the ‘Go’ button is pressed, the graph will begin recording Force v. time data. During this time, you can use

4 the slider to the left, or click and drag the object on top to apply a force. The HELP Button on the bottom right gives helpful tips on working the simulation.

Imagine a 200 kg File Cabinet at rest on the ground. To push it forward, we apply a 600N force for 5 seconds Assuming there were no friction, how far would the cabinet have gone in that time? Fnet=ma 600= 200 x a

d= vit + ½ at^2 d= 37.5m

a=3m/s^2 Test this out in the Simulation. Select the File Cabinet Object, and press GO on the left hand side. Apply a 600N force, and see how far the cart moves. About 37 meters

Testing Static and Kinetic Friction 1. Imagine a 200 kg File Cabinet at rest on the ground. To push it forward, we apply a 600N force for 5 seconds Assuming there were no friction, how far would the cabinet have gone in that time? a. Test this out in the Simulation. Select the File Cabinet Object, and press GO on the left hand side. Apply a 600N force, and see how far the cart moves. Same answers as above 2. Press the Clear button. Now, Press the Go button to begin collecting data. Slowly increased the applied force via dragging the cabinet, or by using the slider next to the graph, and stop just after the object begins to move. (Include screenshots of the completed graph) a. What happened to the friction force? Why did this happen? Force applied became greater than the friction force, allowing for file cabinet to start moving. This happens because the person applying the force keeps increasing the magnitude of force in the opposite direction of friction, so once the applied force exceeds static friction, object is able to move and continue to move easier. b. Based off this graph, calculate the coefficient of static friction.

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S

= 589.1/(200*9.81)= 0.30

3. Clear the previous experiment. Now press go, and push the cabinet again until it begins moving. As it moves, play with the force (increase or decrease). How does the friction force change? (include screenshots of the completed graph) a. From this graph, calculate the coefficient of kinetic friction. Ff= usmg 392N= k (200kg)(9.81) 0.20

4. This time, switch to another object from the list. Repeat steps 2 and 3 with the new object, and calculate the coefficients of static and kinetic friction.

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Calculation for Refrigerator:

2745/(400kg x 9.8)= 0.7 (static friction coefficient) 1960/)400kg x 9.8)= = 0.5 (kinetic friction coefficient) Are the coefficients the same compared to the cabinet? If they are not, why might that be? Coefficients are not the same because the mass of the objects are not the same. Both static and kinetic coefficients are greater that the file cabinet for refrigerator. The gravitational force pulling the fridge down is dependent on mass and when the mass is larger, the weight is greater allowing for there to be a need for a larger applied force to overcome static friction. As the refrigerator is in motion as its being pushed more force is still required to keep it in motion due to its larger mass, allowing for kinetic coefficient to be higher that file cabinet as well.

5. This is an example that can be done at home. Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? a. This latter situation is particularly important for drivers to note, especially after a light rain shower. Why?

The Inclined Ramp 1. Let’s explore how Newton's laws and friction apply to inclined surfaces. Open the Inclined Ramp simulation, which can be found here.

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2. Upon opening the simulation, set the ramp angle to 0°, using the slider on the right. Describe the forces on the file cabinet while it is on the ramp. There is a Normal force directed up, gravitational force directed down, friction force directed to opposite direction of force applied in the horizontal directions and vice versa. If on a ramp with no incline and no applied force, only gravitation and normal forces are acting. 3. Reset the scenario using the RESET on the upper right-hand side. Now press go, and slowly lift the ramp. As you lift the ramp what do you notice happening to the forces on the file cabinet? Describe each change. Is there anything that does not change? The gravitational and frictional forces stay the same until it reaches the required angle, then the gravitational and normal force increases in magnitude as friction decreases in magnitude. a. At what angle does the cart begin to move? About 17 degrees (~16.88). 4. This time replace the cabinet with the 300kg crate, and repeat step 3. Are there any notable differences? Yes, firstly, the gravitational and normal forces are much greater in magnitude because the mass of the crate is much larger. The frictional force is also greater because the F Normal is larger. Due to this, the angle required for crate to move will be much larger( ~36.3 degreees), because the gravitational force of the crate is pulling down with a greater force on the ramp along with a greater frictional force, requiring a greater angle to actually get crate to move.

Post-Lab Questions (Show your work) 1. A 1200kg car is driving along a straight road at a speed of 50 m/s, when it brakes sharply. The resulting friction brings the car to rest. What is the frictional force if the coefficient of friction is 0.6, and how far does the car travel before it comes to rest? m=1200kh uk= 0.6. Ff= (0.6)(1200 x 9.8) = 7056N vi=50m/s. Fd= 1/2mv2-1/2mv2 Vf =0 m/s 7056N d = 0-1,500,000 d= 212.58

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2. A 12kg slab is being dragged across a dirt path with an applied force of 300N, and has a net acceleration of 5m/s^2. What is the coefficient of static friction between the slab and the ground? us= F-ma / mg us= (300N)-(12kg)(5m/s2) / (12 x 9.8) coefficient of static friction= 2.04

3. Imagine a 60 kg skydiver free-falling to earth, when they pull their parachute. After pulling the parachute, they descend 46 meters in 4 seconds. Draw a free body diagram labelling all the forces on the skydiver. The relative magnitudes of all forces should be drawn to scale. What is the force of air resistance on the diver? d=vit+1/2at2 a= 2d/t2a= (2)(46)/ 42 a= 5.75m/s2 Fair = mg-ma (60)(9.8) - (60) (5.75) = 243N 4.

A toddler exerts a horizontal force of 20 N to pull a mass of 50 kg on a horizontal surface whose coefficient of static friction is 0.1. Find the static friction force and explain how you obtain your answer.

Fg= mg FN=mg = (50x9.8) = 490N Ff= uFN Ffs = (0.1)(490) = 49N Fapplied = 20N < 49N so , Fapplied= F friction = 20N because if F applied is less than usFN then friction force acts equal to force applied until force applied exceeds maximum force of static friction. 5. A 10 kg mass is placed on an incline as shown. The coefficient of static friction is 0.5. Will this mass slide? Find the frictional force and explain how you obtain your answer.

F N =Mg cos20o

Mg sin20o

20o

FN= mgcos(20) FN = (10)(9.8)(cos(20))= 92.08 Ffr= 0.5 x 92.08= 46.04N

Mg

F=mgsin(20) F= 10 x 9.8 x sin(20)= 33.51N 33.51N...