PHYS 11 _ Newton’s Third Law

Lindsay Lim Block D Nov.11 th 2021 Newton’s Third Law Purpose The purpose of this lab is to gain a better understanding of Newton’s 3 rd Law. We will use an interactive Atwood's Machine simulation that creates three dynamic scenarios involving masses where we will have to use our previous knowledge of Newton’s 1 st and 2 nd law to solve dynamic problems using measurements of distance and time to find acceleration and tension force. Materials Refer to the lab guide (J 1-4) Procedure Refer to the lab guide (J 1-4) Data Collection Atwood’s Machine Part 1 - Scenario 1/2 Scenario 1 Scenario 2 Sketch the situation Free Body Diagram (FBD) Fg = + 29.4 N Fg = - 29. 4N T = - 29. 4N T = +2 9.4 N M 1 M 2 Fg = + 9.8N Fg = - 9.8N T = - 9.8 N T = + 9.8 N M 1 M 2

Lindsay Lim Block D Nov.11 th 2021 Acceleration 0m/s 2 0m/s 2 Tension of rope 29.4N 9.8N If the rope was cut what would happen? If the rope was cut they would both fall at the same time because they have an equal magnitude of force of gravity and tension. Mass (LARGE) F grav = mg m = F grav /g m = 29.4N/9.8m/s m = 3.0 kg (SMALL) F grav = mg m = F grav /g m = 9.8N/9.8m/s m = 1.0 kg With the rope in place what happens? Why? The rope doesn’t move because the magnitude of the F Net (0N)on each side balances each other. How do the masses apply forces on each other? The masses both provide an equal magnitude of tension in the opposite direction on the rope that connects the masses. They also have the same force of gravity acting on them. What is true about these forces? The forces have an equal magnitude of 9.8N, however both the tension forces and gravitational forces on each mass are opposite in direction. For example, the tension force on blue mass is - 9.8N while red tension force is + 9.8N.

Lindsay Lim Block D Nov.11 th 2021 Red Mass 1.0kg Blue Mass 3.0kg Why does this situation behave differently from scenarios 1 and 2? This scenario is different from scenario 1 and 2 because there is movement and acceleration. Although the forces are balanced the masses are not. The weight of the blue mass is heavier and so the F Net of the blue mass will be bigger than the red mass causing an imbalance and the blue mass to fall downwards with an acceleration of 4.9m/s 2 How does the tension in the rope compare to the values in scenarios 1 and 2? The tension in this scenario is different then the others because in this scenario there is movement. In scenario 1 and 2 the tension remained at a constant value of 29.4N and 9.8N because there was no movement or acceleration. In this scenario however the tension on the rope changes as the mass accelerates downward. The tension starts at 29.6N like scenario 1 , but because of the imbalance of masses it accelerates downwards changing the tension to 14.7N and than when at rest on the ground is 9.8N Why does the blue mass move down The blue mass moves down because it has a higher force of gravity of +29.4N than force of tension which is -14.7N. Since the force of gravity is higher the blue mass will accelerate in that direction which is downwards at 4.9m/s 2 F Net equation for the FBD of blue mass F Net = F g - (-T) F Net = ma ma = F g - (-T) - T 2 = ma - F g - T 2 = (3.0kg)(4.9m/s 2 ) - 29.4N - T 2 = (14.7m/s 2 ) - 29.4N - T 2 = -14.7N T 2 = - 14.7N Tension of rope ( blue mass) -14.7N F Net equation for the FBD of b red mass F Net = F g - T F Net = ma ma = F g - T - T= ma - F g - T= (1.0kg)(-4.9m/s 2 ) - 9.8N - T= (-4.9 ) - 9.8N - T= -14.7N T= 14.7N Tension of rope ( red mass) 14.7N

Lindsay Lim Block D Nov.11 th 2021 With the rope in place what happens to the masses? Why do they behave this way? Why does the red mass accelerate up?Why does the blue mass accelerate down? The blue mass moves downwards and causes a reaction force of the red mass moving upwards. This occurs because there is an imbalance of forces. The red mass has a bigger tension force of 14.7N compared to the gravitational force which is 9.8N and so it accelerates upwards. The same thing happens to the blue mass, but in the opposite direction. The blue mass has a higher gravitational force of 29.4N than the opposing tension force of 14.7N and so accelerates downwards. How does the red mass apply a force to the blue mass? How does the blue mass apply a force to the red mass?What must be true about these “forces” ? The red mass and blue mass both apply a force of tension on each other through the rope that are equal in magnitude but opposite in direction.

Lindsay Lim Block D Nov.11 th 2021 a = 4.9m/s 2 Tension of rope 4.9N Red cart mass 1.0kg Blue mass 1.0kg How does the tension in the rope compare to the values in scenario 1 and 2? The tension in the rope is 4.9N which is lower than both scenarios one and two which had a tension of 9.8N and 24.9N. Why does the blue mass move down The blue mass moves down because it has a higher force of gravity of 9.8N than the tension force of 4.9N. This difference in forces causes an imbalance and so the mass will accelerate in the direction of the bigger force which in this case is gravitational force in the downward direction. F Net equation for the FBD of blue mass F Net = F fr - T F Net = ma ma = F fr - (-T) -T= ma - F fr -T= (1kg)(-4.9m/s 2 ) - 0N -T= (-4.9 ) - 0N -T= -4.9N T= - 4.9N Tension of rope ( blue mass ) - 4.9N F Net equation for the FBD of b red mass T= ma - F fr T= (1kg)(-4.9m/s 2 ) - 0N T= (-4.9 ) - 0N T= -4.9N T= 4.9N Tension of rope ( red mass) +4.9N

Lindsay Lim Block D Nov.11 th 2021 With the rope in place what happens to the cart and mass?Why do they behave this way? The rope connects the mass to the cart and so when the blue mass accelerates downwards it causes an equal and opposite reaction and pulls the red mass to the right. They behave this way because each mass has an imbalance of forces and so they accelerate in the direction of the bigger force. The red mass accelerates right because it has a higher tension force of 4.9N than opposing force and the blue mass accelerates downwards because it has a higher gravitational force of 9.8N than tension force. How does the red mass apply a force to the blue mass? How does the blue mass apply a force to the red mass?What must be true about these “forces” ? The red mass and blue mass both apply a force of tension on each other through the rope that are equal in magnitude but opposite in direction. Part 3 Advanced Problem Solving with Newton’s Third Law Scenario 1 Scenario 2 Sketch the situation Free Body Diagram (FBD) Distance 2.06m 2.03m ‘ ‘ 1.0k g 1 . 0 k g T 2 T 1 F g 2 F g 1 F N F f r 1.0k g 3 . 0 k g T 2 T 1 F g 2 F g 1 F N F f r

Lindsay Lim Block D Nov.11 th 2021 Coefficient of Friction F fr = F ? N Normal Force F N F N = F g = mg = (1.0) (9.8) F N =9.8N Acceleration d = 1/2v i t +1/2at 2 2.06m = 0 +1/2a(1.18s) 2 2.06m = 0 +1/2a(1.3924) 2.06m = 0 +a(0.6962) a = 2.96m/s 2 Tension F Net = F g - T 2 F Net = ma ma = F g - T 2 ma = F g - T 2 - T 2 = ma - F g - T 2 = (1kg)(2.96m/s 2 ) - 9.8N ` - T 2 = -6.84N -T 2 = T 1 T 1 =6.84N Force of Friction ma = F fr - T 1 (1.0kg) (2.96m/s 2 ) = F fr 6.84N 2.96 = F fr 6.84N F fr = 3.88N Coefficient of Friction F fr = F N 𝜇 = F fr /F N = 3.88N / 9.8N = 0.40 ? F fr = F ? N Normal Force F N F N = F g = mg = (1.0) (9.8) F N =9.8N Acceleration d = 1/2v i t +1/2at 2 2.03m = 0 +1/2a(0.8s) 2 2.03m = 0 +1/2a(0.64) 2.03m = 0 +a(0.32) a = 6.34m/s 2 Tension F Net = F g - T 2 F Net = ma ma = F g - T 2 ma = F g - T 2 - T 2 = ma - F g - T 2 = (3kg)(6.34m/s 2 ) - 9.8N - T 2 = (19.02 ) - 9.8N - T 2 = 9.22N -T 2 = T 1 T 1 =9.22N Force of Friction ma = F fr - T 1 (1.0kg) (6.34m/s 2 ) = F fr 9.22N 6.34 = F fr 9.22N F fr = 2.88N Coefficient of Friction F fr = F N 𝜇 = F fr /F N = 2.88N / 9.8N = 0.29 ?

Lindsay Lim Block D Nov.11 th 2021 Discussion Questions 1. Summarize your findings from part 1 and provide evidence that supports Newton’s Third Law Newton’s third law states that every action has an equal and opposite reaction. In part 1 we hung two masses of equal weight over a pulley system and studied the movement. What we observed was that the masses did not move. If we use scenario 1 we can see Newton's third law applied on the tension force of +29.4N applied on the rope by the blue mass and an equal and opposite reaction of -29.4N applied by the red mass. These equal forces cancel out and are a contributing factor to why the masses did not move during the simulation. Another example is in scenario 2 the F Net of the blue mass is +19.6N and the opposing F Net of the red mass is -19.6N. This shows Newton's third law because the F Net of each mass are equal in magnitude but opposite in direction and so cancel each other out and that is why there is no movement. Newton’s third law explains why what happens in part 1 occurs.a 2. Summarize the key steps that are required to solve a dynamic problems When it comes to solving a dynamic problem the first thing you should do is draw and label a free body diagram of all your forces and acceleration if it applies. When it comes to dealing with dynamic problems the key is to find acceleration. If not given the mass it can be determined by using the equation F g = mg. Once you determine the mass you can calculate the acceleration like how I did in part 2 by either using the kinematic equation d = 1/2v i t +1/2at 2 or the dynamic equation F Net = ma, both will give you the same answer. Once you have found acceleration you can find either tension or force of gravity using the equations F Net = ma and F Net = Fg + (- T 1 ) + T 2 and combining them to equal ma = Fg +(- T 1 ) + T 2 . Input the number from your free body diagram into the equation and ensure your answer is in correct sig figs. 3. Calculate force of tension and acceleration without distance or time FBD Acceleration F Net = F g + (-T 1 ) F Net = ma 1.0k g Fg = - 9.8N FN = + 9.8N 3.0k g Fg = +29.4 N . T = +7. 4N T = -7. 4N

Lindsay Lim Block D Nov.11 th 2021 Conclusion The point of this lab was to gain a better understanding of Newton’s third law: Every action force has an equal and opposite reaction force. In part 1 of the lab there were two hanging masses of equal weight over a pulley and what I observed was that no movement occurred. Using Newton’s third law I can explain why this happened. If you separate the masses and look at individual forces you can see they are equal in magnitude. In scenario 1 both masses had a force of tension of (29.4N) and a force of gravity of (29.4N), however each mass's force was opposite in direction compared to each other and themselves. If you look at the tension forces they are equal in magnitude, however they are also opposite in direction with the red mass being +29.4N and the blue mass being -29.4N. In math terms they being opposites means they cancel each other out. This applies for the force of gravity, F Net and on each mass tension vs force of gravity. So using Newton’s third law of motion we can understand that the masses don’t move because for every action force of +29.4N there is an equal and opposite reaction force of -29.4N that cancels each other out causing no movement. In part 2, it was the same scenario but the masses had different weights and unlike part 1 I observed movement. In simple terms Newton’s third law could be shown by the blue force (action force) going down at an acceleration of 4.9m/s 2 and so the red mass has a reaction force of accelerating 4.9m/s 2 upwards. This action and reaction can be explained by an imbalance of forces. The red mass has a higher force of tension than force of gravity and so it has a F Net of 4.9N upwards in contrast the blue mass has a bigger force of gravity downwards than force of tension and so has a F Net of 14.7N downwards. Part 3 the same thing occurs, but this time one of the masses is on a table. Although there is now added gravity force and normal force they are on a different axis and can be canceled out. The exact thing happens in part 2 in part 3 the red mass has no opposing force to the tension force of 4.9N and so the mass would be pulled right and and the blue mass has a higher force of gravity than force of tension by 4.9N and so would accelerate downwards. This scenario goes through Newton’s third law because the action force of the blue mass accelerating at 4.9m/s 2 downwards causes a reaction force of the red mass accelerating 4.9m/s 2 to the right. During this lab I also developed and practiced solving dynamics by drawing free body diagrams and using both dynamic and kinematic equations to help find tension, F Net and acceleration. Taking what I know from Newton’s first and second law as well as analyzing the free body diagrams I was able to physically see the imbalance of forces and gain a better understanding of why the masses would move the way they do. This lab has helped me gain a better understanding of dynamic equations and forces as well as showed me examples of Newton’s third law: every action force has an equal and opposite reaction force.