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Sir Isaac Newtons baby lays in his crib, staring up at his mobile and thinks to himself: I wonder what the tension is in each of those massless strings.

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Presentation on theme: "Sir Isaac Newtons baby lays in his crib, staring up at his mobile and thinks to himself: I wonder what the tension is in each of those massless strings."— Presentation transcript:

1 Sir Isaac Newtons baby lays in his crib, staring up at his mobile and thinks to himself: I wonder what the tension is in each of those massless strings. Please find the tensions T 1 and T 2, so that baby Newton can sleep. m 1 =3.5kg m 2 =4.6kg T2T2 T1T1 m1m1 m2m2 y x

2 Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Reflection Questions

3 1.What physics principle should we use to solve this problem? (Assume that the mobile is in static equilibrium.) a) a) Newtons 1st Law b) b) Newtons 2nd Law c) c) Newtons 3nd Law

4 Newtons 1st law tells us that an object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Nothing is moving or changing in this situation. Newtons 1st law is irrelevant here. Choice: A Incorrect

5 Choice: B Correct Analyzing the forces acting on the pieces of the mobile and using the given quantities, we will be able to solve this problem. Applying the conditions of static equilibrium on a system involves using Newtons 2nd Law.

6 Choice: C Incorrect Newtons 3rd law tells us that for every action, there is an equal and opposite reaction. This applies to all situations, but it does not help us gain a useful equation to find the tension in the strings.

7 2. What are the conditions for static equilibrium in 2 dimensions? a) b) c) =Torque

8 The net force in the x-direction must also be zero. Choice: A Incorrect

9 The sum of torques through any axis of rotation must be zero. Choice: B Incorrect

10 Choice: C Correct All of these conditions are required for static equilibrium. Notice that in the situation at hand: We chose to calculate torque on the axis defined by the strings. Nothing is rotating. None of the forces cause torques, because none of them have a component that is perpendicular to any possible lever arm. There are NO torques in this problem. *If you choose a point other than on the line of the strings to calculate torque, the torque equation will provide a result, but no new useful equation will be obtained. Notice that in the situation at hand: We chose to calculate torque on the axis defined by the strings. Nothing is rotating. None of the forces cause torques, because none of them have a component that is perpendicular to any possible lever arm. There are NO torques in this problem. *If you choose a point other than on the line of the strings to calculate torque, the torque equation will provide a result, but no new useful equation will be obtained. In General we must consider 3 dimensions: The sum of forces in the z-component must also be zero. We are simplifying this problem to 2 dimensions. In General we must consider 3 dimensions: The sum of forces in the z-component must also be zero. We are simplifying this problem to 2 dimensions.

11 3. Which pair of freebody diagrams is correct for the two pieces of the mobile? a) b) c) T1T1 T2T2 T2T2 T2T2 m2gm2g T1T1 T1T1 m2gm2g m2gm2g T2T2 m1gm1g m1gm1g T2T2 m2gm2g

12 These diagrams depict all of the forces acting on the two objects correctly. Choice: A Correct

13 The star supports the weight of the moon that is hanging from it by a massless string. There should be more than one force acting on the star in the negative y-direction. Choice: B Incorrect

14 Choice: C Incorrect There is a gravitational force acting on the star in the negative y-direction. Remember, the stars mass is m 1.

15 4. From our freebody diagrams, we see that the x-components of all of the forces are zero, so this condition does not provide us with a useful equation. We can move on to the forces with y- components. Which of the following expressions do we get after correctly applying Newtons 2nd Law (the first condition for equilibrium) to the y-components of the forces acting on the star (m 1 )? a) b) c)

16 Choice: A Incorrect Since the stars mass is m 1, the gravitational force pulling down on it has a magnitude of m 1 g. g=9.8m/s 2

17 Choice: B Correct We can find this from looking at the freebody diagram from the previous question.

18 T 1 acts in the opposite direction of T 2 and m 1 g. Choice: C Incorrect Correct Answer: Note that one can misidentify the directions of forces in free body diagrams and still obtain correct answers from the mathematics. A negative result indicates that you chose the force (or its component) incorrectly. It is beneficial to carefully correspond your equations to your freebody diagrams to avoid mistakes and to become more systematic when problem solving.

19 5. Which of the following relations do we get after correctly applying Newtons 2nd Law (the first condition for equilibrium in the y-direction) to the moon (m2)? a) b) c)

20 The only string that is connected to the moon has tension T 2. Choice: A Incorrect

21 This comes directly from the freebody diagram that we found in the previous question. Choice: B Correct

22 The moon has mass m 2. Check the correct freebody diagram for the moon from the previous question and try again. Choice: C Incorrect

23 6. Insert the known quantities to solve for T 2. Answer =3.5kg =4.6kg

24

25 7. Insert the known quantities to solve for T 1. Answer =3.5kg =4.6kg =45.0N

26 Notice that since T 2 =m 2 g T 1 can also be expressed as: T 1 =(m 1 +m 2 )g Notice that since T 2 =m 2 g T 1 can also be expressed as: T 1 =(m 1 +m 2 )g

27 Reflection Questions: Does it matter what shape the pieces of the mobile are? Why does the top string have more tension? In order to calculate the tensions, must the two strings lie along the same straight line?


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