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Published byKatelyn Andersen Modified over 4 years ago

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**“I wonder what the tension is in each of those massless strings.”**

Sir Isaac Newton’s 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 T1 and T2, so that baby Newton can sleep. m1=3.5kg m2=4.6kg T1 y m1 T2 x m2

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

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**What physics principle should we use to solve this problem**

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

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Choice: A Incorrect Newton’s 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. Newton’s 1st law is irrelevant here.

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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 Newton’s 2nd Law.

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Choice: C Incorrect Newton’s 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.

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**2. What are the conditions for static equilibrium in 2 dimensions?**

=Torque a) b) c)

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**The net force in the x-direction must also be zero.**

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

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**The sum of torques through any axis of rotation must be zero.**

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

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**All of these conditions are required for static equilibrium.**

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. 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.

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**3. Which pair of freebody diagrams is correct for the two pieces of the mobile?**

m2g m1g T1 T2 m2g m1g T2 T1 T2 m2g m2g

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

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Choice: B Incorrect 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.

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Choice: C Incorrect There is a gravitational force acting on the star in the negative y-direction. Remember, the star’s mass is m1.

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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 Newton’s 2nd Law (the first condition for equilibrium) to the y-components of the forces acting on the star (m1)? a) b) c)

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Choice: A Incorrect Since the star’s mass is m1, the gravitational force pulling down on it has a magnitude of m1g. g=9.8m/s2

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

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**T1 acts in the opposite direction of T2 and m1g.**

Choice: C Incorrect T1 acts in the opposite direction of T2 and m1g. 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.

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

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**The only string that is connected to the moon has tension T2.**

Choice: A Incorrect The only string that is connected to the moon has tension T2.

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

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Choice: C Incorrect The moon has mass m2. Check the correct freebody diagram for the moon from the previous question and try again.

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**6. Insert the known quantities to solve for T2.**

=3.5kg =4.6kg Answer

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**7. Insert the known quantities to solve for T1.**

=3.5kg =45.0N =4.6kg Answer

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Notice that since T2=m2g T1 can also be expressed as: T1=(m1+m2)g

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**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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M 1 and M 2 – Masses of the two objects [kg] G – Universal gravitational constant G = 6.67x10 -11 N m 2 /kg 2 or G = 3.439x10 -8 ft 4 /(lb s 4 ) r – distance.

M 1 and M 2 – Masses of the two objects [kg] G – Universal gravitational constant G = 6.67x10 -11 N m 2 /kg 2 or G = 3.439x10 -8 ft 4 /(lb s 4 ) r – distance.

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