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

T2

T1

m1

m2

y

x

• Question 1• Question 2• Question 3• Question 4• Question 5

• Question 6• Question 7• Reflection Questions

1. 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

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.

Choice: A

Incorrect

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.

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.

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

a)

b)

c)

€

Fy =0, ∑ τ = 0∑

€

Fx =0, Fy = 0, ∑∑

€

Fx =0, Fy = 0, ∑∑ τ = 0∑

=Torque=Torque

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

Choice: A

Incorrect

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

Choice: B

Incorrect

Choice: C

CorrectAll 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.

€

∑ =0

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.

€

Fz∑ =0

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

a)

b)

c)

T1 T2

T2

T2

m2g

T1

T1

m2g

m2g

T2

m1g

m1gT2

m2g

These diagrams depict all of the forces acting on the two objects

correctly.

Choice: A

Correct

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

Choice: C

Incorrect

There is a gravitational force acting on the star in the negative y-direction.

Remember, the star’s mass is m1.

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)

€

T1 =m2g+ T2

€

T1 =m1g+ T2

€

T2 =m1g+ T1

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

g=9.8m/s2

Choice: B

Correct

€

Fy =T1 −T2 −m1g=0∑

T1 =m1g+ T2

We can find this from looking at the freebody diagram from the previous question.

T1 acts in the opposite direction of T2 and m1g.

Choice: C

Incorrect

€

Fy =T1 −T2 −m1g=0∑

T1 =m1g+ T2

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.

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)

€

T1 =m2g

€

T2 =m2g

€

T2 =m1g

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

Choice: A

Incorrect

This comes directly from the freebody diagram that we found in the previous question.

Choice: B

Correct

€

Fy =T2 −m2g=0∑

T2 =m2g

The moon has mass m2.

Check the correct freebody diagram for the moon from the previous

question and try again.

Choice: C

Incorrect

6. Insert the known quantities to solve for T2.

Answer

=3.5kg

=4.6kg

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T2 =m2g

T2 =4.6kg(9.8m /s 2 )

T2 = 45N

7. Insert the known quantities to solve for T1.

Answer

=3.5kg

=4.6kg

=45.0N

€

T1 =m1g+ T2

T1 =3.5kg(9.8m /s 2 ) + 45N

T1 = 79N

Notice that since T2=m2g

T1 can also be expressed as:

T1=(m1+m2)g

Notice that since T2=m2g

T1 can also be expressed as:

T1=(m1+m2)g

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?