16.4 Newton's third law, equilibrium and connected particles — Learn

Prerequisites: Newton's first and second laws (Sections 16.1 and 16.2); solving two linear equations at once (Section 2.5).

By the end of this section you can: state Newton's third law; say why an action–reaction pair does not cancel; write a pair of equations for two bodies joined by a string; and solve them for the acceleration and the tension.

1. Newton's third law

Newton's third law says that if body A exerts a force on body B, then body B exerts a force of the same size on body A, in the opposite direction.

The two forces are called an action–reaction pair. A pair has three features.

• The two forces are equal in size and opposite in direction.
• They act on different bodies.
• They are of the same type. A contact force is answered by a contact force, and a pull by a pull.

Because the two forces act on different bodies, they do not cancel. Only forces on the same body can cancel.

Example 1. A book rests on a table. Describe the action–reaction pair between the book and the table.

The book pushes down on the table, and the table pushes up on the book with the same size of force. These two forces act on different bodies, so they are a third-law pair.

The weight of the book and the normal reaction on the book are not a third-law pair. Both act on the book, and they are different types of force.

2. Equilibrium

A particle is in equilibrium when the resultant force on it is zero. The forces that act at one point are called concurrent. In equilibrium, the forces can be drawn head to tail to form a closed triangle.

For two or three forces in a plane, the method is the same as anywhere else: add the forces as vectors, or resolve them in two perpendicular directions. The full method for coplanar equilibrium is in Section 16.5.

3. Connected particles

Two bodies joined by a string are connected particles. A question usually tells you that the string is light and inextensible and that the pulley is smooth.

• Light means the string has negligible mass.
• Inextensible means the string does not stretch, so the two bodies move with accelerations of the same size.
• Smooth pulley means no friction at the pulley, so the tension is the same throughout the string.

The method has four steps.

1. Draw each body on its own, with the forces on that body only.
2. Choose the direction of motion of each body as positive.
3. Write $F = ma$ for each body.
4. Solve the two equations together for the acceleration and the tension.

Example 2. Masses of $3$ kg and $5$ kg hang from the ends of a light inextensible string over a smooth pulley. The system is released from rest. Find the acceleration and the tension. Take $g = 9.8$ m s$^{-2}$.

The $5$ kg mass moves down and the $3$ kg mass moves up. Let the acceleration have size $a$ m s$^{-2}$.

For the $5$ kg mass, taking downwards as positive:

For the $3$ kg mass, taking upwards as positive:

Add the two equations. The tension cancels:

Substitute back into the second equation:

Example 3. A block of mass $4$ kg rests on a smooth horizontal table. It is joined by a light inextensible string over a smooth pulley at the table edge to a hanging mass of $1$ kg. The system is released. Find the acceleration and the tension. Take $g = 9.8$ m s$^{-2}$.

The hanging weight drives the system, and the mass moved is the total of the two masses. For the whole system:

For the block on the table the only horizontal force is the tension, so

Example 4. A crate of mass $45$ kg rests on a smooth ramp. A light inextensible string runs parallel to the ramp, over a smooth pulley at the top, to a hanging mass of $20$ kg. The system is released. The component of the crate's weight down the ramp is $136.3$ N. Find the acceleration. Take $g = 9.8$ m s$^{-2}$.

The hanging weight is $20 \times 9.8 = 196$ N, which is larger than $136.3$ N, so the crate accelerates up the ramp. The resultant force is

and the mass moved is $45 + 20 = 65$ kg. So

Watch out. Three mistakes are common.

Treating an action–reaction pair as balanced forces. The two forces of a third-law pair act on different bodies, so they never cancel. The book pushes the table down and the table pushes the book up; one force acts on the table and the other on the book. Re-check: to cancel, two forces must act on the same body.

Using different accelerations. Because the string is inextensible, both bodies have the same size of acceleration. Write one symbol, $a$, and use it in both equations.

Forgetting that the tension is the same on both sides. A light string over a smooth pulley has the same tension everywhere. Example 2 checks this by substituting the acceleration back into either equation.

Progress check

Return to the chapter problem. The crate of mass $45$ kg is on the ramp at $18^\circ$. The string from the crate passes over a smooth pulley and a hanging mass of $20$ kg pulls it.

The component of the crate's weight down the ramp is $136.3$ N.

1. Write the equation of motion for the hanging mass, moving down.
2. Find the resultant force on the whole system.
3. Find the acceleration of the crate.
4. Explain why the tension in the string is less than the hanging weight.

You should now be able to

• state Newton's third law and identify an action–reaction pair;
• explain why a third-law pair does not cancel;
• describe a particle in equilibrium;
• set up $F = ma$ for each of two connected bodies;
• solve the two equations for the acceleration and the tension;
• handle masses over a pulley, a mass on a table, and a mass on a smooth ramp.