How to Calculate Contact Forces Between Three Pushed Blocks

Three blocks, A, B, and C, are touching in a line on a horizontal surface. You push block A to the right. The blocks accelerate together, but the force between A and B is not necessarily the same as the force between B and C.

Here’s the important question: which contact force is larger?

Before calculating, think about what each interface has to do. The contact force between A and B must accelerate both B and C. The contact force between B and C only needs to accelerate C. So, if there is no friction and the blocks all accelerate together, the A-B contact force must be larger.

This gives us the key method for finding contact forces between three pushed blocks:

1. Treat all three blocks as one system to calculate their acceleration.
2. Isolate the blocks on one side of the first contact and apply Newton’s second law.
3. Repeat for the second contact.

The trick is choosing the right system before writing your equations. Let’s build that method carefully.

01Why the external push is not the contact force

Imagine an experiment designed to investigate the forces between three blocks.

Place blocks A, B and C in contact on a horizontal, low-friction track. At the left end, connect a horizontal actuator to block A through a force sensor. At the other end, position a motion sensor facing a reflector attached to block C.

When the actuator pushes A, the force travels through the contacting blocks. The motion sensor measures the acceleration of C.

If the blocks remain in contact, they all have the same acceleration.

The first experimental design measures two things:

• The external force applied to A.
• The acceleration of the three-block system.

However, it cannot directly tell us the force at either internal contact.

Why not?

Because when we treat all three blocks as one system, the contact forces between the blocks are internal forces. They cancel in equal-and-opposite pairs.

The equation for the entire system only tells us about the resultant external force.

To measure the individual contact forces directly, we could revise the apparatus by inserting calibrated compression force sensors between A and B, and between B and C.

Alternatively, we can calculate those forces by isolating smaller systems. This second approach is particularly useful in HSC exam questions because it does not require us to know the individual contact forces beforehand.

02Define the system and the contact force pairs

Before writing any equations, identify the three blocks and the interactions between them.

Let:

• \(m_A\) = mass of block A, in kilograms.
• \(m_B\) = mass of block B, in kilograms.
• \(m_C\) = mass of block C, in kilograms.
• \(P\) = external force pushing block A to the right, in newtons.
• \(a\) = common acceleration, in metres per second squared.

We will initially assume:

• The surface is horizontal and frictionless.
• The blocks remain in contact.
• The blocks move together to the right.
• Right is the positive horizontal direction.

There are two separate contact interactions.

At the A-B interface:

A pushes B to the right, while B pushes A to the left.

These forces have equal magnitudes and opposite directions. This is Newton’s third law.

Call their common magnitude \(F_{AB}\).

At the B-C interface:

B pushes C to the right, while C pushes B to the left.

Again, these forces have equal magnitudes and opposite directions.

Call their common magnitude \(F_{BC}\).

Notice something important: Newton’s third law tells us that the two forces at the same interface have equal magnitudes. It does not say that the forces at different interfaces are equal.

In other words:

\[ F_{A\text{ on }B}=F_{B\text{ on }A} \]

in magnitude, but this does not mean:

\[ F_{AB}=F_{BC} \]

The two interfaces generally transmit different forces.

Choosing the system

This is the most important decision in the problem.

We can select different groups of blocks as our system.

Why does this work?

When we select all three blocks, both contact interactions occur inside our system. They cancel as internal force pairs.

When we select B and C together, the B-C interaction is internal, but the force exerted by A on B acts across the system boundary. Therefore, \(F_{AB}\) is an external force on this smaller system.

Finally, when we select C alone, the force exerted by B on C is external.

This is how we isolate the individual contact forces.

03The experimental prediction: different forces at different interfaces

Consider an illustrative experiment with the following values:

• Block A has a mass of 1 kg.
• Block B has a mass of 2 kg.
• Block C has a mass of 3 kg.
• A horizontal force of 30 N pushes A to the right.
• The track is frictionless.

First, calculate the acceleration of the entire system.

The total mass is:

\[ m_{\text{total}}=1+2+3=6\text{ kg} \]

Using Newton’s second law:

\[ a=\frac{F_{\text{net}}}{m_{\text{total}}} \]

\[ a=\frac{30}{6} \]

\[ \boxed{a=5\text{ m s}^{-2}} \]

Now isolate C.

The only horizontal force accelerating C is the contact force from B:

\[ F_{BC}=m_Ca \]

\[ F_{BC}=3(5) \]

\[ \boxed{F_{BC}=15\text{ N}} \]

Next, isolate B and C together:

\[ F_{AB}=(m_B+m_C)a \]

\[ F_{AB}=(2+3)(5) \]

\[ \boxed{F_{AB}=25\text{ N}} \]

If we inserted the two additional force sensors into the experiment, the ideal predicted readings would be:

These are theoretical predictions, not reported experimental measurements.

The unequal contact forces support the idea that each interface transmits the force needed to accelerate the blocks beyond it.

In an actual experiment, sensor masses, track friction, and alignment would need to be accounted for. The idealised calculations assume those effects are negligible.

04The general method for finding contact forces

For three blocks pushed from the left, with friction opposing their rightward motion, we can write three useful equations.

1. Treat all three blocks as one system

The external push must overcome friction on all three blocks and provide their acceleration.

\[ P-f_A-f_B-f_C = (m_A+m_B+m_C)a \]

Here, \(f_A\), \(f_B\), and \(f_C\) are the magnitudes of the friction forces on each block.

Rearranging:

\[ a= \frac{P-f_A-f_B-f_C} {m_A+m_B+m_C} \]

If the surface is frictionless, all three friction forces are zero.

2. Isolate blocks B and C

The force at the A-B interface must accelerate both B and C, while overcoming any friction acting on those two blocks.

\[ F_{AB}-f_B-f_C = (m_B+m_C)a \]

Therefore:

\[ \boxed{ F_{AB}=(m_B+m_C)a+f_B+f_C } \]

3. Isolate block C

The force at the B-C interface must accelerate C and overcome its friction.

\[ F_{BC}-f_C=m_Ca \]

Therefore:

\[ \boxed{ F_{BC}=m_Ca+f_C } \]

These equations are based on a rightward acceleration and friction acting to the left.

The important decision rule is:

To calculate a contact force, isolate the blocks that the contact force needs to accelerate.

If you can identify those blocks, the equation becomes much easier to construct.

07What happens when the push comes from the other direction?

So far, we’ve pushed A from the left.

What happens if we push C from the right instead?

Consider the same three masses:

• A = 2 kg.
• B = 3 kg.
• C = 5 kg.

A 40 N force pushes C to the left on a frictionless surface.

The total mass is still 10 kg, so the acceleration has a magnitude of:

\[ a=\frac{40}{10}=4\text{ m s}^{-2} \]

However, the contact forces are now different.

First, isolate A.

The only horizontal force accelerating A to the left is the contact force from B.

\[ F_{AB}=m_Aa \]

\[ F_{AB}=2(4) \]

\[ \boxed{F_{AB}=8\text{ N}} \]

Next, isolate A and B together.

The contact force from C must accelerate both blocks to the left.

\[ F_{BC}=(m_A+m_B)a \]

\[ F_{BC}=(2+3)(4) \]

\[ \boxed{F_{BC}=20\text{ N}} \]

The results have reversed:

This makes sense.

When you push from the left, the contact force closest to the push must accelerate the blocks further along the row.

When you push from the right, the same principle applies in the opposite direction.

The useful rule is not that the left contact force is always larger. It is that the contact force must accelerate the blocks on the far side of that interface from the applied push.

08What if the blocks are in equilibrium?

There is another misconception worth fixing.

A block can experience a contact force even when its acceleration is zero.

Imagine three blocks A, B, and C pressed against a wall. A horizontal force of 30 N pushes A towards the wall. All three blocks remain stationary.

The acceleration is zero, so Newton’s second law gives:

\[ F_{\text{net}}=ma=0 \]

For block C, the wall must exert a 30 N force to the left to balance the contact force from B.

Therefore:

\[ F_{BC}=30\text{ N} \]

For block B, the force from A must balance the force from C.

Therefore:

\[ F_{AB}=30\text{ N} \]

So:

\[ \boxed{F_{AB}=F_{BC}=30\text{ N}} \]

The contact forces are not zero just because the blocks are stationary.

This is why you should always start with Newton’s second law:

\[ \sum F=ma \]

For equilibrium, simply substitute \(a=0\).

Do not assume that every force disappears.

09Questions and solutions

Try these questions before reading each solution. They start with standard HSC applications and gradually introduce more demanding situations.

Question 1

Three blocks A, B, and C have masses of 1 kg, 2 kg, and 3 kg respectively. They are touching on a frictionless horizontal surface.

A horizontal force of 24 N pushes block A to the right.

Calculate:

a) The acceleration of the three blocks.

b) The contact force between A and B.

c) The contact force between B and C.

Solution 1

The acceleration is \(4\text{ m s}^{-2}\), the A-B contact force is 20 N, and the B-C contact force is 12 N.

Step 1: Calculate the total mass.

\[ m_{\text{total}}=1+2+3=6\text{ kg} \]

Step 2: Calculate the acceleration.

\[ F_{\text{net}}=ma \]

\[ 24=6a \]

\[ \boxed{a=4\text{ m s}^{-2}} \]

Step 3: Isolate block C.

\[ F_{BC}=m_Ca \]

\[ F_{BC}=3(4) \]

\[ \boxed{F_{BC}=12\text{ N}} \]

Step 4: Isolate blocks B and C.

\[ F_{AB}=(m_B+m_C)a \]

\[ F_{AB}=(2+3)(4) \]

\[ \boxed{F_{AB}=20\text{ N}} \]

The A-B interface transmits a greater force because it accelerates a combined mass of 5 kg, whereas the B-C interface only accelerates 3 kg.

Question 2

Three blocks have masses of 2 kg, 3 kg, and 5 kg. They are touching on a horizontal surface and sliding to the right.

A horizontal force of 70 N pushes block A to the right.

The friction forces acting on A, B, and C are 4 N, 6 N, and 10 N respectively.

Calculate:

a) The acceleration of the system.

b) The contact force between A and B.

c) The contact force between B and C.

Solution 2

The acceleration is \(5\text{ m s}^{-2}\), the A-B contact force is 56 N, and the B-C contact force is 35 N.

Step 1: Calculate the total external horizontal force.

The total friction is:

\[ f_{\text{total}}=4+6+10=20\text{ N} \]

Therefore:

\[ F_{\text{net}}=70-20=50\text{ N} \]

Step 2: Calculate the acceleration.

The total mass is:

\[ m_{\text{total}}=2+3+5=10\text{ kg} \]

Using Newton’s second law:

\[ 50=10a \]

\[ \boxed{a=5\text{ m s}^{-2}} \]

Step 3: Isolate block C.

The contact force from B must overcome the 10 N friction force and accelerate C.

\[ F_{BC}-10=5(5) \]

\[ F_{BC}=25+10 \]

\[ \boxed{F_{BC}=35\text{ N}} \]

Step 4: Isolate blocks B and C.

The contact force from A must overcome the friction on both blocks.

\[ F_{AB}-6-10=(3+5)(5) \]

\[ F_{AB}-16=40 \]

\[ \boxed{F_{AB}=56\text{ N}} \]

Step 5: Check block A.

\[ F_{\text{net}}=70-4-56 \]

\[ F_{\text{net}}=10\text{ N} \]

And:

\[ m_Aa=2(5)=10\text{ N} \]

The answer is consistent.

Remember that the contact force must overcome the friction acting on the blocks beyond the interface. Ignoring friction on B or C will underestimate the contact forces.

Question 3

Three blocks A, B, and C have masses of 2 kg, 3 kg, and 5 kg respectively. They are touching on a rough horizontal surface and are sliding to the right.

A force of 25 N is applied to block A at an angle of \(37^\circ\) above the horizontal.

The coefficient of kinetic friction between each block and the surface is 0.20.

Take gravitational acceleration as \(g=10\text{ m s}^{-2}\).

Calculate:

a) The horizontal acceleration of the blocks.

b) The contact force between A and B.

c) The contact force between B and C.

Solution 3

The acceleration is \(0.30\text{ m s}^{-2}\), the A-B contact force is 18.4 N, and the B-C contact force is 11.5 N.

The important detail is that the applied force has both horizontal and vertical components. Its upward component reduces the normal force on A, which changes the friction acting on that block.

Step 1: Resolve the applied force into components.

Using the angle of \(37^\circ\):

\[ P_x=25\cos37^\circ \]

Using the 3-4-5 triangle approximation:

\[ P_x=25(0.8)=20\text{ N} \]

The vertical component is:

\[ P_y=25\sin37^\circ \]

\[ P_y=25(0.6)=15\text{ N} \]

So the applied force has:

• A horizontal component of 20 N to the right.
• A vertical component of 15 N upwards.

Step 2: Calculate the normal reaction on each block.

For block A, there is no vertical acceleration.

Therefore:

\[ N_A+15-m_Ag=0 \]

\[ N_A+15=2(10) \]

\[ \boxed{N_A=5\text{ N}} \]

For block B:

\[ N_B=m_Bg \]

\[ N_B=3(10)=30\text{ N} \]

For block C:

\[ N_C=m_Cg \]

\[ N_C=5(10)=50\text{ N} \]

Step 3: Calculate the friction on each block.

Using:

\[ f_k=\mu_kN \]

For A:

\[ f_A=0.20(5)=1\text{ N} \]

For B:

\[ f_B=0.20(30)=6\text{ N} \]

For C:

\[ f_C=0.20(50)=10\text{ N} \]

Total friction:

\[ f_{\text{total}}=1+6+10=17\text{ N} \]

Step 4: Calculate the acceleration.

Only the horizontal component of the applied force contributes directly to horizontal acceleration.

\[ F_{\text{net}}=20-17=3\text{ N} \]

The total mass is 10 kg.

\[ 3=10a \]

\[ \boxed{a=0.30\text{ m s}^{-2}} \]

Step 5: Calculate the B-C contact force.

Isolate C:

\[ F_{BC}-10=5(0.30) \]

\[ F_{BC}=1.5+10 \]

\[ \boxed{F_{BC}=11.5\text{ N}} \]

Step 6: Calculate the A-B contact force.

Isolate B and C:

\[ F_{AB}-6-10=(3+5)(0.30) \]

\[ F_{AB}-16=2.4 \]

\[ \boxed{F_{AB}=18.4\text{ N}} \]

The upward component of the applied force reduces the normal reaction on A. This reduces friction on A, which changes the acceleration and both contact forces.

The tempting mistake is to use the full 25 N as the horizontal applied force. Only 20 N acts horizontally.

Question 4

Three blocks A, B, and C have masses of 2 kg, 3 kg, and 5 kg respectively. They are arranged in a horizontal row on a frictionless surface.

Block C is touching a rigid wall.

A horizontal force of 30 N pushes block A towards the wall. The blocks remain stationary.

Calculate:

a) The contact force between A and B.

b) The contact force between B and C.

c) The force exerted by the wall on C.

Explain why the contact forces are not zero, even though the acceleration is zero.

Solution 4

The A-B contact force is 30 N, the B-C contact force is 30 N, and the wall exerts a 30 N force to the left on C.

The blocks are in equilibrium, so their net horizontal force must be zero.

Step 1: Consider the entire system.

Since the blocks are stationary:

\[ a=0 \]

Therefore:

\[ \sum F_x=0 \]

The applied 30 N force must be balanced by a 30 N reaction from the wall.

Step 2: Isolate block C.

Block C experiences:

• A contact force from B towards the right.
• A wall reaction towards the left.

For equilibrium:

\[ F_{BC}-F_{\text{wall}}=0 \]

Therefore:

\[ F_{BC}=30\text{ N} \]

Step 3: Isolate block B.

Block B experiences:

• A force from A towards the right.
• A force from C towards the left.

For equilibrium:

\[ F_{AB}-F_{BC}=0 \]

Therefore:

\[ F_{AB}=30\text{ N} \]

The contact forces are non-zero because they balance one another and transmit the applied force to the wall.

The misconception is that zero acceleration means zero force. Newton’s second law actually says that zero acceleration means zero net force.

Individual forces can still be large.

Question 5

Three blocks A, B, and C have masses of 2 kg, 3 kg, and 5 kg respectively.

They are initially touching in a horizontal row on a frictionless surface. All three blocks are initially at rest.

Two horizontal forces are switched on simultaneously and maintained:

• A 20 N force pushes block A to the right.
• A 50 N force pulls block C to the right.

There are no strings or other connections between the blocks. They can separate, but cannot pull one another.

Determine:

a) Whether all three blocks remain in contact.

b) The acceleration of each block after any separation occurs.

c) The contact force between A and B.

d) The contact force between B and C.

Explain why assuming all three blocks have the same acceleration gives an impossible result.

Solution 5

Blocks A and B remain in contact, but C separates from B. Blocks A and B accelerate rightwards at \(4\text{ m s}^{-2}\), while C accelerates rightwards at \(10\text{ m s}^{-2}\). The A-B contact force is 12 N, and the B-C contact force becomes zero.

This is the most important conceptual challenge in the set.

It is tempting to assume that all three blocks must have the same acceleration because they are initially touching.

But contact between solid objects can push, not pull. We need to check whether the assumed arrangement is physically possible.

Step 1: Test the assumption that all three blocks remain in contact.

If all three blocks accelerate together, the total external force is:

\[ F_{\text{net}}=20+50=70\text{ N} \]

The total mass is:

\[ m_{\text{total}}=2+3+5=10\text{ kg} \]

Therefore:

\[ a=\frac{70}{10} \]

\[ a=7\text{ m s}^{-2} \]

This is the acceleration required if all three blocks remain in contact.

Step 2: Check whether the B-C contact force is physically possible.

Isolate C.

If all blocks shared the calculated acceleration, Newton’s second law would give:

\[ F_{BC}+50=5(7) \]

\[ F_{BC}+50=35 \]

\[ \boxed{F_{BC}=-15\text{ N}} \]

The negative sign means that B would need to exert a force of 15 N to the left on C.

But this would require B to pull C towards itself.

Two ordinary blocks in contact cannot do that. Their contact force is compressive: one block can push the other, but cannot pull it backwards.

Therefore, the assumption that all three blocks remain in contact is impossible.

C separates from B.

Step 3: Calculate the acceleration of A and B together.

After C separates, A and B remain in contact.

The only external horizontal force acting on this two-block system is the 20 N force on A.

\[ F_{\text{net}}=(m_A+m_B)a_{AB} \]

\[ 20=(2+3)a_{AB} \]

\[ \boxed{a_{AB}=4\text{ m s}^{-2}} \]

Step 4: Calculate the acceleration of C.

Once C separates, the only horizontal force acting on it is the 50 N applied pull.

Therefore:

\[ F_{\text{net}}=m_Ca_C \]

\[ 50=5a_C \]

\[ \boxed{a_C=10\text{ m s}^{-2}} \]

C accelerates faster than A and B, so the gap between B and C increases.

Step 5: Calculate the A-B contact force.

Isolate block B.

The only horizontal force accelerating B is the contact force from A.

\[ F_{AB}=m_Ba_{AB} \]

\[ F_{AB}=3(4) \]

\[ \boxed{F_{AB}=12\text{ N}} \]

A quick check using block A gives:

\[ 20-12=2(4) \]

\[ 8=8 \]

The result is consistent.

Step 6: Determine the B-C contact force.

Since C has separated from B, there is no longer a contact interaction between them.

Therefore:

\[ \boxed{F_{BC}=0\text{ N}} \]

The surprising part is that the A-B contact force changes when C separates.

In the hypothetical all-together calculation, the A-B contact force would have been 6 N. But that calculation depended on all three blocks sharing an acceleration of \(7\text{ m s}^{-2}\).

Once C separates, A and B form a different system. Their acceleration changes to \(4\text{ m s}^{-2}\), and their contact force becomes 12 N.

You cannot keep the contact forces from a model that you have already shown to be impossible.

This is why checking the physical meaning of your calculated forces matters just as much as getting the algebra right.

10Where to go next

The method for three pushed blocks is really an application of choosing the correct system and applying Newton’s second law.

Once you can isolate groups of blocks and identify the forces crossing their boundaries, you can use the same reasoning for connected objects, including systems involving tension.

There is one important difference: a string can pull, whereas two ordinary blocks in contact can only push. That distinction becomes particularly useful when checking whether a calculated contact force is physically possible.

For a refresher on the underlying interactions, see Contact Forces and Field Forces in HSC Physics.