A Hand Pushes Three Identical Bricks

A hand pushes three identical bricks is a classic illustration used to explore how forces, mass, and friction interact in everyday motion. By examining this simple setup, students can grasp Newton’s laws, understand the difference between static and kinetic friction, and see how the arrangement of objects influences the effort required to move them. The following explanation breaks down each component of the scenario, provides step‑by‑step calculations, and offers practical insights that connect theory to real‑world experience.

Understanding the Scenario

Imagine three bricks of equal mass m resting on a horizontal surface. A person exerts a horizontal push F on the leftmost brick, and the bricks are either stacked vertically or placed side‑by‑side in a row. The goal is to determine the acceleration of the system and the internal forces that each brick experiences. Although the situation appears straightforward, several factors—such as the contact area between bricks, the coefficient of friction, and whether the bricks move together or slip relative to one another—can significantly affect the outcome.

Forces Involved

External Forces

1. Applied push (F) – The force exerted by the hand on the first brick. 2. Weight of each brick (W = mg) – Acts downward through the center of mass.
2. Normal reaction from the floor (N_floor) – Balances the total weight of the bricks when they are not accelerating vertically.
3. Frictional force from the floor (f_floor) – Opposes the motion; its magnitude depends on the normal force and the coefficient of kinetic friction (μ_k) once sliding begins.

Internal Forces When the bricks are in contact, each pair exerts equal and opposite forces on each other (Newton’s third law). These internal forces are crucial for transmitting the push through the stack or row, but they cancel out when considering the system as a whole.

Newton’s Second Law Applied to the Whole System

Treating the three bricks as a single rigid body (assuming they move together without slipping) simplifies the analysis. The total mass of the system is

[ M_{\text{total}} = 3m . ]

The net horizontal force acting on the system is the applied push minus the kinetic frictional force from the floor:

[F_{\text{net}} = F - f_{\text{floor}} . ]

According to Newton’s second law,

[ F_{\text{net}} = M_{\text{total}} , a , ]

where a is the common acceleration. Solving for a gives

[ a = \frac{F - f_{\text{floor}}}{3m}. ]

If the push is just enough to overcome static friction, the bricks remain at rest (a = 0). Once the applied force exceeds the maximum static friction (f_{s,\max} = \mu_s N_{\text{floor}}), motion begins and kinetic friction takes over.

Role of Friction Between Bricks and Floor

The frictional force from the floor is calculated as

[ f_{\text{floor}} = \mu_k N_{\text{floor}} = \mu_k (3mg), ]

assuming the bricks are on a uniform surface and the normal force equals the total weight. The coefficient of kinetic friction (μ_k) depends on the materials in contact (e.g., brick on concrete vs. brick on wood). A higher μ_k means a larger opposing force, reducing acceleration for a given push.

If the bricks are stacked, the normal force on the floor remains the same (still 3mg), but the contact area between the bottom brick and the floor does not change the frictional magnitude in the simple Amontons‑Coulomb model. However, real‑world deviations can occur due to surface deformation or interlocking, which are beyond the basic model but worth noting in advanced discussions.

Internal Force Distribution

Even though internal forces cancel for the whole system, they determine how each brick feels the push. Consider the bricks placed side‑by‑side and moving together. Draw a free‑body diagram for each brick:

• Brick 1 (leftmost): Experiences the applied push F to the right, a contact force from brick 2 (C_{12}) to the left, and friction f₁ from the floor.
• Brick 2: Feels a contact force from brick 1 (C_{21} = C_{12}) to the right, a contact force from brick 3 (C_{23}) to the left, and friction f₂.
• Brick 3 (rightmost): Experiences a contact force from brick 2 (C_{32} = C_{23}) to the right and friction f₃.

Because the bricks accelerate together, the net force on each equals m a. Solving the set of equations yields:

[ \begin{aligned} F - C_{12} - f_1 &= m a,\ C_{12} - C_{23} - f_2 &= m a,\ C_{23} - f_3 &= m a . \end{aligned} ]

Adding the three equations recovers the system‑level expression F - (f₁+f₂+f₃) = 3 m a, confirming consistency. The contact forces turn out to be:

[ C_{12} = 2 m a + f_1 + f_2,\qquad C_{23} = m a + f_3 . ]

Thus, the leftmost brick bears the largest internal load, while the rightmost brick only needs to overcome its own friction and provide the acceleration for its mass.

Effect of Stacking the Bricks

When the bricks are stacked vertically, the analysis changes slightly because the normal force between bricks now supports weight, not horizontal motion. Assuming no slipping between bricks, the stack still moves as a single unit with total mass 3m. The frictional force from the floor remains μ_k (3mg), and the acceleration formula stays the same:

[ a = \frac{F - \mu_k (3mg)}{3m}. ]

However, internal forces now include normal reactions that counteract the weight of the bricks above. The bottom brick must support the weight of the two bricks above it (2mg), increasing the normal force on the floor‑brick interface only if the stack tilts or if there is additional horizontal adhesion. In a perfectly level scenario, the vertical normal forces do not influence horizontal motion, but they are essential for preventing the stack from toppling if the push is applied off‑center.

Practical Experiment: Measuring Acceleration

A simple classroom experiment can verify the theoretical predictions:

1. Materials: Three identical house‑bricks, a smooth horizontal surface (e.g., a polished wooden table), a spring scale or force gauge, a stopwatch, and a measuring tape.
2. Procedure: - Attach the spring scale to the leftmost brick to measure the applied force F.
   • Pull the scale at a constant force (read directly from the scale) and release the bricks from rest.
   • Use the stopwatch to record the time t it takes for the bricks to travel a known distance d.
   • Compute acceleration using a = 2d / t² (assuming constant acceleration from rest).
3. Analysis: Compare the experimental a with the theoretical value obtained

From the measured distance d and time t one obtains an experimental acceleration aₑₓₚ = 2d/t². Repeating the trial — say five pulls at the same nominal force — provides a mean value and a standard deviation that quantifies random timing and distance errors. To isolate the contribution of kinetic friction, the experiment can be performed with varying applied forces (by adjusting the spring scale reading) while keeping the bricks identical. Plotting aₑₓₚ versus F yields a straight line whose slope should equal 1/(3m) and whose intercept on the F‑axis gives the total frictional resistance fₜₒₜ = f₁+f₂+f₃. A linear regression therefore provides two independent checks: the slope validates the inertial term 3m, and the intercept yields an estimate of the effective coefficient of kinetic friction μₖ,ₑff = fₜₒₜ/(3 m g).

When the bricks are stacked, the same procedure applies, but the normal force between bricks now supports the weight of the layers above. If the push is applied precisely through the centre of mass of the stack, the horizontal dynamics remain unchanged and the measured acceleration should match the value obtained for the side‑by‑side arrangement. Any systematic reduction in aₑₓₚ for the stacked case indicates either an unintended torque (causing a slight tilt and thus an increase in the normal force on the floor) or inter‑brick slip that dissipates energy. Observing such a deviation offers a practical illustration of how internal normal forces, while irrelevant to pure horizontal translation in an ideal rigid‑body model, become crucial when the load path is not perfectly aligned.

By comparing the side‑by‑side and stacked results, students can see that the internal contact forces derived earlier (C₁₂ = 2 ma + f₁ + f₂, C₂₃ = ma + f₃) are indeed larger for the leftmost brick, which must accelerate both its own mass and the masses to its right while overcoming friction. The rightmost brick, conversely, only needs to overcome its own friction and provide the acceleration for its single mass. The experiment thus makes tangible the abstract notion that internal forces redistribute the external push throughout a multi‑body system.

Conclusion
The analysis shows that, whether the bricks are placed side‑by‑side or stacked, the horizontal acceleration of the assembly is governed solely by the net external force minus the total kinetic‑friction drag, divided by the total mass. Internal contact forces adjust to ensure each brick experiences the same acceleration, with the leftmost brick bearing the greatest load. A straightforward tabletop experiment—measuring distance and time for a known applied force—allows students to verify the predicted linear relationship between force and acceleration, estimate the effective coefficient of friction, and observe how internal forces manifest in the motion of a composite object. Agreement between theory and measurement reinforces the utility of Newton’s second law applied to both the system as a whole and its individual constituents, while any discrepancies highlight real‑world effects such as misalignment, slip, or deformation that are worth investigating further.