How to Find the Minimum Force Needed to Move a Crate

A 20 kg crate rests on a level floor. Its coefficient of static friction is 0.40. You pull horizontally with a force of 50 N, but the crate doesn’t move.

How large is the friction force? Is it 50 N, or is it already at its maximum value of 78.4 N?

It’s 50 N. Static friction adjusts to match the applied force, up to a maximum. As you pull harder, friction increases with you until it reaches 78.4 N. At that point, the crate is on the verge of moving.

For a horizontal pull on a level floor, the minimum force needed to start moving the crate is therefore:

\[ F_{\min}=\mu_s mg \]

Using the values above:

\[ F_{\min}=0.40\times20\times9.8 \]

\[ \boxed{F_{\min}=78.4\text{ N}} \]

This is the answer for a horizontal pull under these conditions. If you pull at an angle, push downwards, or the crate is already moving, the calculation changes.

Let’s build the method so you know exactly which equation to use in each situation.

01Understand the difference between static and limiting friction

The most important idea is that static friction is not a constant force.

Imagine trying to slide a heavy crate across the floor. Initially, you push gently. The crate stays still.

You push harder. Still nothing.

You push harder again. The crate is still stationary, but the friction force opposing your push has increased.

Static friction behaves like an adjustable grip. It provides whatever opposing force is needed to prevent slipping, up to a maximum value.

That maximum is called limiting static friction.

The static friction inequality

The magnitude of static friction is described by:

\[ \boxed{0\leq f_s\leq\mu_s N} \]

Where:

• \(f_s\) is the actual static friction force, measured in newtons (N).
• \(\mu_s\) is the coefficient of static friction, which has no units.
• \(N\) is the normal force, measured in newtons (N).

The maximum possible static friction is:

\[ \boxed{f_{s,\max}=\mu_s N} \]

Notice the difference between these equations.

The inequality describes the actual static friction force. The second equation describes its maximum possible value.

You should not automatically write \(f_s=\mu_sN\) whenever an object is stationary.

For example, suppose a crate has a maximum static friction of 80 N.

The friction force matches the applied force until it reaches its limit.

This is why finding the maximum static friction is essential when calculating the force needed to start moving a crate.

If you want to explore this distinction further, see How to Find Static Friction When a Block Stays at Rest.

Finding the normal force

Before calculating friction, you need the normal force.

The normal force is the support force exerted by the surface, acting perpendicular to that surface.

For a crate resting on a horizontal floor, with no vertical component of the applied force, the vertical forces are:

• Weight, \(mg\), acting downwards.
• Normal force, \(N\), acting upwards.

Since the crate has no vertical acceleration:

\[ \sum F_y=0 \]

Therefore:

\[ N-mg=0 \]

So:

\[ \boxed{N=mg} \]

Here, \(m\) is the crate’s mass in kilograms, and \(g=9.8\text{ m s}^{-2}\) is the gravitational field strength near Earth’s surface.

Substituting into the limiting static friction equation gives:

\[ f_{s,\max}=\mu_s mg \]

For a horizontal pull, the applied force must reach this maximum to initiate motion.

Therefore:

\[ \boxed{F_{\min}=\mu_s mg} \]

This formula assumes:

• The crate is on a horizontal surface.
• The applied force is horizontal.
• There are no other horizontal forces.
• The crate remains in contact with the floor.
• The coefficient of static friction is known.

Change any of these conditions, and you may need a different calculation.

02What if the force is applied at an angle?

A horizontal pull is the simplest case. Realistically, you might pull a crate using a rope angled upwards.

This changes the calculation because the applied force now has two components.

The horizontal component helps move the crate forwards. The upward component partially lifts the crate, reducing the normal force.

And remember: friction depends on the normal force.

Resolving the applied force

Let:

• \(F\) be the magnitude of the applied pulling force.
• \(\theta\) be the angle above the horizontal.
• \(F\cos\theta\) be the horizontal component.
• \(F\sin\theta\) be the upward component.

The normal force is now smaller than the crate’s weight.

Using vertical equilibrium:

\[ N+F\sin\theta-mg=0 \]

Rearranging:

\[ \boxed{N=mg-F\sin\theta} \]

At the point where the crate is about to move, the horizontal component of the applied force equals the maximum static friction:

\[ F\cos\theta=\mu_sN \]

Substitute the expression for the normal force:

\[ F\cos\theta=\mu_s(mg-F\sin\theta) \]

Expand:

\[ F\cos\theta=\mu_smg-\mu_sF\sin\theta \]

Collect the terms containing \(F\):

\[ F(\cos\theta+\mu_s\sin\theta)=\mu_smg \]

Therefore, the minimum pulling force is:

\[ \boxed{ F_{\min}= \frac{\mu_smg} {\cos\theta+\mu_s\sin\theta} } \]

This is the equation to use when a rope pulls upwards at an angle.

Notice that you cannot simply use \(F_{\min}=\mu_smg\). The upward component changes the normal force, while the horizontal component determines how much of the pull actually opposes friction.

What if you push downwards instead?

Now imagine pushing a crate with a force angled downwards.

The vertical component pushes the crate harder against the floor, increasing the normal force.

In this situation:

\[ N=mg+F\sin\theta \]

At the threshold of motion:

\[ F\cos\theta=\mu_s(mg+F\sin\theta) \]

Rearranging gives:

\[ \boxed{ F_{\min}= \frac{\mu_smg} {\cos\theta-\mu_s\sin\theta} } \]

A downward push generally requires more force than a horizontal push because it increases the normal force and therefore increases the maximum static friction.

This is an important distinction in HSC Physics questions: always resolve the applied force before calculating friction.

03What changes once the crate starts moving?

So far, we’ve considered a stationary crate.

But what happens once you finally overcome static friction?

The friction model changes.

When two surfaces slide against each other, we use kinetic friction.

In the HSC model, its magnitude is:

\[ \boxed{f_k=\mu_kN} \]

Where:

• \(f_k\) is the kinetic friction force.
• \(\mu_k\) is the coefficient of kinetic friction.
• \(N\) is the normal force.

Unlike static friction, which adjusts to match the force needed to prevent motion, kinetic friction is modelled as having a constant magnitude for a given normal force and pair of surfaces.

This is an approximation for dry sliding surfaces over the conditions considered in typical HSC problems.

The coefficient of kinetic friction is usually smaller than the coefficient of static friction:

\[ \mu_k<\mu_s \]

This means that once the crate starts moving, the friction opposing it often decreases.

That creates an interesting situation.

You might need 80 N to start moving a crate, but only 55 N to keep it moving at constant speed.

Why? Because starting the motion requires overcoming limiting static friction. Maintaining constant speed requires balancing kinetic friction.

If the applied force remains unchanged after the crate starts moving, there may now be an unbalanced force, causing the crate to accelerate.

The next step is to apply Newton’s second law:

\[ \boxed{\sum F_x=ma} \]

For a horizontal pull to the right, taking right as positive:

\[ F-f_k=ma \]

This distinction is also useful when studying deceleration and stopping distance. See How to Calculate the Coefficient of Friction from Stopping Distance.

04How can you measure the minimum force experimentally?

Suppose a question gives you the mass of a crate but doesn’t provide its coefficient of static friction.

You cannot calculate the exact minimum force from mass alone. You also need information about the contact between the crate and the surface.

One way to obtain this information is through a simple friction experiment.

Setting up the experiment

Place a crate on a level horizontal board.

Attach a force sensor to the front of the crate using a cord. Keep the cord horizontal and connect the sensor to a data logger so that the pulling force can be recorded.

Measure the mass of the crate using a balance.

Now gradually increase the pulling force.

What should the sensor measure?

Before the crate moves, the sensor reading increases as you pull harder.

Why?

The crate is stationary, so the horizontal forces must balance. The static friction force therefore matches the applied pulling force.

Eventually, the sensor records a maximum reading just before the crate starts sliding.

This peak represents the limiting static friction.

If you continue pulling while maintaining constant speed, the sensor reading settles to a lower value. This represents kinetic friction because the crate now has zero horizontal acceleration.

An illustrative calculation

Suppose a 10.0 kg crate produces the following readings:

Assume the cord is horizontal, the board is level, and the crate moves at constant speed during the second measurement.

Step 1: Calculate the normal force.

\[ N=mg \]

\[ N=10.0\times9.8 \]

\[ N=98.0\text{ N} \]

Step 2: Calculate the coefficient of static friction.

At the peak reading:

\[ f_{s,\max}=\mu_sN \]

Rearranging:

\[ \mu_s=\frac{f_{s,\max}}{N} \]

\[ \mu_s=\frac{39.0}{98.0} \]

\[ \boxed{\mu_s=0.398} \]

Step 3: Calculate the coefficient of kinetic friction.

At constant speed, the horizontal acceleration is zero. Therefore, the applied pulling force balances kinetic friction.

\[ \mu_k=\frac{f_k}{N} \]

\[ \mu_k=\frac{29.5}{98.0} \]

\[ \boxed{\mu_k=0.301} \]

The results are consistent with the kinetic coefficient being smaller than the static coefficient.

Why does experimental technique matter?

A simple spring balance held by hand can introduce errors.

For example, if the cord is slightly angled upwards, its vertical component reduces the normal force. The measured threshold will then differ from the threshold for a horizontal pull.

A sudden jerk can also produce an unreliable peak reading.

A better design uses an aligned force sensor, a level board, and a slow, controlled increase in pulling force. Repeating the trials helps identify inconsistent readings.

The important experimental idea is that the sensor measures the applied force, not friction directly. We infer friction from the measured force because we know the crate’s acceleration and the other forces acting on it.

05Worked examples

06Questions and solutions

The following questions progress from straightforward calculations to situations where you need to combine friction, force components, experimental measurements, and Newton’s second law.

Question 1

A 15.0 kg crate rests on a horizontal concrete floor. The coefficient of static friction between the crate and the floor is 0.32.

A student applies a horizontal force to the crate.

Calculate the minimum force required to start moving it.

Take \(g=9.8\text{ m s}^{-2}\). Assume the crate does not tip.

Solution 1

The minimum horizontal force required is 47.0 N.

The crate is on a horizontal floor, and the applied force has no vertical component. Therefore, the normal force equals the weight.

Step 1: Calculate the normal force.

\[ N=mg \]

\[ N=15.0\times9.8 \]

\[ N=147\text{ N} \]

Step 2: Calculate the limiting static friction.

\[ f_{s,\max}=\mu_sN \]

\[ f_{s,\max}=0.32\times147 \]

\[ f_{s,\max}=47.04\text{ N} \]

Step 3: Identify the minimum applied force.

At the threshold of motion, the horizontal applied force equals limiting static friction.

\[ F_{\min}=f_{s,\max} \]

\[ \boxed{F_{\min}=47.0\text{ N}} \]

The crate will remain stationary for horizontal forces below this threshold. An applied force slightly greater than 47.0 N will initiate sliding in the idealised model.

Question 2

A 22.0 kg crate rests on a horizontal floor. A student pushes the crate using a force directed 15° below the horizontal.

The coefficient of static friction between the crate and floor is 0.30.

Calculate:

(a) The minimum magnitude of the applied force required to start moving the crate.

(b) The normal force at the threshold of motion.

Take \(g=9.8\text{ m s}^{-2}\).

Solution 2

The minimum applied force is 72.8 N, and the normal force at the threshold is 234 N.

The important detail is that the applied force points downwards. Its vertical component increases the normal force, which increases the maximum static friction.

Step 1: Calculate the normal force in terms of the applied force.

The downward component is \(F\sin15^\circ\).

Using vertical equilibrium:

\[ N=mg+F\sin15^\circ \]

Step 2: Apply the limiting static friction condition.

The horizontal component of the applied force must equal maximum static friction:

\[ F\cos15^\circ=\mu_sN \]

Substitute the normal force:

\[ F\cos15^\circ = 0.30(22.0\times9.8+F\sin15^\circ) \]

Step 3: Rearrange and solve for \(F\).

\[ F\cos15^\circ = 64.68+0.30F\sin15^\circ \]

Collect the force terms:

\[ F(\cos15^\circ-0.30\sin15^\circ)=64.68 \]

Therefore:

\[ F= \frac{64.68} {\cos15^\circ-0.30\sin15^\circ} \]

\[ \boxed{F=72.8\text{ N}} \]

Step 4: Calculate the normal force.

\[ N=mg+F\sin15^\circ \]

\[ N=215.6+72.8\sin15^\circ \]

\[ \boxed{N=234.4\text{ N}} \]

The result makes physical sense. The downward push increases the normal force from 215.6 N to 234.4 N.

A tempting mistake would be to calculate \(0.30mg=64.7\text{ N}\) and use that as the minimum applied force. However, this ignores both the increased normal force and the fact that only the horizontal component of the push opposes friction.

Question 3

An 18.0 kg crate rests on a horizontal floor.

The coefficients of friction are:

• Coefficient of static friction: \(\mu_s=0.45\).
• Coefficient of kinetic friction: \(\mu_k=0.30\).

A student applies a horizontal force of 50 N, but the crate remains stationary.

The applied force is then increased to 90 N and maintained at that value.

Calculate:

(a) The actual static friction force when the applied force is 50 N.

(b) The minimum force required to start moving the crate.

(c) The acceleration once the crate starts moving, assuming the 90 N force remains applied.

Take \(g=9.8\text{ m s}^{-2}\).

Solution 3

The answers are:

• Actual static friction at 50 N: 50 N.
• Minimum force to start moving: 79.4 N.
• Acceleration once sliding: 2.06 m s⁻² to the right.

This question tests whether you can distinguish actual static friction, limiting static friction, and kinetic friction.

Step 1: Calculate the normal force.

\[ N=mg \]

\[ N=18.0\times9.8 \]

\[ N=176.4\text{ N} \]

Step 2: Calculate the maximum static friction.

\[ f_{s,\max}=\mu_sN \]

\[ f_{s,\max}=0.45\times176.4 \]

\[ f_{s,\max}=79.38\text{ N} \]

Step 3: Determine the actual static friction at 50 N.

The applied force is below the maximum static friction.

Therefore, static friction can balance the entire applied force.

\[ \boxed{f_s=50\text{ N}} \]

The crate remains stationary.

Notice that the friction force is not 79.38 N here. That is its maximum possible value, not its actual value.

Step 4: Determine whether the crate starts moving at 90 N.

Since:

\[ 90>79.38 \]

The applied force exceeds limiting static friction, so the crate starts sliding.

Step 5: Calculate kinetic friction.

Once sliding begins, use the kinetic friction model:

\[ f_k=\mu_kN \]

\[ f_k=0.30\times176.4 \]

\[ f_k=52.92\text{ N} \]

Step 6: Apply Newton’s second law.

Taking right as positive:

\[ \sum F_x=ma \]

\[ 90-52.92=18.0a \]

\[ 37.08=18.0a \]

\[ \boxed{a=2.06\text{ m s}^{-2}} \]

The crate accelerates to the right at \(2.06\text{ m s}^{-2}\).

The key change occurs when the crate starts sliding. Static friction is replaced by kinetic friction, which is smaller for these surfaces. This creates a net forward force.

Question 4

An 8.0 kg crate is tested on a horizontal board using a force sensor.

The sensor is connected to the crate by a cord maintained at 12° above the horizontal.

The sensor records:

• A peak tension of 29.6 N immediately before the crate starts moving.
• A tension of 23.5 N while the crate slides at constant speed.

Calculate:

(a) The coefficient of static friction.

(b) The coefficient of kinetic friction.

(c) A student calculates the coefficient of static friction using \(29.6/(8.0\times9.8)\). Explain why this calculation is incorrect, and determine whether it overestimates or underestimates the actual coefficient.

Assume the crate remains in contact with the board, the angle is maintained throughout the experiment, and \(g=9.8\text{ m s}^{-2}\).

Solution 4

The coefficients are approximately:

\[ \boxed{\mu_s=0.401} \]

\[ \boxed{\mu_k=0.313} \]

The student’s calculation underestimates the coefficient of static friction because it incorrectly treats the full tension as horizontal and assumes the normal force equals the crate’s weight.

Step 1: Calculate the normal force at the threshold of motion.

The upward component of tension reduces the normal force.

\[ N=mg-T\sin12^\circ \]

Substitute the peak tension:

\[ N=(8.0)(9.8)-29.6\sin12^\circ \]

\[ N=78.4-6.15 \]

\[ N=72.25\text{ N} \]

Step 2: Calculate the horizontal component of tension.

\[ T_x=T\cos12^\circ \]

\[ T_x=29.6\cos12^\circ \]

\[ T_x=28.95\text{ N} \]

Step 3: Calculate the coefficient of static friction.

At the threshold of motion:

\[ T_x=\mu_sN \]

Rearrange:

\[ \mu_s=\frac{T_x}{N} \]

\[ \mu_s=\frac{28.95}{72.25} \]

\[ \boxed{\mu_s=0.401} \]

Step 4: Calculate the normal force while the crate is sliding.

The tension is now 23.5 N, but the cord remains at 12°.

\[ N=mg-T\sin12^\circ \]

\[ N=78.4-23.5\sin12^\circ \]

\[ N=73.51\text{ N} \]

Step 5: Calculate the horizontal component of tension.

\[ T_x=23.5\cos12^\circ \]

\[ T_x=22.99\text{ N} \]

Because the crate moves at constant speed, its horizontal acceleration is zero.

Therefore, the horizontal tension balances kinetic friction:

\[ f_k=T_x \]

Step 6: Calculate the coefficient of kinetic friction.

\[ \mu_k=\frac{f_k}{N} \]

\[ \mu_k=\frac{22.99}{73.51} \]

\[ \boxed{\mu_k=0.313} \]

Step 7: Examine the student’s calculation.

The student’s estimate is:

\[ \mu_{\text{estimate}}= \frac{29.6}{8.0\times9.8} \]

\[ \mu_{\text{estimate}}=0.378 \]

This is lower than the correct value of 0.401.

There are two errors:

• The full tension is not horizontal. Only \(T\cos12^\circ\) contributes to overcoming friction.
• The normal force is not equal to \(mg\), because the upward tension component partially supports the crate.

For these particular values, the reduction in normal force has a greater proportional effect than the reduction in horizontal tension. Consequently, ignoring both corrections produces an underestimate.

This question illustrates why experimental measurements must be interpreted using the actual force geometry.

Question 5

A 20.0 kg lower crate rests on a horizontal floor. A second crate, with a mass of 10.0 kg, rests on top of it.

A horizontal force is applied to the lower crate, pulling it to the right.

The coefficients of friction are:

The force is initially increased until the stack starts moving.

Once the crates are sliding across the floor, the applied force is set to 140 N. It is then increased to 150 N.

Determine:

(a) The minimum force required to start moving the stack.

(b) Whether the upper crate slips relative to the lower crate when the applied force is 140 N. If it does not slip, calculate the acceleration of both crates.

(c) Whether the upper crate slips when the applied force is increased to 150 N. If it does, calculate the acceleration of each crate.

Assume:

• Both crates remain in contact with each other and the floor.
• Neither crate tips.
• The coefficients of friction remain constant.
• The applied force acts horizontally on the lower crate.
• \(g=9.8\text{ m s}^{-2}\).

Solution 5

The stack starts moving at 117.6 N.

At 140 N, the crates move together with an acceleration of 1.73 m s⁻².

At 150 N, the upper crate slips relative to the lower crate. The lower crate accelerates at 2.36 m s⁻², while the upper crate accelerates at 1.47 m s⁻².

This is a more demanding friction problem because there are two separate contact surfaces. Each has its own coefficient of friction, and the friction forces at those surfaces play different roles.

Step 1: Calculate the total mass of the stack.

\[ m_{\text{total}}=20.0+10.0 \]

\[ m_{\text{total}}=30.0\text{ kg} \]

The floor supports the weight of both crates.

Therefore, the normal force between the lower crate and floor is:

\[ N_{\text{floor}}=m_{\text{total}}g \]

\[ N_{\text{floor}}=30.0\times9.8 \]

\[ N_{\text{floor}}=294\text{ N} \]

Step 2: Calculate the minimum force required to start moving the stack.

The limiting static friction between the lower crate and the floor is:

\[ f_{s,\max}=\mu_sN_{\text{floor}} \]

\[ f_{s,\max}=0.40\times294 \]

\[ \boxed{F_{\min}=117.6\text{ N}} \]

The stack starts moving when the applied force exceeds this threshold.

An important detail is that the upper crate does not require a separate friction calculation to determine this starting threshold. While the entire stack is stationary, the upper crate has no horizontal acceleration, so the friction force between the two crates is zero.

The floor friction balances the applied force until it reaches its maximum.

Step 3: Determine the kinetic friction between the lower crate and floor.

Once the stack starts sliding, floor friction becomes kinetic friction.

\[ f_{k,\text{floor}}=\mu_{k,\text{floor}}N_{\text{floor}} \]

\[ f_{k,\text{floor}}=0.30\times294 \]

\[ \boxed{f_{k,\text{floor}}=88.2\text{ N}} \]

This force opposes the motion of the lower crate.

Step 4: Investigate the situation when the applied force is 140 N.

First, assume the crates move together without slipping.

The net horizontal force on the combined system is:

\[ F_{\text{net}}=140-88.2 \]

\[ F_{\text{net}}=51.8\text{ N} \]

Apply Newton’s second law to the combined mass:

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

\[ 51.8=30.0a \]

\[ a=1.73\text{ m s}^{-2} \]

Now check whether the upper crate can actually achieve this acceleration without slipping.

The upper crate is not directly attached to the lower crate. The only horizontal force accelerating it is static friction between the two crates.

The required friction is:

\[ f_{\text{required}}=m_{\text{upper}}a \]

\[ f_{\text{required}}=10.0\times1.73 \]

\[ f_{\text{required}}=17.3\text{ N} \]

Calculate the maximum available static friction between the crates.

The normal force between them equals the weight of the upper crate:

\[ N_{\text{between}}=10.0\times9.8 \]

\[ N_{\text{between}}=98.0\text{ N} \]

Therefore:

\[ f_{s,\max,\text{between}} = 0.20\times98.0 \]

\[ f_{s,\max,\text{between}}=19.6\text{ N} \]

Compare the required and maximum friction:

\[ 17.3<19.6 \]

The available static friction is sufficient.

Therefore, the upper crate does not slip. Both crates accelerate together at:

\[ \boxed{a=1.73\text{ m s}^{-2}} \]

Step 5: Investigate what happens when the force increases to 150 N.

Again, initially assume the crates move together.

The combined acceleration would be:

\[ a=\frac{150-88.2}{30.0} \]

\[ a=2.06\text{ m s}^{-2} \]

For the upper crate to maintain this acceleration, the required interfacial static friction would be:

\[ f_{\text{required}}=m_{\text{upper}}a \]

\[ f_{\text{required}}=10.0\times2.06 \]

\[ f_{\text{required}}=20.6\text{ N} \]

But the maximum available static friction is only 19.6 N.

\[ 20.6>19.6 \]

Static friction cannot supply the force needed to keep the upper crate moving with the lower crate.

Therefore, the upper crate slips relative to the lower crate.

This is the critical change. You can no longer treat both crates as one combined system when calculating their individual accelerations.

Step 6: Calculate the kinetic friction between the two crates.

Once the upper crate slips, use the kinetic coefficient at their interface:

\[ f_{k,\text{between}} = \mu_{k,\text{between}}N_{\text{between}} \]

\[ f_{k,\text{between}}=0.15\times98.0 \]

\[ \boxed{f_{k,\text{between}}=14.7\text{ N}} \]

The lower crate moves to the right underneath the upper crate. Therefore, kinetic friction acts to the right on the upper crate and to the left on the lower crate.

Step 7: Calculate the acceleration of the upper crate.

The only horizontal force acting on the upper crate is the kinetic friction force.

Using Newton’s second law:

\[ f_{k,\text{between}}=m_{\text{upper}}a_{\text{upper}} \]

\[ 14.7=10.0a_{\text{upper}} \]

\[ \boxed{a_{\text{upper}}=1.47\text{ m s}^{-2}} \]

Step 8: Calculate the acceleration of the lower crate.

There are three horizontal forces acting on the lower crate:

• The applied force of 150 N to the right.
• Kinetic friction from the floor, 88.2 N to the left.
• Kinetic friction from the upper crate, 14.7 N to the left.

Taking right as positive:

\[ \sum F_x=m_{\text{lower}}a_{\text{lower}} \]

\[ 150-88.2-14.7=20.0a_{\text{lower}} \]

\[ 47.1=20.0a_{\text{lower}} \]

\[ \boxed{a_{\text{lower}}=2.36\text{ m s}^{-2}} \]

The lower crate accelerates faster than the upper crate.

Both crates are still accelerating to the right, but the lower crate pulls ahead. The upper crate therefore moves backwards relative to the lower crate, even though it continues moving forwards relative to the floor.

This problem exposes an important limitation of treating friction as a single force equal to \(\mu N\). Static friction at each contact adjusts to the force required, up to its own maximum. Once a contact starts sliding, its friction model changes independently of the other contacts.

The key decision in every crate problem is the object’s state: stationary, at the threshold of motion, or already sliding. Use the static friction inequality for the first two situations, and the kinetic friction model for the third. Once you’ve identified the correct friction force, Newton’s second law gives you the acceleration, and you can move on to problems involving stopping distance and changes in motion.