How to Solve Angled Force Problems in HSC Physics

An 8.0 kg crate is sliding across a floor. You pull it with a force of 40 N at 30° above the horizontal, then push it with the same force at 30° below the horizontal. Both forces have exactly the same forward component. Would the crate accelerate at the same rate?

Not if there’s friction. Pulling upwards reduces how hard the crate presses against the floor. Pushing downwards increases it. That changes friction, even though the forward force stays the same.

To solve a force problem when the applied force is at an angle:

1. Choose the object or system you’re analysing.
2. Identify every force acting on it and the interaction producing that force.
3. Choose coordinate axes and resolve the angled force into components.
4. Use the perpendicular forces to find the normal reaction, if the object remains in contact with the surface.
5. Determine friction, then use Newton’s second law to calculate acceleration or test equilibrium.

The order matters. If you calculate friction before accounting for the vertical component of the applied force, you can get the acceleration wrong even when all your trigonometry is correct.

01Start with the system, not the triangle

Imagine pulling a heavy school bag along the floor while lifting its handle slightly.

Part of your pull moves the bag forwards. Another part supports some of its weight. A bathroom scale underneath the bag would therefore register a smaller load.

That gives us a useful first model: an upward pull can reduce the force exerted by the floor.

But the scale reading alone doesn’t tell us the bag’s acceleration. We still need the forward force, friction, and mass.

Which forces actually act on the crate?

Choose the crate as your system. The Earth, floor, and pulling rope are outside it.

If the crate is sliding right, four forces act on it:

Here, \(m\) is the crate’s mass in kilograms, \(g\) is gravitational acceleration in \(\mathrm{m\,s^{-2}}\), and all four forces are measured in newtons (N).

The normal reaction is the force exerted perpendicular to the contacting surface. On a horizontal floor, it acts vertically upwards.

Notice that every force in the table acts on the crate.

Newton’s third law tells us that each interaction also produces an equal and opposite force on another object. For example, the floor pushes upwards on the crate, and the crate pushes downwards on the floor.

Those two forces do not cancel in the crate’s free-body diagram because they act on different objects.

The crate’s weight and normal reaction can balance, but they’re not a Newton’s third-law pair.

02Resolve the angled force into components

Suppose the applied force \(F\) makes an angle \(\theta\) above the horizontal.

Imagine drawing a right-angled triangle with \(F\) as its hypotenuse.

The horizontal component tells us how strongly the force acts along the floor. The vertical component tells us how strongly it acts away from the floor.

Using trigonometry:

\[ F_x=F\cos\theta \]

\[ F_y=F\sin\theta \]

Here:

• \(F_x\) is the horizontal component of the applied force, in newtons.
• \(F_y\) is its vertical component, in newtons.
• \(F\) is the magnitude of the original applied force, in newtons.
• \(\theta\) is measured from the horizontal.

With right and up chosen as positive, an upward-right pull has positive \(F_x\) and \(F_y\).

A downward-right push has positive \(F_x\), but negative \(F_y\).

The components aren’t two extra forces. They’re two mathematical descriptions of the same applied force.

Why the angle changes the normal reaction

Suppose the crate remains on a horizontal floor without accelerating vertically.

Newton’s second law in the vertical direction becomes:

\[ \sum F_y=0 \]

For an upward pull:

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

Rearranging:

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

The upward pull helps support the crate, so the floor doesn’t need to push as hard.

For a downward push:

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

Therefore:

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

The downward push adds to the load supported by the floor.

So the normal reaction is not automatically equal to \(mg\).

These equations apply only while the crate maintains contact with the horizontal floor and has zero vertical acceleration.

Why this affects friction

For a crate already sliding across a surface, the standard HSC kinetic-friction model is:

\[ f_k=\mu_k N \]

Here, \(f_k\) is the magnitude of kinetic friction in newtons, \(\mu_k\) is the dimensionless coefficient of kinetic friction, and \(N\) is the normal reaction in newtons.

Under this model, a larger normal reaction produces greater kinetic friction.

Taking right as positive, Newton’s second law horizontally is:

\[ F\cos\theta-f_k=ma_x \]

where \(a_x\) is the horizontal acceleration in \(\mathrm{m\,s^{-2}}\).

For an upward pull, while the crate remains in contact and slides right:

\[ a_x= \frac{F\cos\theta-\mu_k N}{m} \]

For a downward push, the same horizontal equation applies, but \(N\) is larger.

This explains why two forces with identical horizontal components can produce different accelerations.

03An experiment to test the explanation

A useful way to investigate this is to compare the crate’s acceleration when the applied force points above and below the horizontal.

The apparatus

Consider an experiment using a crate on a horizontal bench.

A motion sensor sits at the left end, facing a reflector attached to the crate’s left side. It records the crate’s position as the crate travels to the right. Acceleration can be estimated from the change in velocity over time.

A rope is attached to a low hook on the crate’s right side. A spring balance, connected in line with the rope, measures the pulling force. The rope extends upwards and rightwards to the person applying the force.

For the downward-pushing trial, the rope is replaced with a rigid rod and an inline compression force sensor. The rod transmits a force directed downwards and rightwards.

An angle guide is used to keep the applied force at the chosen angle.

The experiment should use the same crate, surface, and applied force magnitude in both trials.

The crate should already be sliding when acceleration measurements begin. Otherwise, static friction introduces another variable.

We also assume that the crate doesn’t tip or rotate appreciably, and that the applied force remains approximately constant during each measurement.

Why the first experimental design isn’t enough

An obvious first attempt would be to pull horizontally, then pull at 30° above the horizontal, keeping the spring balance at 40 N.

Suppose the acceleration changes.

That observation alone wouldn’t show whether the change came from friction.

Why?

Because changing the rope from 0° to 30° changes two things simultaneously:

• The horizontal component of the applied force.
• The normal reaction, which affects friction.

A better experiment compares a 30° upward pull with a 30° downward push.

Both forces then have the same horizontal component. Their vertical components have equal magnitudes but opposite directions.

This helps isolate the effect of changing the normal reaction.

What would the experiment show?

The calculations give these illustrative predictions:

These values are calculated predictions for the stated idealised conditions, not measurements from a real experiment.

If experimental readings were close to these predictions, they would support the model that kinetic friction depends on the normal reaction.

But we can strengthen the test.

Repeat the comparison on a surface with negligible friction.

If \(\mu_k\) is approximately zero, both cases should have nearly the same acceleration:

\[ \begin{aligned} a_x&=\frac{F\cos30^\circ}{m}\\ &=\frac{34.64}{8.0}\\ &=4.33\ \mathrm{m\,s^{-2}} \end{aligned} \]

The predicted difference between upward pulling and downward pushing disappears.

This second test helps distinguish two explanations: a change in the forward component, or a change in friction.

It also shows the limitation of the first intuition. Pulling upwards isn’t automatically better because the force is angled. It helps when the reduction in friction is large enough to matter.

If the surface is frictionless, tilting a fixed-magnitude force away from the horizontal reduces its forward component without providing a friction benefit.

Real measurements may also differ because friction coefficients are approximate, the applied force varies, or the crate tips slightly. The simple model assumes constant kinetic friction coefficients, steady sliding, and no significant rotation.

04What changes when the force acts on a slope?

The same method works on a ramp, but the coordinate axes should change.

Instead of horizontal and vertical, choose:

• The \(x\)-axis parallel to the ramp, positive uphill.
• The \(y\)-axis perpendicular to the ramp, positive away from its surface.

This choice makes the normal reaction act entirely along the \(y\)-axis.

The applied force and weight must then be resolved relative to the ramp.

05Decide whether the object is at rest, sliding, or losing contact

The calculations so far assumed the object was already sliding.

That assumption affects the friction equation.

A stationary object experiences static friction, which adjusts to oppose the tendency for relative sliding. Its magnitude is not automatically \(\mu_sN\).

The maximum possible static friction is:

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

Here, \(\mu_s\) is the coefficient of static friction.

If the friction required to maintain equilibrium exceeds this maximum, the object cannot remain stationary under the assumed conditions.

There’s another condition worth checking: the floor can push the crate upwards, but it cannot pull the crate downwards to hold it in place.

Therefore, a contact force must satisfy:

\[ N\geq 0 \]

If your calculation gives a negative normal reaction, the assumption of continued contact is physically impossible.

For example, increasing the upward component of a pull eventually risks lifting the crate off the floor.

Once the crate loses contact, the vertical acceleration need not be zero. The equation \(N+F_y-mg=0\) is no longer valid.

This is why the force diagram needs to match the physical situation before the equations are solved.

06Questions and solutions

The questions move from straightforward force resolution to situations where the assumed motion or contact conditions change.

Use \(g=9.8\ \mathrm{m\,s^{-2}}\) throughout. Treat ropes and rods as ideal force transmitters, neglect air resistance, and assume the objects do not tip or rotate.

Question 1

A 6.0 kg suitcase rests on a smooth horizontal floor. A person pulls it with a constant force of 20 N at 60° above the horizontal.

Calculate:

a. The horizontal component of the pulling force.

b. The suitcase’s horizontal acceleration.

c. The normal reaction exerted by the floor.

Assume the suitcase remains in contact with the floor.

Solution 1

The horizontal component is 10.0 N, the suitcase accelerates right at \(1.67\ \mathrm{m\,s^{-2}}\), and the normal reaction is 41.5 N.

Step 1: Calculate the horizontal component.

\[ \begin{aligned} F_x&=F\cos60^\circ\\ &=20(0.5)\\ &=10.0\ \mathrm{N} \end{aligned} \]

Step 2: Calculate horizontal acceleration.

The floor is smooth, so friction is negligible.

\[ \begin{aligned} a_x&=\frac{F_x}{m}\\ &=\frac{10.0}{6.0}\\ &=1.67\ \mathrm{m\,s^{-2}} \end{aligned} \]

Step 3: Find the normal reaction.

The vertical component is:

\[ F_y=20\sin60^\circ=17.32\ \mathrm{N} \]

Because the suitcase has no vertical acceleration:

\[ \begin{aligned} N&=mg-F_y\\ &=6.0(9.8)-17.32\\ &=41.48\ \mathrm{N} \end{aligned} \]

Therefore, \(N=41.5\ \mathrm{N}\) to three significant figures.

The important distinction is that the vertical component affects the normal reaction but doesn’t contribute directly to horizontal acceleration.

Using the full 20 N as a horizontal force would overestimate the acceleration.

Question 2

A 12 kg equipment case is initially at rest on a rough horizontal floor.

A cable pulls it at 35° above the horizontal. The coefficients of friction are:

\[ \mu_s=0.40,\qquad \mu_k=0.30 \]

a. Calculate the critical applied force at which the case is just about to slide right.

b. Estimate its horizontal acceleration immediately after sliding begins, assuming the applied force is held only slightly above this critical value.

Assume the case remains in contact with the floor.

Solution 2

The critical force is approximately 44.9 N. Immediately after sliding begins, the limiting acceleration is approximately \(0.77\ \mathrm{m\,s^{-2}}\) to the right.

Step 1: Set up the limiting static-friction condition.

At the point of impending rightward motion, the horizontal pull equals the maximum static friction:

\[ F\cos35^\circ=\mu_sN \]

Because the pull is angled upwards:

\[ N=mg-F\sin35^\circ \]

Substitute:

\[ F\cos35^\circ =0.40(mg-F\sin35^\circ) \]

Using \(mg=117.6\ \mathrm{N}\):

\[ F\cos35^\circ =0.40(117.6-F\sin35^\circ) \]

Collect the terms involving \(F\):

\[ F(\cos35^\circ+0.40\sin35^\circ)=47.04 \]

\[ \begin{aligned} F&=\frac{47.04}{1.049}\\ &=44.9\ \mathrm{N} \end{aligned} \]

At exactly this value, the case is on the verge of sliding. A force just above it initiates motion in the ideal model.

Step 2: Find the normal reaction at the critical force.

Using the unrounded critical force:

\[ \begin{aligned} N&=117.6-F\sin35^\circ\\ &\approx117.6-25.73\\ &=91.87\ \mathrm{N} \end{aligned} \]

Step 3: Switch to kinetic friction after motion begins.

Once the case slides, the friction magnitude approaches:

\[ \begin{aligned} f_k&=\mu_kN\\ &=0.30(91.87)\\ &=27.56\ \mathrm{N} \end{aligned} \]

The horizontal pulling component is approximately 36.75 N.

Therefore:

\[ \begin{aligned} a_x&=\frac{36.75-27.56}{12}\\ &=0.766\ \mathrm{m\,s^{-2}} \end{aligned} \]

The acceleration is approximately \(0.77\ \mathrm{m\,s^{-2}}\) right.

The tempting mistake is to continue using maximum static friction after the case starts moving.

Static friction determines when motion begins. Kinetic friction determines the resistance once sliding has started.

Because \(\mu_k<\mu_s\), the resisting force drops as the case transitions into sliding, allowing a positive acceleration even when the applied force is only slightly above the starting threshold.

Question 3

A 6.0 kg crate initially rests on a ramp inclined at 25° above the horizontal.

A cable attached to the crate makes an angle of 15° above the ramp surface and pulls uphill.

The coefficients of friction are:

\[ \mu_s=0.40,\qquad\mu_k=0.25 \]

The cable initially pulls with 35 N.

a. Determine whether the crate stays at rest. Calculate the magnitude and direction of the actual friction force.

b. The pulling force is increased suddenly to 55 N. Determine whether the crate begins moving and calculate its acceleration immediately after it starts sliding.

Assume the crate maintains contact with the ramp.

Solution 3

At 35 N, the crate remains stationary, with static friction of 8.96 N acting downhill. At 55 N, it starts sliding uphill and accelerates at \(3.09\ \mathrm{m\,s^{-2}}\).

Step 1: Calculate the component of weight downhill.

\[ \begin{aligned} W_x&=mg\sin25^\circ\\ &=6.0(9.8)\sin25^\circ\\ &=24.85\ \mathrm{N} \end{aligned} \]

The component of weight into the ramp is:

\[ \begin{aligned} W_y&=mg\cos25^\circ\\ &=53.29\ \mathrm{N} \end{aligned} \]

Step 2: Examine the initial 35 N pull.

The uphill cable component is:

\[ F_x=35\cos15^\circ=33.81\ \mathrm{N} \]

Without friction, the net force along the ramp would be:

\[ 33.81-24.85=8.96\ \mathrm{N} \]

This would tend to move the crate uphill.

Therefore, static friction must act downhill to prevent motion.

The normal reaction is:

\[ \begin{aligned} N&=53.29-35\sin15^\circ\\ &=44.23\ \mathrm{N} \end{aligned} \]

Maximum static friction is:

\[ \begin{aligned} f_{s,\max}&=0.40(44.23)\\ &=17.69\ \mathrm{N} \end{aligned} \]

Only 8.96 N is needed for equilibrium, which is less than 17.69 N.

The crate remains stationary.

The actual static friction is 8.96 N downhill, not 17.69 N.

Step 3: Test equilibrium when the pull becomes 55 N.

The uphill force component becomes:

\[ 55\cos15^\circ=53.13\ \mathrm{N} \]

The normal reaction becomes:

\[ \begin{aligned} N&=53.29-55\sin15^\circ\\ &=39.06\ \mathrm{N} \end{aligned} \]

Maximum static friction is now:

\[ f_{s,\max}=0.40(39.06)=15.62\ \mathrm{N} \]

To maintain equilibrium, friction would need to oppose:

\[ 53.13-24.85=28.28\ \mathrm{N} \]

But the maximum available static friction is only 15.62 N.

Therefore, the crate must begin sliding uphill.

Step 4: Calculate acceleration after sliding begins.

Kinetic friction is:

\[ \begin{aligned} f_k&=0.25(39.06)\\ &=9.76\ \mathrm{N} \end{aligned} \]

Taking uphill as positive:

\[ \begin{aligned} a_x&=\frac{53.13-24.85-9.76}{6.0}\\ &=3.09\ \mathrm{m\,s^{-2}} \end{aligned} \]

The crate accelerates uphill.

The subtle point is that the cable changes two conditions simultaneously. Increasing its force provides more uphill drive while reducing the normal reaction and the maximum available static friction.

Both changes make uphill sliding more likely.

Question 4

A 6.0 kg crate is already sliding right across a horizontal floor. The coefficient of kinetic friction is 0.25.

A cable pulls the crate at 60° above the horizontal. The cable can exert a constant force of any specified magnitude while maintaining this direction.

An engineer wants the crate to accelerate horizontally at \(6.0\ \mathrm{m\,s^{-2}}\) while remaining on the floor.

a. Calculate the cable force predicted by the usual floor-sliding equations. Explain whether this result is physically possible.

b. Determine the greatest horizontal acceleration possible while the crate remains in contact with the floor.

c. The cable force is instead set to 72 N. Calculate the horizontal and vertical accelerations immediately after this force is applied.

Solution 4

A horizontal acceleration of \(6.0\ \mathrm{m\,s^{-2}}\) is impossible while the crate remains on the floor. The contact model predicts a cable force of 70.8 N, but that would require a negative normal reaction. The greatest horizontal acceleration at the contact boundary is \(5.66\ \mathrm{m\,s^{-2}}\).

With a 72 N pull, the crate instead lifts off, accelerating horizontally at \(6.0\ \mathrm{m\,s^{-2}}\) and vertically upwards at \(0.592\ \mathrm{m\,s^{-2}}\).

Step 1: Write the equations assuming continued contact.

Vertically:

\[ N=mg-T\sin60^\circ \]

Horizontally:

\[ T\cos60^\circ-\mu_kN=ma_x \]

Substitute for \(N\):

\[ T\cos60^\circ -\mu_k(mg-T\sin60^\circ) =ma_x \]

Collect the terms involving \(T\):

\[ T(\cos60^\circ+\mu_k\sin60^\circ) =m(a_x+\mu_kg) \]

For the requested \(a_x=6.0\ \mathrm{m\,s^{-2}}\):

\[ \begin{aligned} T&=\frac{6.0(6.0+2.45)}{0.7165}\\ &=70.8\ \mathrm{N} \end{aligned} \]

Step 2: Test the normal reaction.

\[ \begin{aligned} N&=6.0(9.8)-70.8\sin60^\circ\\ &\approx58.8-61.3\\ &=-2.5\ \mathrm{N} \end{aligned} \]

This is physically impossible for an ordinary floor contact.

The floor can push upwards, but it cannot exert a downward normal reaction to hold the crate against it.

The assumed floor-sliding model has failed.

Step 3: Find the contact boundary.

At the instant contact is lost:

\[ N=0 \]

Thus:

\[ T\sin60^\circ=mg \]

\[ \begin{aligned} T&=\frac{58.8}{\sin60^\circ}\\ &=67.9\ \mathrm{N} \end{aligned} \]

At this boundary, kinetic friction approaches zero because \(f_k=\mu_kN\).

The limiting horizontal acceleration is:

\[ \begin{aligned} a_{x,\max}&=\frac{67.9\cos60^\circ}{6.0}\\ &=5.66\ \mathrm{m\,s^{-2}} \end{aligned} \]

Therefore, a horizontal acceleration of 6.0 m/s² cannot occur while contact is maintained at this angle.

Step 4: Analyse the actual 72 N force.

Since:

\[ 72\sin60^\circ>58.8 \]

the upward cable component exceeds the weight.

The crate lifts away from the floor.

Once contact is lost:

\[ N=0,\qquad f_k=0 \]

Horizontally:

\[ \begin{aligned} a_x&=\frac{72\cos60^\circ}{6.0}\\ &=6.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

Vertically:

\[ \begin{aligned} a_y&=\frac{72\sin60^\circ-58.8}{6.0}\\ &=0.592\ \mathrm{m\,s^{-2}} \end{aligned} \]

The crate accelerates upwards as well as right.

The tempting mistake is to accept the 70.8 N calculation without checking its physical assumptions.

Newton’s second law was used correctly algebraically, but the proposed contact condition was impossible.

The calculation is only valid when the free-body diagram describes a physically possible state.

Question 5

Two identical 5.0 kg crates, A and B, are on separate, identical rough horizontal tracks. They cannot interact with one another.

The coefficients of friction for both crates are:

\[ \mu_s=0.70,\qquad\mu_k=0.25 \]

Each crate is pushed to the right by a rigid rod exerting a constant force of 120 N at 60° below the horizontal.

Initially:

• Crate A is at rest.
• Crate B is already sliding right at \(1.0\ \mathrm{m\,s^{-1}}\).

Both rods maintain their forces and directions for 1.0 s.

a. Determine the motion of each crate during this second. Calculate B’s acceleration, final velocity, and displacement.

b. Someone suggests increasing the force on A until it starts moving. Prove whether any force magnitude applied at 60° below the horizontal could start A sliding right from rest, under the stated friction model.

c. After 1.0 s, both rods are rotated to 30° below the horizontal without changing their 120 N force magnitudes. Determine whether A begins sliding. Once both crates are sliding, compare their horizontal accelerations.

Assume the friction model remains valid, the crates do not tip, and the rods can transmit arbitrarily large forces when required for the theoretical argument.

Solution 5

Crate A remains stationary for the first second. Crate B accelerates right at \(4.35\ \mathrm{m\,s^{-2}}\), travels 3.18 m, and reaches \(5.35\ \mathrm{m\,s^{-1}}\).

No force magnitude at 60° below horizontal can start A moving under the stated model. However, changing the direction to 30° below horizontal allows A to start sliding. Once both crates are sliding, their accelerations are equal at \(15.3\ \mathrm{m\,s^{-2}}\).

This happens because the direction of the applied force changes how quickly the normal reaction, and therefore friction, increases.

Step 1: Calculate the force components at 60°.

For either crate:

\[ \begin{aligned} F_x&=120\cos60^\circ\\ &=60.0\ \mathrm{N} \end{aligned} \]

The downward component is:

\[ \begin{aligned} |F_y|&=120\sin60^\circ\\ &=103.92\ \mathrm{N} \end{aligned} \]

The normal reaction is:

\[ \begin{aligned} N&=mg+F\sin60^\circ\\ &=49.0+103.92\\ &=152.92\ \mathrm{N} \end{aligned} \]

Step 2: Determine whether A moves.

Maximum static friction is:

\[ \begin{aligned} f_{s,\max}&=0.70(152.92)\\ &=107.04\ \mathrm{N} \end{aligned} \]

The rightward component of the applied force is only 60.0 N.

Therefore, static friction can balance it.

Crate A remains stationary, with an actual static-friction force of 60.0 N to the left.

Its displacement during the first second is zero.

Step 3: Determine B’s acceleration and motion.

B is already sliding right, so kinetic friction applies.

\[ \begin{aligned} f_k&=0.25(152.92)\\ &=38.23\ \mathrm{N} \end{aligned} \]

The net horizontal force is:

\[ F_{\text{net}}=60.0-38.23=21.77\ \mathrm{N} \]

Therefore:

\[ \begin{aligned} a_x&=\frac{21.77}{5.0}\\ &=4.35\ \mathrm{m\,s^{-2}} \end{aligned} \]

Using the constant-acceleration equation \(v=u+at\), where \(u\) is initial velocity, \(v\) is final velocity, and \(t\) is time:

\[ \begin{aligned} v&=1.0+4.35385(1.0)\\ &=5.35\ \mathrm{m\,s^{-1}} \end{aligned} \]

B’s displacement, \(s\), is:

\[ s=ut+\frac12at^2 \]

\[ \begin{aligned} s&=1.0(1.0)\\ &\quad+\frac12(4.35385)(1.0)^2\\ &=3.18\ \mathrm{m} \end{aligned} \]

Despite having the same mass, applied force, and surface, the crates behave differently because their initial motion states are different.

Step 4: Prove whether increasing the 60° push can start A moving.

Let the applied force magnitude be any non-negative value \(F\).

To start A sliding right, the horizontal component must exceed maximum static friction:

\[ F\cos60^\circ>\mu_sN \]

For a downward push:

\[ N=mg+F\sin60^\circ \]

Substitute:

\[ F\cos60^\circ >0.70(mg+F\sin60^\circ) \]

Rearrange:

\[ F(\cos60^\circ-0.70\sin60^\circ) >0.70mg \]

Using \(m=5.0\ \mathrm{kg}\):

\[ F(0.500-0.70\times0.866)>34.3 \]

Therefore:

\[ -0.106F>34.3 \]

This inequality cannot be satisfied for any non-negative \(F\).

Increasing the push also increases the normal reaction. At this steep downward angle, the maximum available static friction increases faster than the forward component of the push.

Within the idealised friction model, no magnitude of this 60° downward push can start A sliding right.

Yet B, which was already sliding, accelerates right because its lower kinetic-friction coefficient allows the forward force to exceed kinetic friction.

That is the central surprise: the same applied force can fail to start motion but still accelerate an object that is already sliding.

Step 5: Rotate the force to 30° below horizontal.

Now the horizontal force component is:

\[ \begin{aligned} F_x&=120\cos30^\circ\\ &=103.92\ \mathrm{N} \end{aligned} \]

The downward component is:

\[ F_y=-120\sin30^\circ=-60.0\ \mathrm{N} \]

The normal reaction becomes:

\[ N=49.0+60.0=109.0\ \mathrm{N} \]

Maximum static friction for A is:

\[ \begin{aligned} f_{s,\max}&=0.70(109.0)\\ &=76.3\ \mathrm{N} \end{aligned} \]

But the forward applied component is 103.92 N.

Since:

\[ 103.92>76.3 \]

A can no longer remain stationary.

It begins sliding right.

Step 6: Compare the accelerations once both crates are sliding.

Both crates now experience kinetic friction:

\[ \begin{aligned} f_k&=0.25(109.0)\\ &=27.25\ \mathrm{N} \end{aligned} \]

Their horizontal accelerations are:

\[ \begin{aligned} a_x&=\frac{103.92-27.25}{5.0}\\ &=15.3\ \mathrm{m\,s^{-2}} \end{aligned} \]

Both crates have the same acceleration once they are sliding.

Their velocities are different because B was already moving, but their masses and external forces are now identical.

The final lesson isn’t simply that a smaller angle gives a larger horizontal component.

The deeper idea is that an angled applied force affects both the driving force and the contact conditions. Those effects determine whether an object remains at rest, starts moving, continues sliding, or loses contact altogether.

07What to check in your next force problem

Before substituting numbers into \(\sum F=ma\), make three decisions: which object is your system, which interactions act on it, and which directions are physically constrained.

Then resolve only the forces that need resolving. The perpendicular equation tells you the contact conditions; the parallel equation tells you the motion.

If a calculated acceleration points opposite to the object’s existing velocity, that doesn’t automatically mean the solution is wrong. Acceleration describes how velocity changes, not necessarily the direction of travel. The distinction is explored in Can a Force Act Opposite to Motion?.

And when a problem introduces several touching objects, the next useful skill is changing the system boundary without confusing internal and external forces. See How to Calculate Contact Forces Between Three Pushed Blocks.

The equations stay familiar. The skill is deciding which equations actually describe the physical situation.