How to Decide Which Way Friction Points on an Inclined Plane

Put a block on a rough inclined plane and leave it alone. It doesn’t move. Which way does friction point?

Up the slope. Gravity has a component pulling the block downhill, so static friction acts uphill to prevent it from sliding.

Now attach a rope and pull the same block uphill. Friction might point downhill instead. Tilt the rope further, and friction could even reverse direction while the block remains stationary.

The key is this: friction acts opposite to the relative sliding, or the tendency for relative sliding, between the surfaces in contact. It does not automatically point downhill, oppose the applied force, or oppose the block’s acceleration.

To determine its direction, examine the forces parallel to the incline first. Then calculate the normal reaction, because that determines how much friction the surfaces can supply.

01The direction of friction depends on the tendency to slip

Imagine placing a book on a sloping desk.

If you remove friction from the situation mentally, which way would the book start moving?

It would slide downhill because gravity has a component acting along the slope. Friction must therefore act uphill to prevent that movement.

Now imagine someone pulling the book uphill with a rope. If the pull is strong enough, the book would tend to slide uphill without friction. Static friction must now act downhill.

The surface hasn’t changed. The angle hasn’t changed. Only the other forces have changed.

This gives us a reliable decision rule:

There are two types of friction to distinguish.

Static friction acts when the surfaces are not sliding relative to each other. Its direction opposes the direction in which the surfaces would otherwise begin to slip.

Kinetic friction acts when the surfaces are sliding relative to each other. Its direction opposes that sliding.

This distinction matters because a stationary block has no actual direction of motion. You cannot determine its friction direction by looking at its velocity. You must work out its tendency to move.

A useful mental test

Before drawing a friction arrow, temporarily remove friction from your force diagram.

• If the remaining forces produce a tendency to move the block uphill, friction must point downhill.
• If they produce a tendency to move the block downhill, friction must point uphill.
• If the remaining forces already balance along the slope, no friction is required.

This is more reliable than memorising a friction arrow for every inclined-plane diagram.

02Resolve weight into components along the slope

Before applying the decision rule, you need to understand how gravity acts on an incline.

Choose two perpendicular axes:

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

The block’s weight, \(mg\), still acts vertically downwards. Tilting the surface does not change the direction of gravity.

However, we can resolve weight into two components.

The component parallel to the incline is:

\[ W_x=mg\sin\theta \]

It points downhill.

The component perpendicular to the incline is:

\[ W_y=mg\cos\theta \]

It points into the surface.

Here:

• \(m\) is the mass in kilograms.
• \(g\) is gravitational acceleration, taken as \(9.8\text{ m s}^{-2}\).
• \(\theta\) is the angle of the incline above the horizontal.

Notice that the parallel component uses sine, while the perpendicular component uses cosine.

A useful way to remember this is that the angle \(\theta\) appears between the weight vector and the inward perpendicular direction. The perpendicular component is therefore the adjacent side of the component triangle.

What does this tell us about friction?

Consider a block resting on an incline with no other applied forces.

Gravity’s parallel component acts downhill:

\[ mg\sin\theta \]

The block would therefore slide downhill if friction disappeared.

Static friction must act uphill.

If the block remains stationary, the friction force balances the parallel component of weight:

\[ f_s=mg\sin\theta \]

This does not mean static friction always equals \(mg\sin\theta\). It does so in this particular situation because there are no other forces acting parallel to the incline.

For a more complete explanation of resolving forces and drawing inclined-plane free-body diagrams, see Forces on Inclined Planes: Components, Normal Force and Friction.

03When does the normal reaction change?

This is where many otherwise correct solutions go wrong.

Students often write:

\[ N=mg\cos\theta \]

This is correct only under a particular condition: there must be no other forces with components perpendicular to the incline, and the block must have zero acceleration perpendicular to the surface.

The normal reaction is the perpendicular contact force exerted by the surface.

It adjusts to balance the perpendicular forces while the block remains in contact with the incline.

Case 1: No additional applied force

If gravity is the only force with a perpendicular component, then:

\[ N=mg\cos\theta \]

Case 2: A force pulls the block above the incline

Suppose a rope pulls the block uphill at an angle \(\alpha\) above the surface.

The tension \(F\) now has two components:

\[ F_x=F\cos\alpha \]

\[ F_y=F\sin\alpha \]

The perpendicular component points away from the incline.

It helps support the block against gravity, so the surface does not need to push as hard.

Therefore:

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

The normal reaction decreases.

Case 3: A force pushes the block into the incline

Now reverse the situation.

An applied force pushes the block uphill but also into the surface, at an angle \(\alpha\) below the plane.

Its perpendicular component adds to the perpendicular component of weight.

Therefore:

\[ \boxed{N=mg\cos\theta+F\sin\alpha} \]

The normal reaction increases.

Compare the three situations

The last row is particularly important.

A force can be very large and still leave the normal reaction unchanged if it acts entirely parallel to the incline.

Why should you care about the normal reaction?

Because the maximum static friction depends on it.

Static friction adjusts to whatever value is needed to prevent slipping, up to a maximum:

\[ f_s\leq\mu_sN \]

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

When sliding occurs, the kinetic friction magnitude is:

\[ f_k=\mu_kN \]

Here, \(\mu_k\) is the coefficient of kinetic friction.

Do not confuse the two.

Static friction is not automatically equal to \(\mu_sN\). That value is its maximum possible magnitude, reached when slipping is just about to occur.

This gives us an important distinction:

• The forces parallel to the incline determine the required direction of friction.
• The normal reaction determines the maximum friction available.

If the calculated static friction requirement exceeds \(\mu_sN\), the block cannot remain stationary.

One final limitation: if your calculation predicts \(N\leq0\), the block cannot maintain contact with the surface under that assumed arrangement. The normal reaction cannot pull the block towards the incline. You must instead analyse the motion after contact is lost.

04An investigation that makes friction reverse direction

A simple adjustable-ramp investigation makes the changing direction of static friction easier to understand.

This is a proposed classroom investigation. The numerical values below are illustrative predictions from the force model, not reported experimental measurements.

The original apparatus

Place a \(2.0\text{ kg}\) block on a rough adjustable ramp set at \(20^\circ\).

Attach a spring balance to the uphill side of the block. Connect it using a string that runs parallel to the inclined surface.

The spring balance measures the tension pulling the block uphill.

What should happen if the spring balance initially reads zero?

The block would tend to slide downhill under gravity. Static friction must point uphill.

Now gradually increase the tension.

The block can remain stationary, but the amount of friction required changes.

Taking uphill as positive, the horizontal force balance along the incline is:

\[ T+f_s-mg\sin\theta=0 \]

Rearranging:

\[ f_s=mg\sin\theta-T \]

This equation gives the signed friction force. A positive result means uphill friction, while a negative result means downhill friction.

Following the changing friction

First calculate the normal reaction:

\[ N=mg\cos20^\circ \]

\[ N=2.0(9.8)\cos20^\circ \]

\[ N=18.4\text{ N} \]

The maximum static friction, using an illustrative coefficient of \(\mu_s=0.50\), is:

\[ f_{s,\max}=0.50(18.4) \]

\[ f_{s,\max}=9.21\text{ N} \]

The gravitational component along the slope is:

\[ mg\sin20^\circ=6.70\text{ N} \]

We can now predict what happens as the tension increases.

At \(T=6.70\text{ N}\), the tension exactly balances the downhill component of gravity.

No friction is needed.

At \(T=10.0\text{ N}\), the tension is larger than the downhill gravitational component. The block would tend to move uphill without friction, so static friction must act downhill.

At approximately \(15.9\text{ N}\), static friction reaches its maximum. The block is just about to slide uphill.

The investigation predicts a complete reversal of static friction while the block remains stationary.

What the original design cannot test

The string is parallel to the incline, so it has no perpendicular component.

The normal reaction stays constant throughout the investigation:

\[ N=mg\cos20^\circ \]

This is useful for isolating the change in friction direction, but it cannot demonstrate how an angled force changes the normal reaction.

Revising the apparatus

To investigate the normal reaction, raise the string so that it pulls at an angle above the incline.

Support the ramp on a suitable force sensor that measures the perpendicular contact load, accounting for the ramp’s own weight.

The spring balance measures the tension, while the perpendicular sensor measures the changing normal reaction.

The predicted relationship becomes:

\[ N=mg\cos\theta-T\sin\alpha \]

As the string pulls further away from the surface, the normal reaction decreases.

Consequently, the maximum available static friction also decreases.

The two designs answer different questions:

• The parallel pull isolates how friction direction changes.
• The angled pull demonstrates how an applied force changes the normal reaction and therefore the available friction.

05A reliable method for deciding friction direction

When you encounter an inclined-plane question in an HSC Physics exam, use this sequence.

Step 1: Draw the axes.

Choose one axis parallel to the slope and another perpendicular to it.

Positive uphill is usually convenient.

Step 2: Resolve the weight.

Write:

\[ W_x=-mg\sin\theta \]

\[ W_y=-mg\cos\theta \]

The negative signs follow from our choice of positive uphill and positive away from the surface.

Step 3: Consider every other applied force.

Resolve any tension, push, or applied force into parallel and perpendicular components.

Do not assume an applied force acts entirely along the incline.

Step 4: Determine the tendency to slip.

For a stationary block, add the parallel components of all forces except friction.

• A net tendency uphill means friction acts downhill.
• A net tendency downhill means friction acts uphill.
• A zero resultant means friction may be zero.

If the block is already sliding, use its relative sliding direction to determine kinetic friction.

Step 5: Calculate the normal reaction.

Use the perpendicular force balance.

Do not automatically substitute \(mg\cos\theta\) if another force has a perpendicular component.

Step 6: Check whether static friction is sufficient.

Calculate the friction required to maintain equilibrium and compare its magnitude with:

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

If the required friction is too large, the block slips. Then use kinetic friction in the direction opposite the sliding.

The important point is to determine direction before calculating the friction magnitude. You should not simply write \(f=\mu N\) and hope the sign works itself out.

08Questions and solutions

Question 1

A \(3.0\text{ kg}\) block rests on a rough inclined plane at \(25^\circ\) to the horizontal. There are no other applied forces.

The coefficient of static friction is \(0.60\).

Calculate:

a) The normal reaction.

b) The direction and magnitude of static friction required to keep the block stationary.

c) Whether the block remains at rest.

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

Solution 1

The block remains stationary, with static friction acting uphill at \(12.4\text{ N}\) and a normal reaction of \(26.6\text{ N}\).

a) Normal reaction

There are no other perpendicular forces, so:

\[ N=mg\cos\theta \]

\[ N=3.0(9.8)\cos25^\circ \]

\[ \boxed{N=26.6\text{ N}} \]

b) Direction and magnitude of friction

Without friction, gravity would make the block slide downhill.

Static friction therefore acts uphill.

For equilibrium:

\[ f_s=mg\sin\theta \]

\[ f_s=3.0(9.8)\sin25^\circ \]

\[ \boxed{f_s=12.4\text{ N uphill}} \]

c) Check whether the block remains stationary

The maximum static friction is:

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

\[ f_{s,\max}=0.60(26.6) \]

\[ f_{s,\max}=15.96\text{ N} \]

Since:

\[ 12.4<15.96 \]

static friction is sufficient to prevent sliding.

The block remains at rest. The friction force is below its maximum possible value.

Question 2

A \(4.0\text{ kg}\) block rests on a rough inclined plane at \(30^\circ\) to the horizontal. The plane rises towards the right.

A horizontal force of \(20\text{ N}\) pushes the block towards the right, towards the raised end of the incline.

The coefficient of static friction is \(0.50\).

Determine:

a) The components of the horizontal force parallel and perpendicular to the incline.

b) The normal reaction.

c) The direction and magnitude of the required static friction.

d) Whether the block remains stationary.

Solution 2

The block remains stationary. Static friction acts uphill with a magnitude of \(2.28\text{ N}\), and the normal reaction is \(43.95\text{ N}\).

a) Resolve the horizontal force

The horizontal force makes an angle of \(30^\circ\) with the direction of the incline.

Its parallel component is:

\[ F_x=20\cos30^\circ \]

\[ F_x=17.32\text{ N} \]

This component acts uphill.

Its perpendicular component is:

\[ F_y=20\sin30^\circ \]

\[ F_y=10.0\text{ N} \]

This component acts into the incline.

b) Calculate the normal reaction

The weight also acts into the incline:

\[ mg\cos30^\circ =4.0(9.8)\cos30^\circ \]

\[ mg\cos30^\circ=33.95\text{ N} \]

The horizontal push adds another \(10.0\text{ N}\) into the surface.

Therefore:

\[ N=mg\cos30^\circ+20\sin30^\circ \]

\[ N=33.95+10.0 \]

\[ \boxed{N=43.95\text{ N}} \]

c) Determine the friction direction and magnitude

The gravitational component downhill is:

\[ mg\sin30^\circ =4.0(9.8)\sin30^\circ \]

\[ mg\sin30^\circ=19.60\text{ N} \]

The horizontal push contributes \(17.32\text{ N}\) uphill.

Without friction, the resultant would be:

\[ 19.60-17.32=2.28\text{ N downhill} \]

The block therefore tends to move downhill.

Static friction must act uphill, with a magnitude of:

\[ \boxed{f_s=2.28\text{ N}} \]

d) Check whether the block remains stationary

The maximum static friction is:

\[ f_{s,\max}=0.50(43.95) \]

\[ f_{s,\max}=21.98\text{ N} \]

Since:

\[ 2.28<21.98 \]

the block remains stationary.

The important detail is that the horizontal force has two effects. It helps pull the block uphill, but it also pushes the block into the incline, increasing the normal reaction.

A student who resolves only the parallel component misses half the physics.

Question 3

A \(6.0\text{ kg}\) block is initially stationary on a rough incline at \(25^\circ\) to the horizontal.

A rope pulls the block uphill at \(30^\circ\) above the surface with a tension of \(45\text{ N}\).

The coefficients of friction are:

\[ \mu_s=0.35,\qquad \mu_k=0.25 \]

Calculate:

a) The normal reaction.

b) The direction and magnitude of the static friction that would be required to maintain equilibrium.

c) Whether the block starts moving.

d) Its acceleration once it begins sliding.

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

Solution 3

The block slides uphill with an acceleration of approximately \(1.07\text{ m s}^{-2}\). The required static friction is downhill, but it is not large enough to prevent motion.

a) Calculate the normal reaction

The weight component into the surface is:

\[ mg\cos25^\circ =6.0(9.8)\cos25^\circ \]

\[ mg\cos25^\circ=53.29\text{ N} \]

The outward component of tension is:

\[ F\sin30^\circ=45\sin30^\circ \]

\[ F\sin30^\circ=22.50\text{ N} \]

Therefore:

\[ N=53.29-22.50 \]

\[ \boxed{N=30.79\text{ N}} \]

b) Calculate the required static friction

The tension component uphill is:

\[ F\cos30^\circ=45\cos30^\circ \]

\[ F\cos30^\circ=38.97\text{ N} \]

The gravitational component downhill is:

\[ mg\sin25^\circ =6.0(9.8)\sin25^\circ \]

\[ mg\sin25^\circ=24.85\text{ N} \]

Without friction, the resultant is:

\[ 38.97-24.85=14.12\text{ N uphill} \]

Therefore, static friction would need to act downhill with a magnitude of:

\[ \boxed{f_{s,\text{required}}=14.12\text{ N}} \]

c) Determine whether the block starts moving

The maximum static friction is:

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

\[ f_{s,\max}=0.35(30.79) \]

\[ f_{s,\max}=10.78\text{ N} \]

The required friction is \(14.12\text{ N}\), which exceeds the maximum available value.

The block cannot remain stationary. It begins sliding uphill.

d) Calculate the acceleration

Kinetic friction now acts downhill:

\[ f_k=\mu_kN \]

\[ f_k=0.25(30.79) \]

\[ f_k=7.70\text{ N} \]

Using uphill as positive:

\[ \sum F_x=ma \]

\[ 38.97-24.85-7.70=6.0a \]

\[ 6.42=6.0a \]

\[ \boxed{a=1.07\text{ m s}^{-2}\text{ uphill}} \]

The rope reduces the normal reaction, which reduces the available friction. This is why the block starts sliding even though the coefficient of static friction is not particularly small.

Question 4

A \(5.0\text{ kg}\) block is sliding uphill on a rough incline at \(20^\circ\) to the horizontal.

Its initial speed is \(6.0\text{ m s}^{-1}\). An applied force that was previously pulling it uphill is suddenly removed.

The coefficients of friction are:

\[ \mu_k=0.20,\qquad \mu_s=0.45 \]

Determine:

a) The direction and magnitude of kinetic friction while the block is sliding uphill.

b) The acceleration of the block.

c) The distance travelled uphill before it stops.

d) Whether the block slides back downhill or remains stationary after stopping. Justify your answer.

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

Solution 4

The block decelerates while moving uphill, travels approximately \(3.47\text{ m}\), and then remains stationary. It does not slide back downhill because static friction can support it.

a) Determine the kinetic friction

The block is sliding uphill.

Kinetic friction must therefore act downhill, regardless of whether the block is speeding up or slowing down.

The normal reaction is:

\[ N=mg\cos20^\circ \]

\[ N=5.0(9.8)\cos20^\circ \]

\[ N=46.05\text{ N} \]

The kinetic friction is:

\[ f_k=\mu_kN \]

\[ f_k=0.20(46.05) \]

\[ \boxed{f_k=9.21\text{ N downhill}} \]

b) Calculate the acceleration

Gravity also has a component acting downhill:

\[ mg\sin20^\circ =5.0(9.8)\sin20^\circ \]

\[ mg\sin20^\circ=16.76\text{ N} \]

Both gravity and kinetic friction act downhill.

The total downhill force is:

\[ F_{\text{net}}=16.76+9.21 \]

\[ F_{\text{net}}=25.97\text{ N} \]

Taking uphill as positive:

\[ \sum F_x=ma \]

\[ -25.97=5.0a \]

\[ \boxed{a=-5.19\text{ m s}^{-2}} \]

The negative sign indicates acceleration downhill.

c) Calculate the stopping distance

The block initially travels uphill at \(6.0\text{ m s}^{-1}\), with acceleration of \(-5.19\text{ m s}^{-2}\).

Use:

\[ v^2=u^2+2as \]

At the stopping point, \(v=0\).

\[ 0=6.0^2+2(-5.19)s \]

\[ 10.38s=36 \]

\[ \boxed{s=3.47\text{ m}} \]

The block travels \(3.47\text{ m}\) further uphill before stopping.

d) Determine what happens after the block stops

This is the interesting part.

While the block is moving uphill, kinetic friction acts downhill.

However, once the block stops, we must reconsider the situation. We can no longer assume that kinetic friction continues acting downhill.

Without friction, gravity would make the stationary block slide downhill. Static friction must therefore act uphill.

The required static friction is:

\[ f_s=mg\sin20^\circ \]

\[ f_s=16.76\text{ N} \]

The maximum static friction is:

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

\[ f_{s,\max}=0.45(46.05) \]

\[ f_{s,\max}=20.72\text{ N} \]

Since:

\[ 16.76<20.72 \]

static friction is sufficient to hold the block.

The block remains stationary after stopping.

This question exposes an important misconception: friction does not have to keep pointing in the same direction throughout a problem. When the block changes from sliding to resting, the friction model changes, and its direction can reverse.

Question 5

A \(4.0\text{ kg}\) block rests on a rough incline at \(30^\circ\) to the horizontal.

A rope pulls the block uphill with a constant tension of \(32\text{ N}\). The rope initially runs parallel to the incline, but its angle is gradually increased so that it pulls further away from the surface.

Let \(\alpha\) represent the angle between the rope and the incline.

The coefficients of friction are:

\[ \mu_s=0.50,\qquad \mu_k=0.30 \]

Assume:

• The block is initially stationary.
• The tension remains constant at \(32\text{ N}\) as the rope rotates.
• The incline is fixed.
• The block remains in contact with the incline.
• The angle is increased slowly until slipping begins, and is then held at \(65^\circ\).
• Use \(g=9.8\text{ m s}^{-2}\).

Determine:

a) The initial direction and magnitude of static friction when \(\alpha=0^\circ\), and whether the block can remain stationary.

b) The angle at which the required static friction becomes zero.

c) The first angle beyond this point at which static equilibrium fails. State the direction in which the block begins to slide.

d) At \(\alpha=65^\circ\), calculate the normal reaction, the direction and magnitude of kinetic friction, and the block’s acceleration.

This problem tests two effects happening simultaneously. As the rope rotates, its uphill component decreases, but its outward component increases. The friction direction and the maximum available friction therefore change together.

Solution 5

The friction initially acts downhill, becomes zero at approximately \(52.2^\circ\), and then reverses to act uphill. Static equilibrium first fails at approximately \(59.2^\circ\), after which the block slides downhill.

At \(65^\circ\), the normal reaction is approximately \(4.95\text{ N}\), kinetic friction acts uphill with a magnitude of \(1.48\text{ N}\), and the acceleration is approximately \(1.15\text{ m s}^{-2}\) downhill.

a) Determine the initial friction direction and magnitude

When \(\alpha=0^\circ\), the rope pulls parallel to the incline.

The uphill tension is:

\[ T=32\text{ N} \]

The downhill component of weight is:

\[ mg\sin30^\circ =4.0(9.8)\sin30^\circ \]

\[ mg\sin30^\circ=19.60\text{ N} \]

Without friction, the resultant force would be:

\[ 32-19.60=12.40\text{ N uphill} \]

The block would tend to move uphill.

Therefore, static friction acts downhill with a required magnitude of:

\[ f_s=12.40\text{ N} \]

The normal reaction is:

\[ N=mg\cos30^\circ \]

\[ N=4.0(9.8)\cos30^\circ \]

\[ N=33.95\text{ N} \]

The maximum static friction is:

\[ f_{s,\max}=0.50(33.95) \]

\[ f_{s,\max}=16.97\text{ N} \]

Since:

\[ 12.40<16.97 \]

the block can remain stationary.

b) Find the angle at which friction becomes zero

As the rope rotates upwards, its component parallel to the incline decreases.

The uphill component of tension is:

\[ T_x=32\cos\alpha \]

For friction to become zero, the parallel component of tension must exactly balance the downhill component of weight.

Therefore:

\[ 32\cos\alpha=19.60 \]

Rearranging:

\[ \cos\alpha=\frac{19.60}{32} \]

\[ \alpha=\cos^{-1}(0.6125) \]

\[ \boxed{\alpha=52.2^\circ} \]

At this angle, the block has no tendency to slide in either direction along the surface.

Static friction is zero.

Importantly, the block is still stationary. Zero friction does not mean that the block must be moving.

c) Find the first angle at which static equilibrium fails

Beyond \(52.2^\circ\), the rope’s uphill component becomes smaller than the downhill component of gravity.

The block now tends to slide downhill.

Static friction must therefore act uphill.

The required static friction is:

\[ f_{s,\text{required}} =mg\sin30^\circ-32\cos\alpha \]

However, the normal reaction is also changing.

The outward component of tension is \(32\sin\alpha\), so:

\[ N=mg\cos30^\circ-32\sin\alpha \]

Substituting the mass and incline angle:

\[ N=33.95-32\sin\alpha \]

At the point where static equilibrium first fails, the required static friction reaches its maximum:

\[ f_{s,\text{required}}=\mu_sN \]

Substitute:

\[ 19.60-32\cos\alpha = 0.50(33.95-32\sin\alpha) \]

Solving using a calculator gives:

\[ \boxed{\alpha\approx59.2^\circ} \]

The static friction is now at its maximum and acts uphill.

The block is just about to slide downhill.

The static condition remains satisfied between \(0^\circ\) and \(52.2^\circ\); the limiting condition is reached later, on the uphill-friction branch.

d) Analyse the block at \(65^\circ\)

The block has now started sliding downhill, so kinetic friction acts uphill.

First calculate the normal reaction:

\[ N=33.95-32\sin65^\circ \]

\[ N=33.95-29.00 \]

\[ \boxed{N=4.95\text{ N}} \]

Next, calculate the kinetic friction:

\[ f_k=\mu_kN \]

\[ f_k=0.30(4.95) \]

\[ \boxed{f_k=1.48\text{ N}} \]

It acts uphill because the block is sliding downhill.

Now calculate the tension component parallel to the incline:

\[ T_x=32\cos65^\circ \]

\[ T_x=13.52\text{ N} \]

The downhill component of weight remains:

\[ mg\sin30^\circ=19.60\text{ N} \]

Taking uphill as positive:

\[ \sum F_x=ma \]

\[ 13.52+1.48-19.60=4.0a \]

\[ -4.60=4.0a \]

\[ \boxed{a\approx-1.15\text{ m s}^{-2}} \]

The negative sign means the acceleration is downhill.

There are two tempting mistakes here.

The first is to assume friction must always point downhill because the rope pulls uphill. That is true initially, but not after the rope rotates far enough. At \(65^\circ\), the rope’s parallel component is only \(13.52\text{ N}\), while gravity’s parallel component is \(19.60\text{ N}\). The block therefore slides downhill, and kinetic friction acts uphill.

The second mistake is to keep using:

\[ N=mg\cos30^\circ \]

That would give a normal reaction of \(33.95\text{ N}\), rather than the actual \(4.95\text{ N}\).

The rope is now pulling strongly away from the incline. This greatly reduces the normal reaction and, consequently, the maximum available friction.

The block can therefore slide downhill even though the rope is still pulling it uphill.

That is the central lesson: friction direction depends on the tendency or direction of relative sliding, while the normal reaction controls how much friction the surfaces can provide.

Once you can determine both independently, you can tackle the next step in inclined-plane problems: deciding whether a block remains in static equilibrium or starts accelerating, then applying Newton’s second law along the slope.