How to Calculate the Force Needed to Hold a Block on a Slope
Learn how to resolve weight along an incline and calculate the force needed to hold a stationary block. Includes worked HSC Physics examples covering angled pulls, normal reactions, and static friction.
A 5.0 kg block sits on a smooth ramp inclined at 30° to the horizontal. Its weight is 49 N. Does a rope need to pull upwards with 49 N to keep it stationary?
No. If the rope runs parallel to the ramp, it needs only **24.5 N of tension acting up the slope**.
The reason is that the rope only needs to balance the component of gravity pulling the block down the ramp. The perpendicular component is balanced by the normal reaction from the surface.
For a smooth slope, the required holding force is:
\[
\boxed{F=mg\sin\theta}
\]
where \(m\) is the block's mass, \(g\) is gravitational acceleration, and \(\theta\) is the angle of the slope above the horizontal.
But there is an important catch. If the rope pulls at an angle above the slope, it also changes the normal reaction. If the surface has friction, that change can affect how much force is actually needed to hold the block.
Let's build a method that handles all three situations.
[IMAGE: An illustrative physical scene of a heavy rectangular crate resting on a sloping loading ramp, held stationary by a taut rope running parallel to the ramp towards an anchor uphill. Show the ramp angle clearly, with the crate, rope, and ground in their correct spatial relationship.]
## 1. Resolve the weight along the slope
The first mistake to avoid is treating the block's entire weight as a force pulling it down the ramp.
Weight always acts vertically downwards, regardless of the surface underneath the block. However, we can split this force into two perpendicular components that match the geometry of the slope.
Choose two axes:
- The \(x\)-axis runs parallel to the slope, with positive direction uphill.
- The \(y\)-axis runs perpendicular to the slope, with positive direction away from the surface.
This makes the problem much easier because we can analyse motion along the slope and contact with the surface separately.
[IMAGE: An accurate free-body diagram of a rectangular block on a ramp rising towards the right at angle theta. Show the vertical weight vector mg acting downwards from the block's centre. Resolve it into mg sin(theta) acting down the slope and mg cos(theta) acting perpendicular into the slope. Show the outward normal reaction N perpendicular to the surface. Label the positive uphill x-axis and outward y-axis. Include the right-angle relationship between the axes and the angle theta.]
### The two components of weight
The weight of the block is:
\[
W=mg
\]
where:
- \(W\) is weight in newtons (N).
- \(m\) is mass in kilograms (kg).
- \(g\) is gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\).
Resolving weight into the two chosen directions gives:
**Parallel to the slope:**
\[
W_x=mg\sin\theta
\]
This component acts down the slope. It is the part of gravity that the rope or another applied force must balance.
**Perpendicular to the slope:**
\[
W_y=mg\cos\theta
\]
This component acts into the surface. It is balanced by the normal reaction, provided there is no other perpendicular force or acceleration.
Notice that the sine and cosine have different jobs. The component parallel to the slope uses sine, while the perpendicular component uses cosine.
### What is the normal reaction?
The normal reaction, \(N\), is the contact force exerted by the surface on the block.
It always acts perpendicular to the surface and away from it.
For a stationary block on a smooth incline, with an applied force parallel to the slope:
\[
\sum F_y=0
\]
Therefore:
\[
N-mg\cos\theta=0
\]
Giving:
\[
\boxed{N=mg\cos\theta}
\]
This is an important result, but it is not a universal formula for every block on a slope.
The normal reaction equals \(mg\cos\theta\) only when the other forces have no perpendicular components and the block has no acceleration perpendicular to the surface.
We will see why this qualification matters shortly.
## 2. Calculate the force needed to hold a block stationary
Start with the simplest situation:
- The slope is smooth, so friction is negligible.
- A rope pulls parallel to the slope.
- The block remains stationary.
- There is no acceleration in either direction.
The forces parallel to the slope are:
- Tension \(F\), acting uphill.
- Weight component \(mg\sin\theta\), acting downhill.
Because the block is stationary, the resultant force must be zero.
Taking uphill as positive:
\[
\sum F_x=0
\]
\[
F-mg\sin\theta=0
\]
Rearranging:
\[
\boxed{F=mg\sin\theta}
\]
That is the force required to hold the block stationary under these assumptions.
### Worked example: Holding a block on a smooth 25° slope
A 4.0 kg block rests on a smooth slope inclined at 25° to the horizontal. A rope pulls parallel to the slope, uphill.
Calculate:
1. The tension required to keep the block stationary.
2. The normal reaction exerted by the slope.
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: Setup diagram of a 4.0 kg rectangular block on a smooth ramp inclined at 25 degrees, with a rope attached to the block and running parallel to the surface towards an anchor uphill. Label the mass, slope angle, and rope direction. Do not show force components or calculated answers.]
**Step 1**: Calculate the weight of the block.
\[
W=mg
\]
\[
W=4.0\times9.8=39.2\ \mathrm{N}
\]
**Step 2**: Resolve the weight parallel to the slope.
\[
W_x=mg\sin\theta
\]
\[
W_x=39.2\sin25^\circ
\]
\[
W_x=16.6\ \mathrm{N}
\]
This is the gravitational force pulling the block downhill.
**Step 3**: Apply Newton's first law along the slope.
Since the block is stationary, the tension must balance the downhill component of weight.
\[
F=mg\sin\theta
\]
\[
\boxed{F=16.6\ \mathrm{N}}
\]
The rope needs to provide 16.6 N of tension uphill.
**Step 4**: Calculate the normal reaction.
There are no other forces perpendicular to the surface, so:
\[
N=mg\cos\theta
\]
\[
N=39.2\cos25^\circ
\]
\[
\boxed{N=35.5\ \mathrm{N}}
\]
The block experiences a normal reaction of 35.5 N, perpendicular to and away from the slope.
Notice that the tension and normal reaction are different because they balance different components of weight.
## 3. When does the normal reaction change?
Now imagine that the rope is no longer parallel to the slope.
Instead, it pulls upwards at an angle of 15° above the surface.
You might initially think that the required tension is still \(mg\sin\theta\). But that would ignore part of the rope's force.
The tension now has two components:
- \(F\cos\alpha\), acting uphill along the slope.
- \(F\sin\alpha\), acting away from the slope.
Here, \(\alpha\) is the angle between the rope and the surface.
[IMAGE: Free-body diagram of a block on an inclined plane, with a rope pulling uphill at an angle alpha above the surface. Show the tension vector pointing upwards and away from the slope, its parallel component F cos(alpha), and perpendicular component F sin(alpha). Show mg vertically downwards, its components mg sin(theta) down the slope and mg cos(theta) into the slope, and the normal reaction N pointing outwards. Use clearly separated arrows and correct perpendicular geometry.]
### Resolving the angled tension
For equilibrium parallel to the slope:
\[
F\cos\alpha-mg\sin\theta=0
\]
Therefore:
\[
\boxed{F=\frac{mg\sin\theta}{\cos\alpha}}
\]
The tension must now be larger than the parallel-pull result because only part of it acts uphill.
But something else changes.
For equilibrium perpendicular to the slope:
\[
N+F\sin\alpha-mg\cos\theta=0
\]
Rearranging:
\[
\boxed{N=mg\cos\theta-F\sin\alpha}
\]
The perpendicular component of tension pulls the block away from the surface, reducing the normal reaction.
This gives us an important decision rule:
| Applied force | Effect on normal reaction |
|---|---|
| Force parallel to the slope | No change to \(N\) |
| Force pulling away from the slope | Decreases \(N\) |
| Force pushing into the slope | Increases \(N\) |
The normal reaction is determined by balancing *all* perpendicular forces, not simply by calculating \(mg\cos\theta\) every time.
### Worked example: A rope pulling 15° above the slope
A 6.0 kg block rests on a smooth slope inclined at 35° to the horizontal. A rope pulls the block uphill at 15° above the slope.
Calculate:
1. The tension required to keep the block stationary.
2. The normal reaction.
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: Setup diagram showing a 6.0 kg block on a smooth 35-degree incline, attached to a rope directed uphill at 15 degrees above the surface. Clearly show both angles, the block, and the rope's direction. Do not include resolved force components or answers.]
**Step 1**: Calculate the weight.
\[
W=mg
\]
\[
W=6.0\times9.8=58.8\ \mathrm{N}
\]
**Step 2**: Resolve the forces parallel to the slope.
The parallel component of tension must balance the downhill component of weight.
\[
F\cos15^\circ=mg\sin35^\circ
\]
Rearranging:
\[
F=\frac{58.8\sin35^\circ}{\cos15^\circ}
\]
\[
\boxed{F=34.9\ \mathrm{N}}
\]
**Step 3**: Resolve the forces perpendicular to the slope.
The rope pulls away from the surface, so its perpendicular component reduces the normal reaction.
\[
N+F\sin15^\circ-mg\cos35^\circ=0
\]
Therefore:
\[
N=mg\cos35^\circ-F\sin15^\circ
\]
Substituting:
\[
N=58.8\cos35^\circ-34.9\sin15^\circ
\]
\[
\boxed{N=39.1\ \mathrm{N}}
\]
The normal reaction is 39.1 N.
If the rope had instead pulled parallel to the slope, the normal reaction would have been:
\[
N=58.8\cos35^\circ=48.2\ \mathrm{N}
\]
The angled rope reduces the normal reaction by approximately 9.0 N.
That difference comes entirely from the perpendicular component of tension.
## 4. What changes when the slope has friction?
So far, we have assumed smooth surfaces. Real surfaces can exert friction, which changes the meaning of the force needed to hold a block stationary.
Static friction acts along the surface and adjusts to prevent relative motion, up to a maximum value.
Its maximum magnitude is:
\[
\boxed{f_{s,\max}=\mu_sN}
\]
where:
- \(f_{s,\max}\) is the maximum static friction in newtons.
- \(\mu_s\) is the coefficient of static friction.
- \(N\) is the normal reaction in newtons.
The crucial point is that static friction does not automatically equal \(\mu_sN\).
It only reaches this maximum when the block is on the verge of slipping.
### Can friction hold the block without a rope?
Suppose a block is resting on a rough slope, with no rope or other applied force.
Gravity pulls it downhill with force:
\[
mg\sin\theta
\]
Static friction acts uphill to oppose the tendency to slide.
The block can remain stationary if:
\[
mg\sin\theta\leq\mu_smg\cos\theta
\]
Cancel the common factor \(mg\):
\[
\sin\theta\leq\mu_s\cos\theta
\]
Therefore:
\[
\boxed{\tan\theta\leq\mu_s}
\]
If this condition is satisfied, friction alone can hold the block stationary. No additional holding force is required.
If it is not satisfied, an additional uphill force may be needed.
### What if a rope is already pulling uphill?
Take uphill as positive. If the rope pulls parallel to the slope, the equilibrium equation becomes:
\[
F+f_s-mg\sin\theta=0
\]
Rearranging:
\[
\boxed{f_s=mg\sin\theta-F}
\]
This tells us the actual static friction required for equilibrium.
The block can remain stationary provided:
\[
|mg\sin\theta-F|\leq\mu_sN
\]
Notice the absolute value. If the rope pulls strongly enough, friction can actually act downhill instead of uphill.
That is why assuming friction always acts uphill is another common mistake.
For a rope parallel to the slope, the normal reaction remains \(N=mg\cos\theta\). For an angled rope, however, the normal reaction must first be recalculated before checking the friction limit.
### A quick decision process
Before substituting numbers, work through these decisions:
| Step | What to check | What to do |
|---|---|---|
| 1 | Is the block stationary? | Set the resultant force to zero in each direction. |
| 2 | What is the slope angle? | Resolve weight into \(mg\sin\theta\) and \(mg\cos\theta\). |
| 3 | Is the applied force parallel to the slope? | If not, resolve it into parallel and perpendicular components. |
| 4 | Is the surface rough? | Include static friction and its maximum allowable value. |
| 5 | Is the block still in contact with the slope? | Check that \(N>0\). |
The safest order is to calculate the normal reaction first, then use it to check the maximum available static friction.
## 5. Testing the model with an inclined-plane experiment
There is a straightforward way to investigate whether the holding force really follows \(F=mg\sin\theta\).
The experimental question is simple: as we increase the angle of a ramp, does the force required to hold a stationary block increase according to the sine of that angle?
### Setting up the apparatus
Imagine an adjustable inclined plane with a block resting on its surface.
A spring balance or force sensor is connected to the block by a light string. The string runs parallel to the ramp and pulls uphill.
An angle gauge measures the ramp's angle relative to the horizontal. A digital balance measures the block's mass before the investigation.
The force sensor measures the tension needed to keep the block stationary.
[IMAGE: Detailed side-view experimental apparatus showing an adjustable inclined plane with a rectangular block on its surface. A spring balance is connected directly to the block by a string running parallel to the incline, with the balance held or mounted uphill. Show the adjustable ramp support, horizontal reference, angle gauge measuring theta, and the block's position. Include a small digital balance nearby for measuring the block's mass. Make the physical connections clear.]
### What should happen?
If the ramp is sufficiently smooth and the string remains parallel to the surface, the predicted tension is:
\[
F=mg\sin\theta
\]
The normal reaction should be:
\[
N=mg\cos\theta
\]
For a 0.50 kg block, the weight is:
\[
W=0.50\times9.8=4.90\ \mathrm{N}
\]
At a slope angle of 30°, the predicted holding force is:
\[
F=4.90\sin30^\circ
\]
\[
F=2.45\ \mathrm{N}
\]
The block weighs 4.90 N, yet only 2.45 N of parallel tension is needed.
That is exactly the distinction we want the investigation to test.
### Illustrative predictions
The following values are theoretical predictions, not measurements from an actual experiment.
| Slope angle | Predicted tension (N) | Predicted normal reaction (N) |
|---:|---:|---:|
| 10° | 0.85 | 4.83 |
| 20° | 1.68 | 4.60 |
| 30° | 2.45 | 4.24 |
| 40° | 3.15 | 3.75 |
[IMAGE: Scientific graph with slope angle theta on the horizontal axis and predicted holding tension on the vertical axis for a 0.50 kg block. Plot the theoretical relationship F = mg sin(theta) from 0 to 60 degrees. Show clearly labelled points at 10, 20, 30, and 40 degrees. Include a separate small comparison curve or inset for the normal reaction N = mg cos(theta), showing its decrease as the slope angle increases.]
The predicted tension increases with the slope angle, while the normal reaction decreases.
If the force sensor follows this pattern closely, the results support the resolved-weight model.
### What could go wrong with the original design?
A real ramp is not perfectly smooth. Static friction can contribute to the force balance.
If the string is slightly above the surface, its perpendicular component also changes the normal reaction.
Both effects can produce readings that differ from the ideal predictions.
A better investigation would use a low-friction trolley, carefully align the force sensor with the ramp, and measure the normal reaction separately using a force sensor mounted flush with the surface.
If friction cannot be eliminated, it must be included in the force equations rather than treated as experimental noise.
The experiment illustrates the same principle used in calculations: every force must be resolved in the direction in which it acts.
## Questions and solutions
The following questions progress from straightforward force resolution to situations where friction and the normal reaction interact.
### Question 1
A 3.0 kg block is stationary on a smooth incline at 20° to the horizontal.
A rope runs parallel to the incline and holds the block in place.
Calculate:
a. The tension in the rope.
b. The normal reaction exerted by the incline.
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: Exam-style setup diagram showing a 3.0 kg block on a smooth ramp inclined at 20 degrees, with a rope parallel to the ramp pulling uphill. Label the mass, angle, and rope direction. Do not show force components or answers.]
### Solution 1
The required tension is **10.1 N uphill**, and the normal reaction is **27.6 N perpendicular to the slope**.
Because the block is stationary, the net force is zero in both directions.
**Part a: Tension**
Resolve the weight parallel to the incline:
\[
F=mg\sin\theta
\]
Substitute:
\[
F=3.0\times9.8\times\sin20^\circ
\]
\[
F=29.4\times0.3420
\]
\[
\boxed{F=10.1\ \mathrm{N}}
\]
The tension balances the component of gravity pulling the block downhill.
**Part b: Normal reaction**
Resolve the weight perpendicular to the incline:
\[
N=mg\cos\theta
\]
Substitute:
\[
N=29.4\cos20^\circ
\]
\[
\boxed{N=27.6\ \mathrm{N}}
\]
The normal reaction balances the perpendicular component of weight.
Remember that the tension acts parallel to the incline, so it does not change the normal reaction.
### Question 2
A 4.0 kg block is held stationary on a smooth incline at 28° to the horizontal.
A cable pulls the block uphill at 18° above the surface of the incline.
Calculate:
a. The tension in the cable.
b. The normal reaction.
A student calculates the normal reaction using \(N=mg\cos28^\circ\) and obtains 34.6 N.
Explain why this answer is incorrect for the angled cable.
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: Setup diagram showing a 4.0 kg block on a smooth incline at 28 degrees, with a cable pulling uphill at 18 degrees above the incline. Show the two angles clearly, with the cable pointing away from the surface as well as uphill. Do not show force components or answers.]
### Solution 2
The correct tension is **19.4 N**, and the normal reaction is **28.6 N**. The student's normal reaction is too large because the cable pulls the block away from the surface.
**Part a: Tension**
First calculate the weight:
\[
W=mg=4.0\times9.8=39.2\ \mathrm{N}
\]
The tension is angled, so only its parallel component balances the downhill component of weight.
\[
F\cos18^\circ=mg\sin28^\circ
\]
Rearranging:
\[
F=\frac{39.2\sin28^\circ}{\cos18^\circ}
\]
\[
\boxed{F=19.4\ \mathrm{N}}
\]
**Part b: Normal reaction**
The perpendicular component of tension acts away from the incline.
Therefore:
\[
N+F\sin18^\circ-mg\cos28^\circ=0
\]
Rearranging:
\[
N=mg\cos28^\circ-F\sin18^\circ
\]
Substituting:
\[
N=39.2\cos28^\circ-19.35\sin18^\circ
\]
\[
N=34.61-5.98
\]
\[
\boxed{N=28.6\ \mathrm{N}}
\]
The student's answer of 34.6 N would be correct if the cable ran parallel to the incline.
It is incorrect here because the perpendicular component of tension partially lifts the block away from the surface.
This is why you should always resolve the applied force before calculating the normal reaction.
### Question 3
A 5.0 kg block rests on a rough incline at 25° to the horizontal.
The coefficient of static friction between the block and the incline is 0.50.
A horizontal rope applies a force of 10 N uphill, parallel to the incline.
Determine:
a. The normal reaction.
b. The magnitude and direction of the static friction.
c. Whether the block can remain stationary.
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: Exam-style setup showing a 5.0 kg block on a rough 25-degree incline, with a rope pulling parallel to the slope in the uphill direction. Label the mass, slope angle, and applied 10 N force. Show the rough contact surface. Do not include friction arrows or resolved components.]
### Solution 3
The block can remain stationary. The normal reaction is **44.4 N**, and the actual static friction is **10.7 N acting uphill**.
The key is to distinguish the actual friction from its maximum possible value.
**Part a: Normal reaction**
The rope acts parallel to the surface, so it has no perpendicular component.
Therefore:
\[
N=mg\cos25^\circ
\]
\[
N=5.0\times9.8\times\cos25^\circ
\]
\[
\boxed{N=44.4\ \mathrm{N}}
\]
**Part b: Static friction**
First calculate the component of weight pulling downhill:
\[
W_x=mg\sin25^\circ
\]
\[
W_x=49\sin25^\circ
\]
\[
W_x=20.7\ \mathrm{N}
\]
The rope pulls uphill with 10 N.
Without friction, the block would tend to move downhill because the gravitational component is larger than the applied force.
Static friction must therefore act uphill.
Using equilibrium:
\[
F+f_s-mg\sin\theta=0
\]
Rearranging:
\[
f_s=mg\sin\theta-F
\]
\[
f_s=20.7-10
\]
\[
\boxed{f_s=10.7\ \mathrm{N}}
\]
**Part c: Check whether the block remains stationary**
Calculate the maximum available static friction:
\[
f_{s,\max}=\mu_sN
\]
\[
f_{s,\max}=0.50\times44.4
\]
\[
f_{s,\max}=22.2\ \mathrm{N}
\]
Compare the actual required friction with the maximum:
\[
10.7\ \mathrm{N}<22.2\ \mathrm{N}
\]
Therefore, sufficient static friction is available to keep the block stationary.
\[
\boxed{\text{The block remains stationary.}}
\]
A common mistake is to use 22.2 N as the actual friction.
That value is only the maximum. The block needs just 10.7 N to satisfy equilibrium.
### Question 4
A 6.0 kg block rests on a smooth incline at 30° to the horizontal.
Instead of using a rope parallel to the surface, a student pushes the block horizontally towards the uphill direction.
The block remains stationary.
Calculate:
a. The magnitude of the horizontal applied force.
b. The normal reaction from the incline.
Explain why the normal reaction is greater than \(mg\cos30^\circ\).
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
[IMAGE: Side-view setup diagram showing a 6.0 kg block on a smooth ramp rising towards the right at 30 degrees. A horizontal applied force arrow points rightwards towards the uphill direction, rather than along the slope. Show the horizontal reference and the 30-degree incline angle. Do not show resolved components or answers.]
### Solution 4
The required horizontal force is **33.9 N**, and the normal reaction is **67.9 N**.
The horizontal force has both a parallel component and a perpendicular component. Its perpendicular component pushes the block further into the incline.
**Part a: Horizontal applied force**
Let the horizontal applied force be \(H\).
The angle between the horizontal force and the slope is 30°.
Its parallel component is:
\[
H_x=H\cos30^\circ
\]
For equilibrium along the slope:
\[
H\cos30^\circ=mg\sin30^\circ
\]
Substituting:
\[
H\cos30^\circ=6.0\times9.8\times\sin30^\circ
\]
Rearranging:
\[
H=\frac{58.8\sin30^\circ}{\cos30^\circ}
\]
\[
\boxed{H=33.9\ \mathrm{N}}
\]
**Part b: Normal reaction**
The horizontal force also has a component perpendicular to the surface.
Because the force pushes towards the incline, this component acts into the surface.
Therefore:
\[
N=mg\cos30^\circ+H\sin30^\circ
\]
Substituting:
\[
N=58.8\cos30^\circ+33.9\sin30^\circ
\]
\[
N=50.9+17.0
\]
\[
\boxed{N=67.9\ \mathrm{N}}
\]
The normal reaction is greater than \(mg\cos30^\circ=50.9\ \mathrm{N}\) because the horizontal push adds another perpendicular force into the surface.
This is the opposite effect to a rope pulling away from the incline.
The direction of the applied force matters just as much as its magnitude.
### Question 5
A 4.0 kg block rests on a rough incline at 25° to the horizontal.
The coefficient of static friction is 0.80.
Initially, no rope is attached, and the block remains stationary.
A rope is then attached and pulled at 70° above the surface of the incline. Its uphill component acts parallel to the surface, while its perpendicular component pulls the block away from the incline.
The tension is slowly increased from zero.
Determine:
a. Why the block can initially remain stationary without the rope.
b. The tension at which static equilibrium first becomes impossible.
c. Whether the block begins sliding down the incline or loses contact with the surface first.
Take \(g=9.8\ \mathrm{m\,s^{-2}}\).
Assume the coefficient of static friction remains constant up to the onset of motion, and the rope maintains its direction throughout.
[IMAGE: Spatial setup diagram of a 4.0 kg block on a rough incline rising to the right at 25 degrees. Attach a rope to the block, directed steeply upwards and away from the incline at 70 degrees above the surface. Show the rope's uphill component direction and its large outward inclination through the actual rope geometry, but do not draw resolved force components, friction arrows, or answers.]
### Solution 5
The block initially remains stationary because static friction is sufficient to balance gravity. However, as the rope tension increases, the block eventually becomes less stable and begins to slide **down the incline at a tension of approximately 28.9 N**.
It starts sliding before losing contact with the surface.
This is the most interesting case in the set because increasing an uphill pull does not necessarily make a block more secure.
**Part a: Check the initial equilibrium**
With no rope attached, the only forces parallel to the incline are the downhill component of weight and static friction.
The downhill component is:
\[
W_x=mg\sin25^\circ
\]
\[
W_x=39.2\sin25^\circ
\]
\[
W_x=16.6\ \mathrm{N}
\]
The initial normal reaction is:
\[
N=mg\cos25^\circ
\]
\[
N=39.2\cos25^\circ
\]
\[
N=35.5\ \mathrm{N}
\]
Calculate the maximum static friction:
\[
f_{s,\max}=\mu_sN
\]
\[
f_{s,\max}=0.80\times35.5
\]
\[
f_{s,\max}=28.4\ \mathrm{N}
\]
The block requires only 16.6 N of uphill friction to remain stationary.
Since:
\[
16.6<28.4
\]
the block can remain stationary without a rope.
**Part b: Find the tension at which equilibrium first becomes impossible**
This is where both force equations must be used together.
As the rope pulls away from the surface, the normal reaction decreases.
The normal reaction is:
\[
N=mg\cos\theta-T\sin\alpha
\]
where:
- \(T\) is the rope tension.
- \(\alpha=70^\circ\) is the angle between the rope and the incline.
- \(\theta=25^\circ\) is the incline angle.
Substituting the values:
\[
N=39.2\cos25^\circ-T\sin70^\circ
\]
The block is initially stationary, and the rope is not pulling strongly enough to make it tend uphill.
As the tension increases, the limiting situation occurs when the block is just about to slide downhill.
Static friction therefore acts uphill at its maximum value:
\[
f_s=\mu_sN
\]
Now apply equilibrium parallel to the incline:
\[
T\cos70^\circ+f_s=mg\sin25^\circ
\]
Substitute \(f_s=\mu_sN\):
\[
T\cos70^\circ+\mu_sN=mg\sin25^\circ
\]
Replace \(N\) with its expression:
\[
T\cos70^\circ+
0.80(39.2\cos25^\circ-T\sin70^\circ)
=39.2\sin25^\circ
\]
Collect the terms containing \(T\):
\[
T(\cos70^\circ-0.80\sin70^\circ)
=
39.2(\sin25^\circ-0.80\cos25^\circ)
\]
Rearranging:
\[
T=
\frac{39.2(0.80\cos25^\circ-\sin25^\circ)}
{0.80\sin70^\circ-\cos70^\circ}
\]
Therefore:
\[
\boxed{T=28.9\ \mathrm{N}}
\]
At approximately 28.9 N, the required static friction reaches its maximum.
We can check the normal reaction at this tension:
\[
N=39.2\cos25^\circ-28.9\sin70^\circ
\]
\[
\boxed{N\approx8.34\ \mathrm{N}}
\]
The normal reaction is still positive, so the block remains in contact with the incline at the point where it starts to slip.
[IMAGE: Explanatory free-body diagram for Question 5 at the threshold of downward slipping. Show the block on the 25-degree incline, weight mg vertically down, normal reaction N perpendicular outwards, rope tension T at 70 degrees above the plane, and limiting static friction mu_s N acting uphill. Also show the resolved tension components T cos(alpha) uphill and T sin(alpha) away from the surface. Label the positive axes and preserve the correct angles.]
**Part c: Does the block slide or lose contact first?**
We have already established that static equilibrium fails at:
\[
T=28.9\ \mathrm{N}
\]
Now calculate the tension at which the block would lose contact with the surface.
Loss of contact occurs when:
\[
N=0
\]
Using the perpendicular force equation:
\[
mg\cos25^\circ-T\sin70^\circ=0
\]
Rearranging:
\[
T=\frac{mg\cos25^\circ}{\sin70^\circ}
\]
Substituting:
\[
T=\frac{39.2\cos25^\circ}{\sin70^\circ}
\]
\[
\boxed{T=37.8\ \mathrm{N}}
\]
Compare the two critical tensions:
| Event | Critical tension |
|---|---:|
| Static equilibrium becomes impossible | 28.9 N |
| Block loses contact with the incline | 37.8 N |
The block starts sliding downhill first.
**Why does increasing the tension make the situation worse?**
This is the subtle part.
The rope has two competing effects:
- Its uphill component, \(T\cos70^\circ\), helps prevent downhill sliding.
- Its outward component, \(T\sin70^\circ\), reduces the normal reaction and therefore reduces the maximum available static friction.
In this particular arrangement, the second effect is stronger.
The rope removes the contact force that was helping hold the block in place. Although the rope pulls uphill, it reduces the block's available friction even faster.
A student who assumes that any uphill pull must make the block more stable misses this interaction.
The important lesson is that you cannot assess the effect of an applied force by looking at its parallel component alone. You must also consider how its perpendicular component changes the normal reaction and, consequently, the friction limit.
## Where to go next
Once a block begins sliding, the problem changes from static equilibrium to acceleration.
You can no longer set the resultant force parallel to the incline equal to zero. Instead, use Newton's second law:
\[
\sum F_x=ma
\]
You will also need to distinguish static friction, which acts before slipping, from kinetic friction, which acts during sliding.
After calculating the acceleration, a useful next step is understanding how the block's velocity changes over time. The [tutorgum guide to velocity-time graphs](https://tutorgum.com/learn/resources/hsc-physics-velocity-time-graphs/) covers that next part of the motion problem.