How to Check Whether Static Friction Can Hold a Pulley System
Learn how to determine whether static friction can keep a connected pulley system stationary. Follow worked examples covering horizontal tables, inclined planes, angled ropes, and systems with additional forces.
Attach a block to a hanging mass using a rope that passes over a pulley. The hanging mass has a downward weight, so you might predict that it must fall and pull the block across the table.
But the table can resist that pull through static friction. If the friction is strong enough, both masses can remain stationary.
The key test is whether the friction required to keep the system in equilibrium is less than or equal to the maximum available static friction.
For the standard arrangement, with a horizontal rope, a rough horizontal table, and an ideal pulley, the condition is:
\[ \boxed{m_B \leq \mu_s m_A} \]
Here, \(m_A\) is the mass on the table, \(m_B\) is the hanging mass, and \(\mu_s\) is the coefficient of static friction.
The important part is how we arrive at this condition. You must calculate the friction actually required by the system before comparing it with the maximum friction available.
01Understand the pulley system first
Consider two objects connected by a rope:
- Block A sits on a rough horizontal table.
- Block B hangs vertically over the edge.
- A rope connects the two blocks and passes over a pulley.
There are two competing effects.
The weight of block B pulls it downwards. Because the rope is connected, this also pulls block A towards the pulley.
Static friction between block A and the table acts against the tendency of A to slide. If the friction is sufficient, neither block moves.
Before calculating anything, notice that the two blocks cannot be analysed as though they were independent objects. The rope connects their motion and their forces.
Assumptions about the rope and pulley
For the standard HSC model, we make the following idealisations:
| Component | Assumption | Consequence |
|---|---|---|
| Rope | Massless | Tension is uniform along a straight rope segment. |
| Rope | Inextensible | Both ends move through equal distances. |
| Pulley | Massless | Its rotational inertia is ignored. |
| Pulley | Frictionless axle | Tension has the same magnitude on both sides. |
| Rope and pulley | No slipping | The rope follows the pulley without relative motion. |
| Table | Horizontal | The normal force is \(m_Ag\) if the rope pulls horizontally and there are no other vertical forces. |
The tension, \(T\), is the pulling force transmitted through the rope.
Because the rope and pulley are ideal, the tension has the same magnitude on both sides of the pulley.
This does not mean that the forces point in the same direction. The horizontal rope segment pulls A towards the pulley, while the vertical segment pulls B upwards.
The rope’s inextensibility also means that if the system starts moving, the two blocks have equal acceleration magnitudes. For a stationary system, both accelerations are zero.
02Draw a separate free-body diagram for each block
This is the most important step in solving pulley problems.
Do not begin by treating the entire arrangement as one object. Isolate each block and identify every force acting on it.
Forces on block A
Block A experiences four forces:
- Weight, \(m_Ag\), acting downwards.
- Normal force, \(N\), acting upwards.
- Tension, \(T\), pulling towards the pulley.
- Static friction, \(f_s\), acting away from the pulley.
The friction acts away from the pulley because block A would otherwise tend to slide towards it.
If block A remains stationary, its horizontal and vertical forces must balance.
Taking right as positive:
\[ \sum F_x=0 \]
\[ T-f_s=0 \]
Therefore:
\[ \boxed{f_s=T} \]
This tells us that the actual static friction must equal the rope tension.
Forces on block B
Block B experiences two forces:
- Weight, \(m_Bg\), acting downwards.
- Tension, \(T\), acting upwards.
Taking downwards as positive:
\[ \sum F_y=0 \]
\[ m_Bg-T=0 \]
Therefore:
\[ \boxed{T=m_Bg} \]
We can now combine the two results:
\[ f_s=T=m_Bg \]
This is the key relationship.
The actual static friction required on block A equals the weight of the hanging block.
Notice that we have not yet used the coefficient of static friction. We first need to establish how much friction the system requires.
What about the normal force?
For block A, the rope pulls horizontally, so it has no vertical component.
The vertical forces therefore balance:
\[ N-m_Ag=0 \]
Hence:
\[ \boxed{N=m_Ag} \]
The normal force is equal to the weight of block A under these particular assumptions.
This result will change if the rope pulls at an angle or the table is inclined.
03Compare the required friction with the maximum available friction
Static friction adjusts to whatever magnitude is needed to prevent relative sliding, provided it does not exceed its maximum.
The maximum static friction is:
\[ f_{s,\max}=\mu_sN \]
Here:
- \(f_{s,\max}\) is the maximum static friction, 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).
For block A:
\[ N=m_Ag \]
Therefore:
\[ f_{s,\max}=\mu_sm_Ag \]
We already established that the friction required is:
\[ f_s=m_Bg \]
For the system to remain stationary, the required friction cannot exceed the maximum available friction.
\[ m_Bg\leq\mu_sm_Ag \]
Cancel \(g\):
\[ \boxed{m_B\leq\mu_sm_A} \]
This is the condition for static equilibrium in the standard horizontal pulley arrangement.
Three possible outcomes
| Comparison | What happens? |
|---|---|
| \(m_B<\mu_sm_A\) | Static friction can hold the system stationary. |
| \(m_B=\mu_sm_A\) | The system is in limiting equilibrium, just about to slip. |
| \(m_B>\mu_sm_A\) | Static friction is insufficient to maintain equilibrium. |
There is a common mistake worth avoiding here.
You might see \(\mu_sN\) and immediately use it as the actual friction force. That is incorrect unless the system is just about to slip.
For example, suppose the maximum static friction is 30 N, but the rope only pulls with 18 N. The actual friction is 18 N, not 30 N.
The maximum is a limit, not a force that must always act.
If you need to refresh this distinction, see Static and Kinetic Friction Explained for HSC Physics and Why Static Friction Isn’t Always at Its Maximum.
04Internal and external forces in a pulley system
There is another detail that can cause confusion in more complicated pulley questions: the difference between internal and external forces.
Whether a force is internal or external depends on which objects you choose to include in your system.
Consider these different choices:
| Chosen system | Is tension external? | Explanation |
|---|---|---|
| Block A alone | Yes | The rope pulls on block A from outside the system. |
| Block B alone | Yes | The rope pulls on block B from outside the system. |
| Both blocks, excluding the rope | Yes | The rope acts externally on both blocks. |
| Both blocks and the rope | No, for rope-block interactions | The interactions between the rope and blocks are internal to this larger system. |
The forces exerted by the rope on blocks A and B are not a Newton’s third-law pair.
The reaction to the rope pulling block A is block A pulling on the rope. It is not the rope pulling on block B.
Also, the tension forces on A and B point in different directions. You cannot simply cancel them because their magnitudes are equal.
If your system includes the rope, the interactions between the rope and blocks can be treated as internal. However, the table’s friction and normal forces, gravity, and the interaction between the rope and the pulley remain external if the table, Earth, and pulley are outside your chosen system.
The pulley itself is held in place by a support reaction at its axle. This balances the resultant force exerted on the pulley by the rope.
For determining whether friction can hold the system, separate free-body diagrams are usually the safest approach. They make the tension and friction requirements on each object clear.
05Worked example: A block on a horizontal table
A 7.0 kg block A rests on a horizontal table. It is connected over a fixed ideal pulley to a hanging block B of mass 1.5 kg.
The coefficient of static friction between block A and the table is 0.30.
Determine whether the system can remain stationary. If it can, calculate the actual static-friction force.
Take \(g=9.8\text{ m s}^{-2}\).
Step 1
For block B, the downward weight must balance the upward tension.
\[ T=m_Bg \]
Substitute:
\[ T=1.5\times9.8 \]
\[ \boxed{T=14.7\text{ N}} \]
Step 2
There are no other vertical forces acting on A, so:
\[ N=m_Ag \]
\[ N=7.0\times9.8 \]
\[ \boxed{N=68.6\text{ N}} \]
Step 3
\[ f_{s,\max}=\mu_sN \]
Substitute:
\[ f_{s,\max}=0.30\times68.6 \]
\[ \boxed{f_{s,\max}=20.58\text{ N}} \]
Step 4
The actual friction required is equal to the tension:
\[ f_s=T=14.7\text{ N} \]
Compare:
\[ 14.7<20.58 \]
Therefore, static friction is sufficient to hold the system stationary.
The actual friction force is:
\[ \boxed{f_s=14.7\text{ N}} \]
It acts away from the pulley.
Notice that the maximum possible friction is 20.58 N, but the system only requires 14.7 N.
06Worked example: A block on an inclined plane
The previous example was relatively straightforward because the table was horizontal.
Now suppose block A rests on an inclined plane. The rope runs parallel to the slope and passes over a pulley at the top.
An 8.0 kg block A rests on a rough \(30^\circ\) incline. It is connected to a hanging 3.0 kg block B.
The coefficient of static friction between A and the incline is 0.35.
Determine whether static friction can keep the system stationary, and state the direction and magnitude of the friction force.
Take \(g=9.8\text{ m s}^{-2}\).
Step 1
The component of block A’s weight parallel to the incline is:
\[ F_{\parallel}=m_Ag\sin30^\circ \]
\[ F_{\parallel}=8.0\times9.8\times0.5 \]
\[ F_{\parallel}=39.2\text{ N} \]
The weight of block B is:
\[ W_B=m_Bg \]
\[ W_B=3.0\times9.8 \]
\[ W_B=29.4\text{ N} \]
Block A has a downslope gravitational component of 39.2 N, while block B has a downward weight of 29.4 N.
Without friction, A would tend to move down the incline, pulling B upwards.
Therefore, static friction on A must act up the incline.
Step 2
For stationary block B:
\[ T=m_Bg \]
\[ \boxed{T=29.4\text{ N}} \]
Step 3
The normal force balances the component of weight perpendicular to the incline.
\[ N=m_Ag\cos30^\circ \]
\[ N=8.0\times9.8\times\cos30^\circ \]
\[ \boxed{N=67.9\text{ N}} \]
Step 4
Taking up the incline as positive:
\[ T+f_s-m_Ag\sin30^\circ=0 \]
Rearrange:
\[ f_s=m_Ag\sin30^\circ-T \]
Substitute:
\[ f_s=39.2-29.4 \]
\[ \boxed{f_s=9.8\text{ N}} \]
The friction acts up the incline.
Step 5
\[ f_{s,\max}=\mu_sN \]
\[ f_{s,\max}=0.35\times67.9 \]
\[ \boxed{f_{s,\max}=23.8\text{ N}} \]
Since:
\[ 9.8<23.8 \]
static friction is sufficient.
The system can remain stationary, with an actual friction force of 9.8 N acting up the incline.
A useful contrast: What if the hanging mass were heavier?
Suppose block B had a mass of 5.0 kg instead of 3.0 kg.
Its weight would now be:
\[ T=5.0\times9.8=49.0\text{ N} \]
This is greater than the 39.2 N downslope component of A’s weight.
The system would now tend to move in the opposite direction: B downwards and A up the incline.
Consequently, static friction on A would act down the incline, opposing A’s tendency to move upwards.
The required friction would be:
\[ f_s=49.0-39.2 \]
\[ \boxed{f_s=9.8\text{ N down the incline}} \]
The maximum friction remains 23.8 N, so the system could still remain stationary.
This is an important point: friction does not always act opposite to tension. It acts opposite to the tendency of the contacting surfaces to slide.
07A quick comparison of different pulley geometries
The simple condition \(m_B\leq\mu_sm_A\) is useful, but it only applies to the standard horizontal arrangement.
The geometry determines both the required friction and the normal force.
| Arrangement | Required friction | Normal force |
|---|---|---|
| Horizontal table, horizontal rope | \(m_Bg\) | \(m_Ag\) |
| Horizontal table, rope at angle \(\theta\) above horizontal | \(T\cos\theta\) | \(m_Ag-T\sin\theta\) |
| Inclined plane, rope parallel to slope | \(\left|m_Bg-m_Ag\sin\theta\right|\) | \(m_Ag\cos\theta\) |
These expressions assume the arrangements shown, ideal rope and pulley conditions, and no additional forces.
For a rope angled above the horizontal, the upward component of tension reduces the normal force. This reduces the maximum available static friction.
For an inclined plane, only the component of weight parallel to the surface contributes to the tendency to slide down the slope.
Always calculate the normal force from the perpendicular forces before using \(f_{s,\max}=\mu_sN\).
08How could you test this model experimentally?
The same reasoning can be used to design a simple experiment.
The question is whether the maximum hanging mass that can remain stationary increases in proportion to the mass of the block on the table.
The apparatus
Place a block on a rough horizontal bench. Connect it to a mass hanger using a light rope that passes over a low-friction pulley.
Add small masses to the hanger until the block is just about to move.
The total hanging mass includes the hanger itself.
What should happen?
At the threshold of movement, the static friction has reached its maximum.
Therefore:
\[ m_{B,\text{crit}}g=\mu_sm_Ag \]
Rearranging:
\[ \boxed{m_{B,\text{crit}}=\mu_sm_A} \]
For example, suppose the block on the table has a mass of 4.0 kg and the coefficient of static friction is 0.30.
The predicted critical hanging mass is:
\[ m_{B,\text{crit}}=0.30\times4.0 \]
\[ \boxed{m_{B,\text{crit}}=1.2\text{ kg}} \]
The model predicts that a hanging mass of approximately 1.2 kg will bring the block to the threshold of slipping.
If you repeat the experiment using different masses for the table block, the critical hanging mass should increase proportionally.
What could go wrong with the original design?
A real pulley is not perfectly frictionless. Resistance in its bearing can create a difference between the tensions on either side.
This means the tension pulling the table block may not equal the hanging mass’s weight. The experiment would then no longer isolate the table’s static friction as intended.
A better design would use:
- A low-friction pulley.
- A light, taut rope.
- A horizontal rope segment between the block and pulley.
- Small, consistent increases in the hanging mass.
- Repeated trials to identify the onset of movement more reliably.
These changes make the experimental arrangement closer to the ideal model.
The predicted relationship is a model result, not a guarantee that real measurements will fall perfectly on a straight line.
09Questions and solutions
Use the following questions to practise applying the method to different pulley arrangements. Pay particular attention to the direction of friction and the normal force.
Question 1
A 5.0 kg block A rests on a rough horizontal table. It is connected over an ideal pulley to a hanging block B of mass 1.5 kg.
The coefficient of static friction between A and the table is 0.40.
Determine:
a) The tension in the rope if the system remains stationary.
b) The maximum available static friction.
c) Whether static friction can hold the system stationary.
d) The actual friction force acting on A.
Take \(g=9.8\text{ m s}^{-2}\).
Solution 1
Yes, the system can remain stationary. The actual friction required is 14.7 N, which is below the maximum available friction of 19.6 N.
Step 1: Calculate the tension using block B.
For equilibrium:
\[ T=m_Bg \]
\[ T=1.5\times9.8 \]
\[ \boxed{T=14.7\text{ N}} \]
Step 2: Calculate the normal force on A.
Since the rope pulls horizontally:
\[ N=m_Ag \]
\[ N=5.0\times9.8 \]
\[ \boxed{N=49.0\text{ N}} \]
Step 3: Calculate the maximum static friction.
\[ f_{s,\max}=\mu_sN \]
\[ f_{s,\max}=0.40\times49.0 \]
\[ \boxed{f_{s,\max}=19.6\text{ N}} \]
Step 4: Compare the forces.
The actual friction required is equal to the tension:
\[ f_s=T=14.7\text{ N} \]
Since:
\[ 14.7<19.6 \]
static friction is sufficient.
Therefore:
\[ \boxed{f_s=14.7\text{ N}} \]
It acts away from the pulley.
The block does not require the full 19.6 N of available static friction.
Question 2
A 6.0 kg block A rests on a rough horizontal table. It is connected over an ideal pulley to a hanging block B of mass 1.0 kg.
The coefficient of static friction between A and the table is 0.35.
A separate horizontal force of 25 N is applied to block A, pushing it away from the pulley.
Determine:
a) The tension in the rope if the system remains stationary.
b) The direction and magnitude of the static-friction force on A.
c) Whether static friction can hold the system stationary.
Take \(g=9.8\text{ m s}^{-2}\).
Solution 2
Yes, the system can remain stationary. Static friction acts towards the pulley with a magnitude of 15.2 N.
This question tests an important idea: friction does not necessarily oppose the tension.
Step 1: Calculate the tension.
For stationary block B:
\[ T=m_Bg \]
\[ T=1.0\times9.8 \]
\[ \boxed{T=9.8\text{ N}} \]
The rope pulls A towards the pulley with a force of 9.8 N.
Step 2: Determine the tendency of block A to move.
There are two horizontal forces before considering friction:
- Tension: 9.8 N towards the pulley.
- Applied force: 25 N away from the pulley.
The applied force is larger.
The block therefore tends to move away from the pulley, meaning friction must act towards the pulley.
Step 3: Calculate the friction required.
Taking towards the pulley as positive:
\[ T+f_s-25=0 \]
Rearrange:
\[ f_s=25-T \]
\[ f_s=25-9.8 \]
\[ \boxed{f_s=15.2\text{ N}} \]
Step 4: Calculate the maximum static friction.
The normal force is:
\[ N=m_Ag \]
\[ N=6.0\times9.8=58.8\text{ N} \]
Therefore:
\[ f_{s,\max}=0.35\times58.8 \]
\[ \boxed{f_{s,\max}=20.58\text{ N}} \]
Step 5: Check whether equilibrium is possible.
\[ 15.2<20.58 \]
Static friction can provide the required 15.2 N.
The system therefore remains stationary under the ideal assumptions.
Notice that friction acts in the same direction as tension in this example. It is opposing the tendency created by the larger external force, not automatically opposing the rope’s pull.
Question 3
A 4.0 kg block A rests on a rough horizontal table and is connected over an ideal pulley to a hanging block B of mass 1.0 kg.
The rope segment between A and the pulley makes an angle of \(35^\circ\) above the horizontal.
The coefficient of static friction between A and the table is 0.22.
Determine whether the system can remain stationary.
Take \(g=9.8\text{ m s}^{-2}\).
Solution 3
No, the system cannot remain stationary. The required friction is approximately 8.03 N, but the maximum available static friction is only approximately 7.39 N.
The important difference from the previous questions is that tension now has a vertical component.
Step 1: Calculate the tension.
For stationary block B:
\[ T=m_Bg \]
\[ T=1.0\times9.8 \]
\[ \boxed{T=9.8\text{ N}} \]
Step 2: Resolve the tension into components.
The rope is at \(35^\circ\) above the horizontal.
Its horizontal component is:
\[ T_x=T\cos35^\circ \]
\[ T_x=9.8\cos35^\circ \]
\[ \boxed{T_x=8.03\text{ N}} \]
Its vertical component is:
\[ T_y=T\sin35^\circ \]
\[ T_y=9.8\sin35^\circ \]
\[ \boxed{T_y=5.62\text{ N}} \]
The horizontal component tries to pull A towards the pulley. Therefore, friction must act away from the pulley.
The vertical component pulls A upwards, reducing the normal force.
Step 3: Calculate the normal force.
Taking upwards as positive:
\[ N+T\sin35^\circ-m_Ag=0 \]
Therefore:
\[ N=m_Ag-T\sin35^\circ \]
Substitute:
\[ N=(4.0)(9.8)-5.62 \]
\[ \boxed{N=33.58\text{ N}} \]
Step 4: Calculate the maximum static friction.
\[ f_{s,\max}=\mu_sN \]
\[ f_{s,\max}=0.22\times33.58 \]
\[ \boxed{f_{s,\max}=7.39\text{ N}} \]
Step 5: Compare the required friction with the maximum.
For horizontal equilibrium, friction must balance the horizontal component of tension:
\[ f_s=T\cos35^\circ \]
\[ f_s=8.03\text{ N} \]
But:
\[ 8.03>7.39 \]
Therefore, static friction cannot hold the system stationary.
The system cannot remain in static equilibrium under the stated assumptions.
A particularly tempting mistake here is to calculate the normal force as \(m_Ag=39.2\text{ N}\). That would give a maximum friction of 8.62 N and incorrectly suggest that the system could remain stationary.
The upward component of tension changes the normal force, so it must be included.
Question 4
A 10.0 kg block A rests on a rough horizontal table. It is connected over an ideal pulley to a hanging block B of variable mass \(m_B\).
The coefficient of static friction between A and the table is 0.30.
A constant horizontal force of 40.0 N pushes block A away from the pulley.
Determine the full range of hanging masses for which the system can remain stationary.
Also state the direction of friction at each limiting mass.
Take \(g=9.8\text{ m s}^{-2}\).
Solution 4
The system can remain stationary when the hanging mass is between approximately 1.08 kg and 7.08 kg, inclusive in the ideal limiting-equilibrium model.
Unlike the previous questions, we cannot assume that friction acts in only one direction throughout the entire range.
For a sufficiently small hanging mass, the external 40 N force dominates. For a sufficiently large hanging mass, the rope tension dominates.
Friction can oppose either tendency.
Step 1: Calculate the normal force and maximum static friction.
There are no vertical forces other than the weight and normal force on A.
\[ N=m_Ag \]
\[ N=10.0\times9.8 \]
\[ N=98.0\text{ N} \]
The maximum static friction is:
\[ f_{s,\max}=\mu_sN \]
\[ f_{s,\max}=0.30\times98.0 \]
\[ \boxed{f_{s,\max}=29.4\text{ N}} \]
Step 2: Express the tension in terms of the hanging mass.
For stationary block B:
\[ T=m_Bg \]
\[ \boxed{T=9.8m_B} \]
Step 3: Write the horizontal equilibrium equation for A.
Take towards the pulley as positive.
The forces on A are:
- Tension, \(T\), towards the pulley.
- Applied force, 40.0 N, away from the pulley.
- Static friction, \(f_s\), which may act in either direction.
Therefore:
\[ T-40.0+f_s=0 \]
Rearranging:
\[ f_s=40.0-T \]
Substitute the tension:
\[ f_s=40.0-9.8m_B \]
Step 4: Apply the static-friction limit.
The magnitude of actual static friction cannot exceed 29.4 N.
\[ |f_s|\leq29.4 \]
Therefore:
\[ |40.0-9.8m_B|\leq29.4 \]
This gives two inequalities:
\[ -29.4\leq40.0-9.8m_B \]
and
\[ 40.0-9.8m_B\leq29.4 \]
Rearranging the first:
\[ 9.8m_B\leq69.4 \]
\[ m_B\leq7.08\text{ kg} \]
Rearranging the second:
\[ 9.8m_B\geq10.6 \]
\[ m_B\geq1.08\text{ kg} \]
Combining both conditions:
\[ \boxed{1.08\leq m_B\leq7.08\text{ kg}} \]
Step 5: Determine the friction direction at each limit.
At the lower limit:
\[ m_B=1.08\text{ kg} \]
The tension is approximately 10.6 N, while the applied force is 40.0 N.
The block tends to move away from the pulley, so friction acts towards the pulley.
Its magnitude is 29.4 N, the maximum available friction.
At the upper limit:
\[ m_B=7.08\text{ kg} \]
The tension is approximately 69.4 N.
Now the tension dominates, so the block tends towards the pulley. Friction acts away from the pulley, again with a magnitude of 29.4 N.
Between these two limits, static friction adjusts to the force imbalance.
At approximately:
\[ m_B=\frac{40.0}{9.8}=4.08\text{ kg} \]
the tension exactly balances the applied force, so no horizontal friction is required.
This question shows why static friction must be treated as a force that can act in either direction. It is not permanently assigned to oppose tension.
Question 5
Two blocks, A and B, rest on opposite rough inclined planes. They are connected by one continuous light, inextensible rope passing over a fixed ideal pulley at the top.
Both rope segments are parallel to their respective inclined planes.
The following information is provided:
| Quantity | Block A | Block B |
|---|---|---|
| Mass | 8.0 kg | 5.0 kg |
| Incline angle | \(35^\circ\) | \(20^\circ\) |
| Coefficient of static friction | 0.20 | 0.25 |
Take \(g=9.8\text{ m s}^{-2}\).
A student argues that the two blocks can be stationary because each block has its own static friction force, which can adjust independently.
Determine:
a) Whether static equilibrium is possible with the given coefficients of friction.
b) The minimum coefficient of static friction required between A and its incline to make equilibrium possible, keeping everything else unchanged.
c) If the coefficient for A is increased to 0.27, find one possible rope tension and the corresponding friction forces on both blocks.
Explain why the same tension must be used when analysing both blocks.
Solution 5
No, the system cannot remain stationary with the original coefficients. The minimum coefficient required for block A is approximately 0.260.
The important constraint is that the rope has one continuous tension. Each block may have its own friction force, but both blocks must be in equilibrium using the same tension magnitude.
This is where analysing each block independently can lead to an incorrect conclusion.
Step 1: Calculate the weight components parallel to both inclines.
For block A:
\[ W_{A,\parallel}=m_Ag\sin35^\circ \]
\[ W_{A,\parallel}=8.0\times9.8\times\sin35^\circ \]
\[ \boxed{W_{A,\parallel}=44.97\text{ N}} \]
For block B:
\[ W_{B,\parallel}=m_Bg\sin20^\circ \]
\[ W_{B,\parallel}=5.0\times9.8\times\sin20^\circ \]
\[ \boxed{W_{B,\parallel}=16.76\text{ N}} \]
These components act down their respective slopes, away from the pulley.
Step 2: Calculate the normal forces.
For block A:
\[ N_A=m_Ag\cos35^\circ \]
\[ N_A=8.0\times9.8\times\cos35^\circ \]
\[ \boxed{N_A=64.22\text{ N}} \]
For block B:
\[ N_B=m_Bg\cos20^\circ \]
\[ N_B=5.0\times9.8\times\cos20^\circ \]
\[ \boxed{N_B=46.04\text{ N}} \]
Step 3: Calculate the maximum static friction on each block.
For A:
\[ f_{A,\max}=\mu_AN_A \]
\[ f_{A,\max}=0.20\times64.22 \]
\[ \boxed{f_{A,\max}=12.84\text{ N}} \]
For B:
\[ f_{B,\max}=\mu_BN_B \]
\[ f_{B,\max}=0.25\times46.04 \]
\[ \boxed{f_{B,\max}=11.51\text{ N}} \]
Step 4: Find the range of tensions that can hold block A stationary.
Take up the incline, towards the pulley, as positive.
For block A:
\[ T+f_A-W_{A,\parallel}=0 \]
Rearranging:
\[ f_A=W_{A,\parallel}-T \]
Static equilibrium requires:
\[ |f_A|\leq f_{A,\max} \]
Therefore:
\[ |44.97-T|\leq12.84 \]
This gives:
\[ \boxed{32.12\leq T\leq57.81\text{ N}} \]
Block A can remain stationary only if the rope tension lies within this range.
Step 5: Find the range of tensions that can hold block B stationary.
Using the same positive direction, up the incline towards the pulley:
\[ T+f_B-W_{B,\parallel}=0 \]
Therefore:
\[ f_B=W_{B,\parallel}-T \]
Applying the static-friction condition:
\[ |16.76-T|\leq11.51 \]
This gives:
\[ \boxed{5.25\leq T\leq28.27\text{ N}} \]
Step 6: Check whether the two tension ranges overlap.
For the entire system to remain stationary, both blocks must be in equilibrium simultaneously.
Therefore, the tension must satisfy both conditions:
\[ 32.12\leq T\leq57.81 \]
and
\[ 5.25\leq T\leq28.27 \]
There is no overlap.
The minimum tension required to hold A is greater than the maximum tension that B can accommodate.
Therefore:
\[ \boxed{\text{Static equilibrium is impossible.}} \]
The system tends to move with A down its incline and B up its incline.
The student’s mistake is assuming that two independently calculated friction forces automatically produce a valid connected system. They do not. Both blocks are connected by the same rope, so their equilibrium equations must agree on one common tension.
Step 7: Calculate the minimum coefficient required for A.
We now increase \(\mu_A\), keeping the other values unchanged.
For equilibrium to become possible, the lowest tension allowed by block A must be no greater than the highest tension allowed by block B.
Therefore:
\[ W_{A,\parallel}-\mu_AN_A \leq W_{B,\parallel}+f_{B,\max} \]
Rearrange to isolate \(\mu_A\):
\[ \mu_A\geq \frac{W_{A,\parallel}-W_{B,\parallel}-f_{B,\max}} {N_A} \]
Substitute:
\[ \mu_A\geq \frac{44.97-16.76-11.51}{64.22} \]
\[ \mu_A\geq0.260 \]
Hence:
\[ \boxed{\mu_{A,\min}\approx0.260} \]
At this exact limiting value, both blocks can be stationary with a single common tension of approximately 28.27 N. Both friction forces are at their respective maximum magnitudes.
Step 8: Find one possible equilibrium when \(\mu_A=0.27\).
First, calculate A’s new maximum static friction:
\[ f_{A,\max}=0.27\times64.22 \]
\[ f_{A,\max}=17.34\text{ N} \]
The allowable tension range for A is now:
\[ 44.97-17.34\leq T\leq44.97+17.34 \]
\[ 27.63\leq T\leq62.31 \]
Block B still requires:
\[ 5.25\leq T\leq28.27 \]
The two ranges now overlap between approximately 27.63 N and 28.27 N.
Choose a possible tension:
\[ \boxed{T=28.0\text{ N}} \]
For block A:
\[ f_A=44.97-28.0 \]
\[ \boxed{f_A=16.97\text{ N up the incline}} \]
This is below its maximum of 17.34 N.
For block B:
\[ f_B=16.76-28.0 \]
\[ f_B=-11.24\text{ N} \]
The negative sign means the friction acts down the incline.
Therefore:
\[ \boxed{f_B=11.24\text{ N down the incline}} \]
This is below its maximum of 11.51 N.
Both blocks can now remain stationary with the same 28.0 N rope tension.
The deeper lesson is that static equilibrium in a connected system is a simultaneous constraint problem.
Every object must satisfy its own force equations, every friction force must remain within its available limit, and the rope tension must be consistent throughout the ideal rope.
09What to do when static friction cannot hold
Once the required friction exceeds the maximum available value, the system cannot remain stationary under the assumptions of the model.
At that point, you need to analyse the motion rather than continue using static equilibrium.
The next step is to use kinetic friction:
\[ f_k=\mu_kN \]
Then apply Newton’s second law separately to each block:
\[ \sum F=ma \]
Remember that the two blocks have equal acceleration magnitudes because the rope is inextensible, although their acceleration directions depend on the chosen coordinate axes.
Most importantly, do not automatically substitute maximum static friction into an acceleration calculation. Once the surfaces slide, you must use the kinetic-friction model.
That is the natural next step after deciding whether static friction can hold a pulley system still.