What is the maximum tension and friction force for a block held on an incline?

[brotherbobby]

Homework Statement

A block of mass 25 kg is held in place on an inclined plane of angle $30^{\circ}$ as shown in the diagram below. The coefficient of static friction is 0.6. (a) What is the tension in the string? (b) If the string is cut, what is the acceleration of the block? $\mathbf{Textbook\;\;answers : (a) T = 122.5\; \text{N}\; (b) Zero.}$

Relevant Equations

Newton's law : $\Sigma F_x = ma_x$ and maximum static frictional force $f_S^{max} = \mu R$, where $R$ is the reaction normal to the surface.

(a) Ridiculously simple though it looks, I can't see how the string will be tight. One of the two has to be true.

(1) The static friction $f_S = mg \sin\theta = 25\times 10\times \sin 30^{\circ} = 122.5\; \text{N}$. The maximum static friction $f_S = \mu mg \cos \theta = 0.6\times 25\times 10\times \cos 30^{\circ} \approx 130\;\text{N}$. Hence if friction is holding the block in place, the rope is slack and the tension in the rope $\boxed{T=0}$. (This is what I believe will happen).

(2) The string was tied to the block (somehow) and the string-block system placed on the incline. In that case there is no tendency of relative motion between the block and the incline so static friction is zero. The rope is tight and it balances the component of gravity along the incline : $\boxed{T = 122.5\;\text{N}}$. (This is the answer in the book as I mentioned earlier).(b) If the string is cut, friction will keep the block in place and therefore its acceleration $\boxed{a=0}$. (Agrees with the book).

Any help?

 

[haruspex]

There is no guarantee which way the static friction will act. Depending on how the arrangement was created, the tension will be anything from 0 to $mg(\sin(\theta)+\mu\cos(\theta))$

 

[brotherbobby]

There is no guarantee which way the static friction will act. Depending on how the arrangement was created, the tension will be anything from 0 to $mg(\sin(\theta)+\mu\cos(\theta))$

Which makes it vexedly difficult. Can you elaborate on the phrase "depending on how the arrangement was created"?

Imagine one arrangement, which I speak about in (1) above. The block was first carried and kept on the incline - clearly it will remain at rest ($f_S \leq f_S^{max}$). The string was then tied to the block. Am I right in assuming that the string will remain slack ($T=0$) and that static friction ($f_S = mg \sin \theta$) will hold the block in place?

With greater difficulty I can imagine situation (2) above (which is what the author of the text says). The block is tied to the support first, held in air and tugged at so that the string became taut. With the string taut and it's other end furthest from the support, it is placed on the incline. Is the author of the book correct in assuming that in this case the static friction $f_S = 0$ and that tension ($T= mg \sin \theta$) will hold the block in place?

 

[PeroK]

Which makes it vexedly difficult. Can you elaborate on the phrase "depending on how the arrangement was created"?

Imagine one arrangement, which I speak about in (1) above. The block was first carried and kept on the incline - clearly it will remain at rest ($f_S \leq f_S^{max}$). The string was then tied to the block. Am I right in assuming that the string will remain slack ($T=0$) and that static friction ($f_S = mg \sin \theta$) will hold the block in place?

With greater difficulty I can imagine situation (2) above (which is what the author of the text says). The block is tied to the support first, held in air and tugged at so that the string became taut. With the string taut and it's other end furthest from the support, it is placed on the incline. Is the author of the book correct in assuming that in this case the static friction $f_S = 0$ and that tension ($T= mg \sin \theta$) will hold the block in place?

The way I would look at it is that the tension in the string is variable: it can take a range of values. If the tension is just enough to hold the block, then there is no force of static friction; if the tension is not enough to hold the block, then static friction opposes the attempt by the unbalanced force of gravity to move the block down the slope.

You could even have the case where the tension is greater than gravity down the sloppe and the string is trying to pull the block up the slope. In that case static friction acts down the slope, to oppose upward motion.

I don't see how you could make any assumptions. What does "taut" mean in terms of tension? Simply that the tension is non-zero?

 

[hutchphd]

What is the book?

 

[brotherbobby]

What is the book?

Patrick Hamill - Intermediate Dynamics.

 

[brotherbobby]

"I don't see how you could make any assumptions. What does "taut" mean in terms of tension? Simply that the tension is non-zero? "

Yes. Your response was illuminating. Thanks. I did not consider all the possibilities.

 

[haruspex]

Can you elaborate on the phrase "depending on how the arrangement was created"?

Block placed on slope, string not quite taut.
String is damp, slowly dries out and shrinks.

 

[brotherbobby]

Block placed on slope, string not quite taut.
String is damp, slowly dries out and shrinks.

Thank you. The point to take home from all this is that there really isn't an answer to the problem above. Both the (static) friction on the block and the tension in the string are dependent on how the apparatus was set. While I can understand that it's a valid point, it is unsettling. I wonder if you can give me some other examples from elementary mechanics where the history of a system determines the dynamics.

 

[PeroK]

Thank you. The point to take home from all this is that there really isn't an answer to the problem above. Both the (static) friction on the block and the tension in the string are dependent on how the apparatus was set. While I can understand that it's a valid point, it is unsettling. I wonder if you can give me some other examples from elementary mechanics where the history of a system determines the dynamics.

I'm not sure why it should be unsettling. Take an example from rock climbing, where the climber is "resting" on the rope above them, with their feet on a small ledge. You can't tell just by looking at the system how much of the climber's weight they are supporting themselves and how much they are weighting the rope. Until you start measuring things, all you can see is a taut rope and their feet on the ledge.

Or, if someone is trying to climb over a wall and is pulling up on the top of the wall, while others are pushing him up from below. There is no unique solution in terms of how much of the force is generated by the different components.

In general, whenever you have collinear forces you can only determine the sum of the forces.

 

[neilparker62]

If the forces are exactly in equilibrium, then there is presumably zero friction force acting and we have:

T=mgsinθ

That seems to be where textbook answer (a) is coming from.

For part (b) the maximum frictional force exceeds mgsinθ so the block will remain stationary if the string is cut. Hence zero acceleration as per textbook answer (b).

 

[haruspex]

If the forces are exactly in equilibrium, then there is presumably zero friction force

That presumption is invalid. This is the blunder made by the problem setter.

 

[neilparker62]

That presumption is invalid. This is the blunder made by the problem setter.

Ok - if the forces are in equilibrium , which way does non-zero friction act: up the slope or down the slope ? If the block is sitting on a flat plane, my understanding is that when you start pushing on it, the friction force simply increases from zero to maximum as the force is increased. After which the block will start moving.

On the slope if one notches up the tension gradually from T<mgsinθ through T=mgsinθ, to T>mgsinθ, the friction force must at some point transit from one direction to the other.

Happy to have my thinking corrected!

 

[brotherbobby]

I'm not sure why it should be unsettling.

It feels unsettling because really there isn't an answer to the problem. The answer depends on how the system was assembled - that is on its history, which is (usually) unavailable. Think about the problems in physics we have been doing all along. The assumption has been that a given set of conditions uniquely determine an outcome. We never questioned or cared how the set of conditions came about, did we?

I realize we have been wrong. But it's a deep point that is yet to sink in. I suppose a parallel can be drawn from the idea of work and internal energy of a thermodynamic system. The former being path dependent, unlike the latter. I can't tell you how much work you did bringing the system from point A to point B unless I tell you which path you took. Still, the point above is more difficult.

Thank you all for the lesson above.

 

[PeroK]

It feels unsettling because really there isn't an answer to the problem. The answer depends on how the system was assembled - that is on its history, which is (usually) unavailable. Think about the problems in physics we have been doing all along. The assumption has been that a given set of conditions uniquely determine an outcome. We never questioned or cared how the set of conditions came about, did we?

I realize we have been wrong. But it's a deep point that is yet to sink in. I suppose a parallel can be drawn from the idea of work and internal energy of a thermodynamic system. The former being path dependent, unlike the latter. I can't tell you how much work you did bringing the system from point A to point B unless I tell you which path you took. Still, the point above is more difficult.

Thank you all for the lesson above.

There was an interesting thread about non-unique solutions to energy-momentum problems like the Newton's cradle. You might like to take a look if it doesn't disturb you too much(!):

https://www.physicsforums.com/threads/a-nice-instructive-riddle.972212/#post-6181892

 

[haruspex]

if the forces are in equilibrium , which way does non-zero friction act

That's the point - there is no way to decide.

On the slope if one notches up the tension gradually from T<mgsinθ through T=mgsinθ, to T>mgsinθ, the friction force must at some point transit from one direction to the other.

Yes. So?

Try this one: a ball of given mass is pushed upwards between two walls that slope towards each other symmetrically at a given angle. The ball remains in place by friction. How great is the normal force?

 

[neilparker62]

That's the point - there is no way to decide.

Yes. So?

Try this one: a ball of given mass is pushed upwards between two walls that slope towards each other symmetrically at a given angle. The ball remains in place by friction. How great is the normal force?

I'm picking up a smooth sided object by pressing my fingers against either side of it. No μmg involved - is that similar to what you are describing ?

If the friction force transits from one direction to the other, I can't see how we avoid a point at which it is zero ? Quoting from Hyperphysics: "Static friction resistance will match the applied force up until the threshold of motion." From which I infer that if the applied force is zero, so is static friction - that is as shown in the graph at the above reference.

 

[haruspex]

If the friction force transits from one direction to the other, I can't see how we avoid a point at which it is zero

Sure, but that does not mean it is zero. It is any value within the static friction range.

Quoting from Hyperphysics: "Static friction resistance will match the applied force up until the threshold of motion." From which I infer that if the applied force is zero, so is static friction

Yes, but in the question there is a string which may be under tension. Your argument is circular: the tension is exactly what is needed to hold the block in place without friction, so the static friction equals zero, so the tension equals $mg\sin(\theta)$.

 

[jbriggs444]

Or, to put it slightly differently.

Tension applied by a string supporting an object will be whatever it needs to be to prevent the object from falling beyond the string's length -- within the limit imposed by the breaking strength of the string.

Friction supplied by the inclined plane supporting an object will be whatever it needs to be to prevent the object from slipping down -- within the limit imposed by the coefficient of static friction.

If both are present, and the sum of the two limits is more than enough to support the object then the situation is statically indeterminate. Like an ideal four-legged table on an ideal floor. Or the other situations already mentioned.

 

[neilparker62]

Ok - post #2 is clear. Tension in the string is anything within the given range. Friction force: T - mgsinθ with a corresponding range - mgsinθ (ie acting up the slope) to μmgcosθ (when the block is on the point of being pulled upwards).