# Forces and Friction on a Block and a Beam

• masterchiefo
In summary: Which ones are they?The frictional forces that will necessarily slide are the ones acting on the block and the plank. They are in the same direction as the plank, but with opposite signs.
masterchiefo

## Homework Statement

A block of mass M = 1 kg is placed between a vertical wall and the end of A
beam mass m = 10 kg . If μ S = 0.6 A , determine the minimum value of the angle θ min
θ for which the block remains in equilibrium and the coefficient of friction
Static B μ S corresponding to prevent the beam from sliding B.

Picture of the problem in attachment.

## The Attempt at a Solution

Block:
∑Fy = +fsablock -Wblock = 0
∑Fx = +Nblock = 0

Beam:
∑Fx = -fsabeam + NA =0
∑Fy = -Wbeam +NB + fsabeam =0

is this correct? and how can I possibly find the min θ?

thank you.

#### Attachments

• probbb111.PNG
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Last edited:
You seem to be using x as the vertical axis and y as the horizontal, which is a bit confusing.
In general, a static frictional force is not ##N\mu_s##. That is its maximum value.
The minimum angle is the one at which the system is only just stable, i.e. the angle at which (some or all of) the static frictional forces are at their maximum values. So merely by assuming those forces to be ##N\mu_s## you will find the minimum angle.
However, there is a catch here. Consider how the system will move when it does slip. Only those interfaces that will necessarily slide can be assumed to have been at their maximal frictional forces the instant before slipping. Which ones are they?

There are two, equal and opposite, normal forces acting on the block. One of these is equal and opposite to a force on the plank.

Last edited:
masterchiefo
haruspex said:
You seem to be using x as the vertical axis and y as the horizontal, which is a bit confusing.
In general, a static frictional force is not ##N\mu_s##. That is its maximum value.
The minimum angle is the one at which the system is only just stable, i.e. the angle at which the static frictional forces are at their maximum values. So merely by assuming those forces to be ##N\mu_s## you will find the minimum angle.

There are two, equal and opposite, normal forces acting on the block. One of these is equal and opposite to a force on the plank.
Yeah sorry, I had them correct on paper for the axies x/y on the block, I edited the thread for that part.

∑Fy = +fsablock -Wblock = 0
∑Fx = +Nblock -Nplank = 0

Now that should be correct.

But I still don't understand how to use the angle in the situation because in my ∑Fx/∑Fy on both the block and the beam, I don't use any angle.

haruspex said:
You seem to be using x as the vertical axis and y as the horizontal, which is a bit confusing.
In general, a static frictional force is not ##N\mu_s##. That is its maximum value.
The minimum angle is the one at which the system is only just stable, i.e. the angle at which (some or all of) the static frictional forces are at their maximum values. So merely by assuming those forces to be ##N\mu_s## you will find the minimum angle.
However, there is a catch here. Consider how the system will move when it does slip. Only those interfaces that will necessarily slide can be assumed to have been at their maximal frictional forces the instant before slipping. Which ones are they?

There are two, equal and opposite, normal forces acting on the block. One of these is equal and opposite to a force on the plank.

"Consider how the system will move when it does slip. Only those interfaces that will necessarily slide can be assumed to have been at their maximal frictional forces the instant before slipping. Which ones are they?"
EDIT: When it does slip, the block will slide on with the plank and the wall. and the plank will slide on the floor and the block. So the 3 frictional forces on the drawing ?

Last edited:
masterchiefo said:

## Homework Statement

A block of mass M = 1 kg is placed between a vertical wall and the end of A
beam mass m = 10 kg . If μ S = 0.6 A , determine the minimum value of the angle θ min
θ for which the block remains in equilibrium and the coefficient of friction
Static B μ S corresponding to prevent the beam from sliding B.

Picture of the problem in attachment.

## The Attempt at a Solution

Block:
∑Fy = +fsablock -Wblock = 0
∑Fx = +Nblock = 0

Beam:
∑Fx = -fsabeam + NA =0
∑Fy = -Wbeam +NB + fsabeam =0

is this correct? and how can I possibly find the min θ?

thank you.

It's helpful - to me anyway - to show a reasonable sized image.

You need to consider torque to get the angle θ involved. Torque about either point B or about point A, whichever makes solving less complicated.

masterchiefo
SammyS said:

It's helpful - to me anyway - to show a reasonable sized image.

You need to consider torque to get the angle θ involved. Torque about either point B or about point A, whichever makes solving less complicated.
But I don't even have any distance in meter, how am I supposed to consider torque?

masterchiefo said:
But I don't even have any distance in meter, how am I supposed to consider torque?
Maybe use length, L ? Otherwise, pick some length. Either way, it will cancel out.

masterchiefo
SammyS said:
Maybe use length, L ? Otherwise, pick some length. Either way, it will cancel out.
∑MA = +NB*(tan(θ)*L) -fsbplank * L -Wplank * (tan(θ)*L)/2 = 0

L is the vertical length A to B.

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masterchiefo said:
∑MA = +NB*d -fsbplank * L -Wplank * (d/2) = 0

L is the vertical length A to B.
and d is the horizontal length A to B.
SammyS said:
Maybe use length, L ? Otherwise, pick some length. Either way, it will cancel out.
One thing I am wondering.

How is the traction force between the plank and the block? is it Vertical? or same direction as the plank with the angle?
if that is the case, I probably don't need to use torque.

masterchiefo said:
"Consider how the system will move when it does slip. Only those interfaces that will necessarily slide can be assumed to have been at their maximal frictional forces the instant before slipping. Which ones are they?"
EDIT: When it does slip, the block will slide on with the plank and the wall. and the plank will slide on the floor and the block. So the 3 frictional forces on the drawing ?
Certainly the block will slide down the wall.
Is it possible that the base of the plank will not slip? Imagine point B as a hinge.
Is it possible that the plank and block will move together? Imagine point A as a hinge?

masterchiefo
masterchiefo said:
∑Fy = +fsablock -Wblock = 0
There are two frictional forces on the block. Don't (yet) assume they're the same.

masterchiefo
haruspex said:
Certainly the block will slide down the wall.
Is it possible that the base of the plank will not slip? Imagine point B as a hinge.
Is it possible that the plank and block will move together? Imagine point A as a hinge?
Yeah well if I imagine both point as a hinge, block and plank may move together and base of plank won't slip.

But I don't understand because if that is the case then there is no friction in those point ?

haruspex said:
There are two frictional forces on the block. Don't (yet) assume they're the same.
Oh yeah true, do I make both in the same direction or opposite ? I believe its in the same direction.

masterchiefo said:
Yeah well if I imagine both point as a hinge, block and plank may move together and base of plank won't slip.
No, not both at once - one or the other.
Imagining B as a hinge, you can see that the block could slip down while the plank stays still. This means the horizontal force at B might be less than the maximum possible there, i.e. ##F_B<\mu_{sB}N_B##.
Imagining (instead) A as a hinge, the block and plank could both slip down, with the points of contact between block and plank staying together. This means the vertical frictional force between block and plank might be less than the maximum value there, ##F_A<\mu_{sA}N_A##.
But only one of these will be true. So there are two cases to consider.
In both cases, the block will slide against the wall, so just before sliding ##F_{wall}=\mu_{sA}N_A##.
masterchiefo said:
do I make both in the same direction or opposite ? I believe its in the same direction.
Yes, same direction.

masterchiefo
haruspex said:
No, not both at once - one or the other.
Imagining B as a hinge, you can see that the block could slip down while the plank stays still. This means the horizontal force at B might be less than the maximum possible there, i.e. ##F_B<\mu_{sB}N_B##.
Imagining (instead) A as a hinge, the block and plank could both slip down, with the points of contact between block and plank staying together. This means the vertical frictional force between block and plank might be less than the maximum value there, ##F_A<\mu_{sA}N_A##.
But only one of these will be true. So there are two cases to consider.
In both cases, the block will slide against the wall, so just before sliding ##F_{wall}=\mu_{sA}N_A##.

Yes, same direction.
Hey sorry, I understand it as a text, but how do I test the two cases ?

Block:
∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
∑Fx = +Nblock - Nblock1 = 0
Nblock = 8.175N
Nblock1 = 8.175N

Plank:
That woul mean NA = 8.175?

∑Fx = -NA*0.6+ NA =0
-NA*0.6+ NA =0
∑Fy = -98.10plank +NB + NA*06 =0
∑MA = +NB*(tan(θ)*L) -NB*us * L -98.10 * (tan(θ)*L)/2 = 0

Thank you very much

Last edited:
masterchiefo said:
Hey sorry, I understand it as a text, but how do I test the two cases ?

Block:
∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
∑Fx = +Nblock - Nblock1 = 0
Nblock = 8.175N
Nblock1 = 8.175N

Plank:
That would mean NA = 8.175?

∑Fx = -NA*0.6+ NA =0
-NA*0.6+ NA =0
∑Fy = -98.10plank +NB + NA*06 =0
∑MA = +NB*(tan(θ)*L) -NB*us * L -98.10 * (tan(θ)*L)/2 = 0

Thank you very much
Looks like you have some errors regarding the Plank, at least you do for Σ Fx .

Isn't the normal force at A, NA, the same as Nblock ?

masterchiefo
SammyS said:
Looks like you have some errors regarding the Plank, at least you do for Σ Fx .

Isn't the normal force at A, NA, the same as Nblock ?

Block:
∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
∑Fx = +Nblock - Nblock1 = 0
Nblock = 8.175N
Nblock1 = 8.175N

Plank:
Would that mean NA is = 8.175N?
∑Fx = -NB*usb+ 8.175 =0
∑Fy = -98.10plank +NB + 8.175*06 =0
When I solve these 2:
usb= 0.087719
NB = 93.195N

∑MA = +93.195*(tan(θ)*L) -93.195*usb * L -98.10 * (tan(θ)*L)/2 = 0

L = vertical length A to B.
I can't figure out how to solve Torque equation, keeps saying false in my calculator.
Is my torque equation right ?

Thank you very much

It's not a good idea to leap into numerics straight away. Keep it all symbolic until the end.
masterchiefo said:
Block:
∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
##N_A\mu_{sA} + F_A - Mg=0##
If we are considering the case where the plank would stay still, ##N_A\mu_{sA} = F_A##, ##2N_A\mu_{sA} = Mg##.
masterchiefo said:
∑Fx = -NB*usb+ 8.175 =0
##F_B=N_A##
If we are considering the case where the plank slips down, ##F_B=N_B\mu_{sB}##. But to avoid confusion we need to do one case at a time, and we haven't finished the static plank case yet.
masterchiefo said:
∑Fy = -98.10plank +NB + 8.175*06 =0
Check your signs. Which way does the frictional force at A act on the plank?

masterchiefo
haruspex said:
It's not a good idea to leap into numerics straight away. Keep it all symbolic until the end.

##N_A\mu_{sA} + F_A - Mg=0##
If we are considering the case where the plank would stay still, ##N_A\mu_{sA} = F_A##, ##2N_A\mu_{sA} = Mg##.

##F_B=N_A##
If we are considering the case where the plank slips down, ##F_B=N_B\mu_{sB}##. But to avoid confusion we need to do one case at a time, and we haven't finished the static plank case yet.

Check your signs. Which way does the frictional force at A act on the plank?
∑Fy = -98.10plank +NB - 8.175*06 =0

Okay,
but I don't understand why we have to consider if one slips etc if in the problem it says I have to find usb and the min angle when everything is in equilibrium? it doesn't say anything about when it starts moving.

masterchiefo said:
Okay,
but I don't understand why we have to consider if one slips etc if in the problem it says I have to find usb and the min angle when everything is in equilibrium?
I should read the question more carefully. I was taking both coefficients as given.
OK, so we also want ##\mu_{sB}## at its minimum.
masterchiefo said:
it doesn't say anything about when it starts moving.
We need to find the forces etc. when it is just about to slip. But in order to do that we have to think about how it will move when it does slip. Those contacting surfaces that will slide in relation to each other must be the ones that are at limit of static friction before slipping.
So, we can assume that just before it slips the wall/block interface is at limit of friction, and the plank/floor interface is at limit of friction. (If the second were not true we could make ##\mu_{sB}## less.)
For now, we still shouldn't assume that the wall/plank interface is at its limit. We can reconsider that later.
So we have ##N_A\mu_{sA} + F_A - Mg=0##
##F_B=N_B\mu_{sB}##

Now have another go at the equations for Fy for the plank and the moments on the plank.

masterchiefo
haruspex said:
I should read the question more carefully. I was taking both coefficients as given.
OK, so we also want ##\mu_{sB}## at its minimum.

We need to find the forces etc. when it is just about to slip. But in order to do that we have to think about how it will move when it does slip. Those contacting surfaces that will slide in relation to each other must be the ones that are at limit of static friction before slipping.
So, we can assume that just before it slips the wall/block interface is at limit of friction, and the plank/floor interface is at limit of friction. (If the second were not true we could make ##\mu_{sB}## less.)
For now, we still shouldn't assume that the wall/plank interface is at its limit. We can reconsider that later.
So we have ##N_A\mu_{sA} + F_A - Mg=0##
##F_B=N_B\mu_{sB}##

Now have another go at the equations for Fy for the plank and the moments on the plank.
I don't see what is wrong with my FY equation and the moments on the plank. I have a sign mistake? Length mistake on the Moments equation?

∑Fy = -Mg +NB - NA*usa =0
∑MA = +NB*(tan(θ)*L) -NB*usb * L -Mg * (tan(θ)*L)/2 = 0

masterchiefo said:
Block:
∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
∑Fx = +Nblock - Nblock1 = 0
Nblock = 8.175N
Nblock1 = 8.175N

Plank:
Would that mean NA is = 8.175N?
∑Fx = -NB*usb+ 8.175 =0
∑Fy = -98.10plank +NB + 8.175*0.6 =0 ##\quad\quad\ ## You have the wrong sign on the term 8.175×0.6
(actually fs,A ). The block exerts force downward on the plank.
When I solve these 2:
usb= 0.087719
NB = 93.195N

∑MA = +93.195*(tan(θ)*L) -93.195*usb * L -98.10 * (tan(θ)*L)/2 = 0

L = vertical length A to B.
I can't figure out how to solve Torque equation, keeps saying false in my calculator.
Is my torque equation right ?

Thank you very much
I'm just pointing out a sign error you have been committing. Continue using haruspex's guidance.

masterchiefo
masterchiefo said:
I don't see what is wrong with my FY equation and the moments on the plank. I have a sign mistake? Length mistake on the Moments equation?

∑Fy = -Mg +NB - NA*usa =0
∑MA = +NB*(tan(θ)*L) -NB*usb * L -Mg * (tan(θ)*L)/2 = 0
That has fixed the sign error. Your moments equation may have been ok before but it was too hard to interpret with those decimal numbers. It looks ok now, but you can simplify it.
But you seem to be using the same symbol M for both masses. Please use m for the plank.

What can you deduce about theta from the equations you have?

masterchiefo
haruspex said:
That has fixed the sign error. Your moments equation may have been ok before but it was too hard to interpret with those decimal numbers. It looks ok now, but you can simplify it.
But you seem to be using the same symbol M for both masses. Please use m for the plank.

What can you deduce about theta from the equations you have?
When I solve the Moments equation while using the NB that I have previously found by solving the Fx and Fy, I get theta = 8.61564.

What I can deduce from theta? It is the same theta for mg and NB.2
to simplify I could remove the L since what ever L is it will be the same result anyway.

masterchiefo said:
When I solve the Moments equation while using the NB that I have previously found by solving the Fx and Fy, I get theta = 8.61564.
Please post your result for tan theta as an algebraic expression - no numeric substitutions for the variables! Posting your working to that point would be even better.
masterchiefo said:
It is the same theta for mg and NB.2
I don't understand the question.

masterchiefo
haruspex said:
Please post your result for tan theta as an algebraic expression - no numeric substitutions for the variables! Posting your working to that point would be even better.
I don't understand the question.
∑MA = +NB*(tan(θ)) -NB*usb -mg * (tan(θ))/2 = 0
+103.005*(tan(θ)) -103.005*0.079b -98.10 * (tan(θ))/2 = 0
θ = 8.61564
tan(θ = 8.61564) = 0.151515

Not sure if this is what you wanted.

masterchiefo said:
∑MA = +NB*(tan(θ)) -NB*usb -mg * (tan(θ))/2 = 0
+103.005*(tan(θ)) -103.005*0.079b -98.10 * (tan(θ))/2 = 0
θ = 8.61564
tan(θ = 8.61564) = 0.151515

Not sure if this is what you wanted.
In any problem, you are given some variables and asked to determine others. Sometimes the 'givens' are only given symbolically: "A mass m collides with... ". Other times, as here, you are given numeric values. But I strongly recommend that you always work a problem using symbols for the givens, only plugging in numbers at the final step. There are numerous benefits to this, not least, making it much easier for others to follow and to check and to comment on your work. See https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/.

Here you are given two masses, M and m, and a coefficient of friction, ##\mu_{sA}##.
The form of answer I hoped to see was tan(θ) = {an algebraic expression in terms of m, M, and ##\mu_{sA}##}.
Likewise for ##\mu_{sB}##.

masterchiefo
haruspex said:
In any problem, you are given some variables and asked to determine others. Sometimes the 'givens' are only given symbolically: "A mass m collides with... ". Other times, as here, you are given numeric values. But I strongly recommend that you always work a problem using symbols for the givens, only plugging in numbers at the final step. There are numerous benefits to this, not least, making it much easier for others to follow and to check and to comment on your work. See https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/.

Here you are given two masses, M and m, and a coefficient of friction, ##\mu_{sA}##.
The form of answer I hoped to see was tan(θ) = {an algebraic expression in terms of m, M, and ##\mu_{sA}##}.
Likewise for ##\mu_{sB}##.
in terms of M and m? my moment equation only use m .

haruspex said:
In any problem, you are given some variables and asked to determine others. Sometimes the 'givens' are only given symbolically: "A mass m collides with... ". Other times, as here, you are given numeric values. But I strongly recommend that you always work a problem using symbols for the givens, only plugging in numbers at the final step. There are numerous benefits to this, not least, making it much easier for others to follow and to check and to comment on your work. See https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/.

Here you are given two masses, M and m, and a coefficient of friction, ##\mu_{sA}##.
The form of answer I hoped to see was tan(θ) = {an algebraic expression in terms of m, M, and ##\mu_{sA}##}.
Likewise for ##\mu_{sB}##.
tan(θ) = (NB * ##\mu_{sB}##) / (NB-(mg/2))

masterchiefo said:
in terms of M and m? my moment equation only use m .
Sure, but your moment equation involves forces which are not givens. You would have to use your other equations to substitute for those forces. The final equation should express tan theta in terms of m, M and ##\mu_{sA}##. It should not include any non-givens (i.e., the forces).

masterchiefo said:
∑MA = +NB*(tan(θ)) -NB*usb -mg * (tan(θ))/2 = 0
+103.005*(tan(θ)) -103.005*0.079b -98.10 * (tan(θ))/2 = 0
θ = 8.61564
tan(θ = 8.61564) = 0.151515

Not sure if this is what you wanted.
Just so you know things are working towards a solution (however slowly), I do also get ##\displaystyle \ \tan(\theta) = \frac{5}{33} = 0.151515\dots \,, \ ## so that θ ≈ 8.61565°.

However, I know the conditions for which this value holds, and whether it's a minimum or a maximum or neither.

It's probably about the time haruspex wakes, so he may get back with another response, or he may be waiting for your response. I hate to second guess him.

The above quoted response from you was in reply to his request to write an expression for tan(θ) ( coming from the torque equation) which only had symbols, no numeric substitutions for the variables. However, you plugged in all sorts of numeric values.

This post was edited. 33/5 was an error (typo). It was corrected to 5/33 .

Last edited:
SammyS said:
Just so you know things are working towards a solution (however slowly), I do also get ##\displaystyle \ \tan(\theta) = \frac{33}{5} = 0.151515\dots \,, \ ## so that θ ≈ 8.61565°.

However, I know the conditions for which this value holds, and whether it's a minimum or a maximum or neither.

It's probably about the time haruspex wakes, so he may get back with another response, or he may be waiting for your response. I hate to second guess him.

The above quoted response from you was in reply to his request to write an expression for tan(θ) ( coming from the torque equation) which only had symbols, no numeric substitutions for the variables. However, you plugged in all sorts of numeric values.
I also confirm that numeric answer. But I still want to get across the concept of working entirely symbolically since it has so many advantages.
In the present case, I was looking for the answer ##\tan(\theta)=\frac M{\mu_{sA}(M+m)}##.

There is also a subtlety to this problem. You are asked to minimise two variables under one constraint. In general, that might not be possible; minimising one might not permit the minimum of the other. Here, a smaller theta makes it more likely to slip at A, but less likely to slip at B. As a result, a smaller theta permits a smaller ##\mu_{sB}##. So it is OK to minimise theta based on slipping at A, then proceed to minimise ##\mu_{sB}## using that theta. (I was hoping to lead you to this by hints, but couldn't find an effective way to do it.)

PS. SammyS's last post was at 2:45am my time.

haruspex said:
I also confirm that numeric answer. But I still want to get across the concept of working entirely symbolically since it has so many advantages.
In the present case, I was looking for the answer ##\tan(\theta)=\frac M{\mu_{sA}(M+m)}##.

There is also a subtlety to this problem. You are asked to minimise two variables under one constraint. In general, that might not be possible; minimising one might not permit the minimum of the other. Here, a smaller theta makes it more likely to slip at A, but less likely to slip at B. As a result, a smaller theta permits a smaller ##\mu_{sB}##. So it is OK to minimise theta based on slipping at A, then proceed to minimise ##\mu_{sB}## using that theta. (I was hoping to lead you to this by hints, but couldn't find an effective way to do it.)

PS. SammyS's last post was at 2:45am my time.
Yeah sorry,
My teacher only ask us for simple equations and we never had to substitute etc to get one final equation since we use the TI Calculator and solve directly with all 3 equations so I was a bit confused with that you asked.

Also, now that i think about it my usb is 0.079, isn't that too low?.

And thank you very much again, you have helped me a lot.

haruspex said:
...
SammyS said:
...
It's probably about the time haruspex wakes, ...

PS. SammyS's last post was at 2:45am my time.
I was significantly in error regarding your local time! -- Maybe I just couldn't estimate my own local time.

Last edited:
masterchiefo
SammyS said:
I was significantly in error regarding your local time! -- Maybe I just couldn't estimate my own local time.
Hey thank you for helping me as well.
This forums is better than my teachers so far.

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