Forces and Friction on a Block and a Beam

[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.

Homework Equations

The Attempt at a Solution

Block:
∑Fy = +fsa_block -W_block = 0
∑Fx = +N_block = 0

Beam:
∑Fx = -fsa_beam + NA =0
∑Fy = -W_beam +NB + fsa_beam =0

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

thank you.

 

[haruspex]

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.

 

[masterchiefo]

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 = +fsa_block -W_block = 0
∑Fx = +N_block -N_plank = 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.

 

[masterchiefo]

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 ?

 

[SammyS]

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.

Homework Equations

The Attempt at a Solution

Block:
∑Fy = +fsa_block -W_block = 0
∑Fx = +N_block = 0

Beam:
∑Fx = -fsa_beam + NA =0
∑Fy = -W_beam +NB + fsa_beam =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]

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?

 

[SammyS]

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]

Maybe use length, L ? Otherwise, pick some length. Either way, it will cancel out.

∑MA = +NB*(tan(θ)*L) -fsb_plank * L -W_plank * (tan(θ)*L)/2 = 0

L is the vertical length A to B.

 

[masterchiefo]

∑MA = +NB*d -fsb_plank * L -W_plank * (d/2) = 0

L is the vertical length A to B.
and d is the horizontal length A to B.

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.

 

[haruspex]

"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?

 

[haruspex]

∑F_y = +fsa_block -W_block = 0

There are two frictional forces on the block. Don't (yet) assume they're the same.

 

[masterchiefo]

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 ?

 

[masterchiefo]

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.

 

[haruspex]

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 $F_{wall}=\mu_{sA}N_A$.

do I make both in the same direction or opposite ? I believe its in the same direction.

Yes, same direction.

 

[masterchiefo]

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 = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0
∑Fx = +N_block - N_block1 = 0
N_block = 8.175N
N_block1 = 8.175N

Plank:
That woul mean NA = 8.175?

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

Thank you very much

 

[SammyS]

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

Block:
∑Fy = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0
∑Fx = +N_block - N_block1 = 0
N_block = 8.175N
N_block1 = 8.175N

Plank:
That would mean NA = 8.175?

∑Fx = -NA*0.6+ NA =0
-NA*0.6+ NA =0
∑Fy = -98.10_plank +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 F_x .

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

 

[masterchiefo]

Looks like you have some errors regarding the Plank, at least you do for Σ F_x .

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

Block:
∑Fy = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0
∑Fx = +N_block - N_block1 = 0
N_block = 8.175N
N_block1 = 8.175N

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

∑MA = +93.195*(tan(θ)*L) -93.195*us_b * 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

 

[haruspex]

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

Block:
∑Fy = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0

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

∑Fx = -NB*us_b+ 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.

∑Fy = -98.10_plank +NB + 8.175*06 =0

Check your signs. Which way does the frictional force at A act on the plank?

 

[masterchiefo]

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 = -98.10_plank +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 us_b and the min angle when everything is in equilibrium? it doesn't say anything about when it starts moving.

 

[haruspex]

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 us_b 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.

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]

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*us_a =0
∑MA = +NB*(tan(θ)*L) -NB*us_b * L -Mg * (tan(θ)*L)/2 = 0

 

[SammyS]

Block:
∑Fy = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0
∑Fx = +N_block - N_block1 = 0
N_block = 8.175N
N_block1 = 8.175N

Plank:
Would that mean NA is = 8.175N?
∑Fx = -NB*us_b+ 8.175 =0
∑Fy = -98.10_plank +NB + 8.175*0.6 =0 $\quad\quad\$ You have the wrong sign on the term 8.175×0.6

(actually f_s,A ). The block exerts force downward on the plank.​

When I solve these 2:
us_b= 0.087719
NB = 93.195N

∑MA = +93.195*(tan(θ)*L) -93.195*us_b * 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.

 

[haruspex]

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*us_a =0
∑MA = +NB*(tan(θ)*L) -NB*us_b * 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]

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.

 

[haruspex]

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.

It is the same theta for mg and NB.2

I don't understand the question.

 

[masterchiefo]

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*us_b -mg * (tan(θ))/2 = 0
+103.005*(tan(θ)) -103.005*0.079_b -98.10 * (tan(θ))/2 = 0
θ = 8.61564
tan(θ = 8.61564) = 0.151515

Not sure if this is what you wanted.

 

[haruspex]

∑MA = +NB*(tan(θ)) -NB*us_b -mg * (tan(θ))/2 = 0
+103.005*(tan(θ)) -103.005*0.079_b -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]

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 .

 

[masterchiefo]

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))

 

[haruspex]

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).

 

[SammyS]

∑MA = +NB*(tan(θ)) -NB*us_b -mg * (tan(θ))/2 = 0
+103.005*(tan(θ)) -103.005*0.079_b -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 .

 

[haruspex]

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.

 

[masterchiefo]

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 us_b is 0.079, isn't that too low?.

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

 

[SammyS]

...

...
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.

 

[masterchiefo]

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.

 

[masterchiefo]

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?

Hey here, wouldn't the vertical force be the same as the Mg to keep it from slipping?

∑Fy = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0
∑Fx = +N_block - N_block1 = 0
N_block = 8.175N
N_block1 = 8.175N

 

[haruspex]

Hey here, wouldn't the vertical force be the same as the Mg to keep it from slipping?

∑Fy = +N_block*0.6 + N_block1*0.6 - 9.81_block = 0
∑Fx = +N_block - N_block1 = 0
N_block = 8.175N
N_block1 = 8.175N

Sure, but we're past that point, aren't we? We now need to figure out the minimum $\mu_{sB}$. And please, please try to post you working in symbolic form. Pretend that you are not told any of the numerical values. It really makes life much easier for those trying to help you, and you should find it less prone to error.

 

[masterchiefo]

Sure, but we're past that point, aren't we? We now need to figure out the minimum $\mu_{sB}$. And please, please try to post you working in symbolic form. Pretend that you are not told any of the numerical values. It really makes life much easier for those trying to help you, and you should find it less prone to error.

what I mean is that Friction Force = W, or it that wrong? so us_a*Na = W
Didn't w already calculated the us_b same time as the angle ?

 

[haruspex]

what I mean is that Friction Force = W, or it that wrong? so us_a*Na = W

There are two (equal) frictional forces helping the block to stay put.

Didn't w already calculated the us_b same time as the angle ?

Ah yes... you asked whether the value was reasonable. I didn't reply because your working was too full of numbers. My gut feel is that the answer seems reasonable. The angle is quite small, so the horizontal forces at the base of the plank would be small.

 

[SammyS]

There are two (equal) frictional forces helping the block to stay put.

Ah yes... you asked whether the value was reasonable. I didn't reply because your working was too full of numbers. My gut feel is that the answer seems reasonable. The angle is quite small, so the horizontal forces at the base of the plank would be small.

haruspex,

Here's something for you to ponder after waking on 4 July.

There is something about this solution that has been to bother me recently. It began with trying to come to grips with that rather small value for μ_sB . In trying to understand how that comes about, (Again we see the advantage of using symbols rather than numeric quantities.) we see that the normal force at B, N_B is $\displaystyle \ N_B=\frac{Mg}{2}+mg \ .\$ That is to say, at point B, the floor supports the full weight of the plank/beam and half the weight of the block.

Contrast this to the case in which there is no block, just the plank, contacting the wall at A and the floor at B, with friction at each point of contact. Surely friction at A will support some of the weight of the plank. The floor will support only a portion of the plank's weight.

In our case, the one with the block, the weight of the block is much less that that of the plank. I suspect, therefore, the frictional force exerted on the plank at A may actually be upward, somewhat like the case with no block.

I must remind myself, as you have pointed out early in this thread, that:

1) Static friction may not be at its maximum. Often it isn't.
2) We must be careful regarding the direction we assign to the static friction.​

 

[haruspex]

haruspex,

Here's something for you to ponder after waking on 4 July.

There is something about this solution that has been to bother me recently. It began with trying to come to grips with that rather small value for μ_sB . In trying to understand how that comes about, (Again we see the advantage of using symbols rather than numeric quantities.) we see that the normal force at B, N_B is $\displaystyle \ N_B=\frac{Mg}{2}+mg \ .\$ That is to say, at point B, the floor supports the full weight of the plank/beam and half the weight of the block.

Contrast this to the case in which there is no block, just the plank, contacting the wall at A and the floor at B, with friction at each point of contact. Surely friction at A will support some of the weight of the plank. The floor will support only a portion of the plank's weight.

In our case, the one with the block, the weight of the block is much less that that of the plank. I suspect, therefore, the frictional force exerted on the plank at A may actually be upward, somewhat like the case with no block.

I must remind myself, as you have pointed out early in this thread, that:

1) Static friction may not be at its maximum. Often it isn't.
2) We must be careful regarding the direction we assign to the static friction.​

This is a good point, but I think it comes down to the issue I already mentioned, that of whether the two variables can be simultaneously at their minima under the one criterion.
Suppose we fix the angle and reduce $\mu_{sB}$ to its minimum. It may well be that the friction from the block is pushing up on the plank.
But consider what happens if we now reduce the angle. The plank will not need as much friction from the ground, and it will pull down less on the block. On the other hand, the normal force between block and wall is reduced, so could this mean the block now slips? Well, no, because before the block can slip the friction between plank and block must start going the other way, with the plank holding the block up. So we must be able to reduce the angle to some extent without the block slipping.