Forces and Friction on a Block and a Beam

[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.