# Statics - Determine the Reactions on this bent bar levering between two surfaces

• Engineering
• Nova_Chr0n0
Nova_Chr0n0
Homework Statement
Determine the reactions at the smooth contact points A, B, and C on the bar.
Relevant Equations
N/A
The figure is shown below:

Here is my FBD for the figure with assign +x and +y directions

I started off by summing up the forces in the x-direction:

Next is the summing up of the forces in the y-direction:

After this, I solved for the moment at point A: assuming that counter-clockwise is +

Now from equation 1, if I input the value of F_B, i would get F_A = 827.349 N.

Here is where my question starts. When I tried to search for the problem in the internet to double check the answers that I've got, I only got the Force C right. Mainly because most of them choose B as their moment point. Solution below:

With this, I got two different values for Fa and Fb when choosing the moment point. I tried to find solution in the internet where they choose A as their moment point but I didn't finding anything. I tried to check my solution multiple times but I cannot find my mistake here. Is there something wrong with the values in my solution? Or is it wrong that I decided to choose A as my point of moment?

You left out one of the components of the man's force in your moment balance around point A.

Nova_Chr0n0
@Nova_Chr0n0

You have represented the reactions correctly, because surface friction should not be considered ("smooth contact points").
For points A and B, the reactions should be perfectly perpendicular to each flat surface.

For point C, the reaction should be perfectly perpendicular to the bar.

Do you have the full correct response to this problem?
It seems to be a complicated problem the way it is shown.
For stable condition, you either need point C or B, but not both.

Nevertheless, point C seems to be necessary for limiting any sliding of the bar horizontally toward the left.
Perhaps, we should only consider C as a simple horizontal support because of that (otherwise, any vertical reaction in C would need to be subtracted from any vertical reaction in B).

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Nova_Chr0n0
Lnewqban said:
@Nova_Chr0n0

You have represented the reactions correctly, because surface friction should not be considered ("smooth contact points").
For points A and B, the reactions should be perfectly perpendicular to each flat surface.

For point C, the reaction should be perfectly perpendicular to the bar.

Do you have the full correct response to this problem?
It seems to be a complicated problem the way it is shown.
For stable condition, you either need point C or B, but not both.

Nevertheless, point C seems to be necessary for limiting any sliding of the bar horizontally toward the left.
Perhaps, we should only consider C as a simple horizontal support because of that (otherwise, any vertical reaction in C would need to be subtracted from any vertical reaction in B).
To me it seems the reaction at B is (should be) zero.

Chestermiller said:
You left out one of the components of the man's force in your moment balance around point A.
Thank you very much! I didn't notice that I left out the component force of 250 N that is parallel to the bar. I've now gotten the same answer.

Lnewqban said:
@Nova_Chr0n0

You have represented the reactions correctly, because surface friction should not be considered ("smooth contact points").
For points A and B, the reactions should be perfectly perpendicular to each flat surface.

For point C, the reaction should be perfectly perpendicular to the bar.

Do you have the full correct response to this problem?
It seems to be a complicated problem the way it is shown.
For stable condition, you either need point C or B, but not both.

Nevertheless, point C seems to be necessary for limiting any sliding of the bar horizontally toward the left.
Perhaps, we should only consider C as a simple horizontal support because of that (otherwise, any vertical reaction in C would need to be subtracted from any vertical reaction in B).
The solution that is at the last page of the e-book I've got (Mechanics - Hibeller 14th edition) also used moment at B. Here is its solution:

But I've already got the correct solution, as Sir Chestermiller stated that I've left out one force.

Lnewqban
It may be the solution in the book, but does ##N_B## being non-zero make sense? Is the rod ##AB## compressed when it's placed in between the plates before the force is applied? If so... there is a host of other questions that need resolved before the normal reactions could be determined.

If it was slid into place without force fit (as a rigid body - a pure statics problem), then when the force is applied, point ##B## immediately loses contact with the plate below it.

I think the problem is flawed.

erobz said:
It may be the solution in the book, but does ##N_B## being non-zero make sense? Is the rod ##AB## compressed when it's placed in between the plates before the force is applied? If so... there is a host of other questions that need resolved before the normal reactions could be determined.

If it was slid into place without force fit (as a rigid body - a pure statics problem), then when the force is applied, point ##B## immediately loses contact with the plate below it.

I think the problem is flawed.
I don't follow. Can you elaborate. Are you saying that the angle iron could not be placed into its shown position without any forces acting on it?

Chestermiller said:
I don't follow. Can you elaborate. Are you saying that the angle iron could not be placed into its shown position without any forces acting on it?
In order to have normal forces at both ##B## and ##A## it would have to be wedged into place. The rod ##AB## would necessarily be compressed - A forced fit.

erobz said:
In order to have normal forces at both ##B## and ##A## it would have to be wedged into place. The rod ##AB## would necessarily be compressed - A forced fit.
Can't it be rotated and translated into place?

Chestermiller said:
Can't it be rotated and translated into place?
That doesn't help IMO. The rod is rigid (and massless). In order for it to have contact on both sides it would need to be compressed. In the limit as zero clearance is approached, the moment the force was applied ( at the end) , normal force at ##B## would go to zero.

If it the case that the rod ##AB## is compressed then the normal forces are arbitrary at this point without further information detailing the compression, rod material , etc...

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erobz said:
That doesn't help IMO. The rod is rigid (and massless). In order for it to have contact on both sides it would need to be compressed. In the limit as zero clearance is approached, the moment the force was applied ##B## would go to zero.
Are all the force and moment balances satisfied by the given solution? If so, then I can't see what's wrong. The solution is statically determinate. Could the rod have be put in place if it were slightly deformable?

Chestermiller said:
Are all the force and moment balances satisfied by the given solution? If so, then I can't see what's wrong. The solution is statically determinate. Could the rod have be put in place if it were slightly deformable?
The rod without the applied force is statically indeterminant. What are the normal forces on the rod before the load is applied?

The following seems ok as a rigid bar statics problem (EDIT: actually, even this is problematic without friction):

The problem as stated is inconsistent with its own assumptions. Sure it yields "an answer", but I don't buy its uniqueness.

erobz said:
The following seems ok as a rigid bar statics problem (EDIT: actually, even this is problematic without friction):

View attachment 332094
The problem as stated is inconsistent with its own assumptions. Sure it yields "an answer", but I don't buy its uniqueness.
erobz said:
The following seems ok as a rigid bar statics problem (EDIT: actually, even this is problematic without friction):

View attachment 332094
The problem as stated is inconsistent with its own assumptions. Sure it yields "an answer", but I don't buy its uniqueness.
I defer to you. You seem very confident.

Chestermiller said:
You seem very confident.
I'm going to try to make my point once more, assuming I have not been doing that well.

Let's pretend it the rod is in there touching both ##A## and ##B##, we don't care how it got there, and we are not yet applying any load to the end.

When we start to apply the load in question to the end, the normal force at ##B## is going to decrease in magnitude, and the normal force at ##A## is going to increase. So if its calculated that ##N_B = 327 \rm{N} ## after the load is applied (using pure statics), it has then necessarily decreased from some larger value it previously had at loads between zero and ##F##. What value did it have before we applied the load? The only way we can specify that initial value from which it decreased is to specify how much the rod segment ##AB## was initially compressed to get it into the slot (given the properties of the rod). Given a final load of ##F## ( and a rigid bar), the normal force at ##B## must depend on the value it had before the load was applied. The idea that it inexplicably ends at ## 327 \rm{N} ## whether it was put in into the slot under ## 0.001 \rm{N}## of compression or ## 1000 \rm{N} ## has to be a failure of the assumptions.

erobz said:
I'm going to try to make my point once more, assuming I have not been doing that well.

Let's pretend it the rod is in there touching both ##A## and ##B##, we don't care how it got there, and we are not yet applying any load to the end.

When we start to apply the load in question to the end, the normal force at ##B## is going to decrease in magnitude, and the normal force at ##A## is going to increase. So if its calculated that ##N_B = 327 \rm{N} ## after the load is applied (using pure statics), it has then necessarily decreased from some larger value it previously had at loads between zero and ##F##. What value did it have before we applied the load? The only way we can specify that initial value from which it decreased is to specify how much the rod segment ##AB## was initially compressed to get it into the slot (given the properties of the rod). Given a final load of ##F## ( and a rigid bar), the normal force at ##B## must depend on the value it had before the load was applied. The idea that it inexplicably ends at ## 327 \rm{N} ## whether it was put in into the slot under ## 0.001 \rm{N}## of compression or ## 1000 \rm{N} ## has to be a failure of the assumptions.
You are saying that AB was preloaded when it was put into place, and the amount of preload was unspecified. If we specify the amount of preload, then is it possible to determine the load at C without any force? and/or the load at C and the load where the guy's hand is located? I don't know. Please see if you can specify an initial case, not the same as the case in the actual problem.

Chestermiller said:
You are saying that AB was preloaded when it was put into place, and the amount of preload was unspecified.
Yes if both points ##A## and ##B## are in contact after the load was applied , then segment ##AB## starts in a state of compression before any load was applied. If there is still ##327~ \rm{N}## of contact force after the load at the end of the bar is applied which works to decrease the normal force at ##B## ,then it certainly the contact force at ##B## was necessarily larger than ##327 \rm{N}## before the external load was applied. How much larger is an arbitrary function of just how much it was compressed to get it in there, but how much each contact force is altered by the external load would remain invariant.
Chestermiller said:
If we specify the amount of preload, then is it possible to determine the load at C without any force? and/or the load at C and the load where the guy's hand is located?
Chestermiller said:
I don't know. Please see if you can specify an initial case, not the same as the case in the actual problem.
It doesn't seem like that will be a trivial task.

erobz said:
Yes if both points ##A## and ##B## are in contact after the load was applied , then segment ##AB## starts in a state of compression before any load was applied. If there is still ##327~ \rm{N}## of contact force after the load at the end of the bar is applied which works to decrease the normal force at ##B## ,then it certainly the contact force at ##B## was necessarily larger than ##327 \rm{N}## before the external load was applied. How much larger is an arbitrary function of just how much it was compressed to get it in there, but how much each contact force is altered by the external load would remain invariant.

It doesn't seem like that will be a trivial task.

Can it be in that position without any contact loads? I would say minimally it needs contact loads at ##C## and ##B## to be in that position.

erobz said:
Can it be in that position without any contact loads? I would say minimally it needs contact loads at ##C## and ##B## to be in that position.
Why? Does it not fit geometrically, or are you considering the weight of the object?

Chestermiller said:
Why? Does it not fit geometrically, or are you considering the weight of the object?
If it was in that position without any points of contact then the problem would be as I suggested in post #14. You can't have it both ways, there is either clearance in that position or there isn't.

Furthermore, lets say there is no clearance; ##A, B ##,and ##C## are all making contact with the force magnitude of beating fairy wings. When the force is applied, the lever would tend to rotate counterclockwise, pushing the rod into ##A##, and away from ##B##, and somehow according to the results of this problem we are to believe the rod is being driven into both A and B through this action. That the normal force at ##B## grows to ##327 \rm{N}## from zero ( in the limit). Someone better call up Rod Serling, I have a new episode for the Twilight Zone.

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erobz said:
Furthermore, lets say there is no clearance; ##A, B ##,and ##C## are all making contact with the force magnitude of beating fairy wings. When the force is applied, the lever would tend to rotate counterclockwise, pushing the rod into ##A##, and away from ##B##, and somehow according to the results of this problem we are to believe the rod is being driven into both A and B through this action. That the normal force at ##B## grows to ##327 \rm{N}## from zero ( in the limit). Someone better call up Rod Serling, I have a new episode for the Twilight Zone.
So are you saying that the final answer calculated by the OP is incorrect, non-unique, or what (irrespective of how the object got in that position)?

Chestermiller said:
So are you saying that the final answer calculated by the OP is incorrect, non-unique, or what (irrespective of how the object got in that position)?
Non unique. The segment ##AB## was initially compressed (in pure statics that's already handwavy - all members are to be idealized as rigid ). In this problem the bar properties and the amount of compression are whatever they need to be, such that the Normals at ##A## and ##B## end up at their respective values. But if I changed the material of the bar for instance (and nothing else) I could make the values different (Imagine a wooden bar or a steel bar undergoing the necessary compression to fit into the slot).

In the end I would say its underspecified enough that it isn't incorrect. Maybe I'm just an a$$for ranting about it. Last edited: erobz said: Non unique. The segment ##AB## was initially compressed (in pure statics that's already handwavy - all members are to be idealized as rigid ). In this problem the bar properties and the amount of compression are whatever they need to be, such that the Normals at ##A## and ##B## end up at their respective values. But if I changed the material of the bar for instance (and nothing else) I could make the values different (Imagine a wooden bar or a steel bar undergoing the necessary compression to fit into the slot). In the end I would say its underspecified enough that it isn't incorrect. Maybe I'm just an a$$ for ranting about it.
You are not an ass, and I totally respect your abilities and judgment.

On this problem, however, I'm afraid we are going to have to agree to disagree. It seems to me that this problem is statically determinate, assuming that structure is stiff enough to resist substantial deformations. The rigid-body force and moment balances are linear in the reaction forces, and therefore must give a unique solution corresponding to the applied loading.

erobz
@erobz On another topic, have you checked out my post # 52 in the thread on a can full of viscous Newtonian fluid rolling down a ramp?

Chestermiller said:
@erobz On another topic, have you checked out my post # 52 in the thread on a can full of viscous Newtonian fluid rolling down a ramp?
I haven't given it much time yet, Its definitely over my level of training. When I do get around to trying to figure it out I'm sure I'll learn from it!

## What are the basic steps to determine the reactions on a bent bar levering between two surfaces?

The basic steps include: 1) Drawing a free-body diagram of the bent bar, 2) Identifying all forces acting on the bar, including applied loads, reaction forces, and moments, 3) Applying equilibrium equations (sum of forces in horizontal and vertical directions, and sum of moments) to solve for the unknown reactions.

## How do you account for the angle of the bent bar in the calculations?

The angle of the bent bar affects the direction of the reaction forces. It is essential to resolve all forces into their horizontal and vertical components using trigonometric functions (sine and cosine) based on the given angle. These components are then used in the equilibrium equations.

## What are the common assumptions made in these types of statics problems?

Common assumptions include: 1) The bar is rigid and does not deform, 2) The surfaces are frictionless unless stated otherwise, 3) The weight of the bar is either negligible or uniformly distributed, and 4) The reactions at the supports are perpendicular to the surfaces unless friction is considered.

## How do you handle multiple supports or points of contact in the analysis?

When dealing with multiple supports or points of contact, each point of contact will have its own reaction force. You need to include all these reaction forces in your free-body diagram and apply the equilibrium equations to solve for each unknown reaction. The system of equations may become more complex, but the fundamental principles remain the same.

## Can you explain the role of moments in determining reactions on a bent bar?

Moments play a crucial role in determining reactions because they account for the rotational effects of forces. By taking moments about a specific point (usually one of the supports or points of contact), you can eliminate one or more unknown reaction forces from the equations, simplifying the problem. The sum of moments about any point must equal zero for the system to be in equilibrium.

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