# Linking moments with friction force for a sliding block

• ols500
In summary: COG?It is important to check to see if the sum of the moments about the CoG is zero in order to determine the direction of the moment. In this case, it appears that the moments about the COG are in opposite directions, which results in a counterclockwise moment about the COG.
ols500
Homework Statement
Moments
Relevant Equations
Moment= Force x Distance

Hi so with this question, I am really confused. Even from the start. Like it doesn't tell us where the force is acting on. From top of block, middle of block etc? And I know how to determine the direction of moment for something on a pivot, not in this scenario though. The answer is A. They did 200N x 0.75M = 150NM and somehow this is clockwise.

Hi @ols500 and welcome to PF.

You have correctly drawn the force of friction ##\vec f##. Draw the position vector ##\vec r## from the CoG to the point of application ##\vec f## and find the torque ##\vec r \times \vec f## the usual way. Does it matter exactly where at the bottom of the block the force is applied for this calculation?

Hi, what's torque? I am doing A level physics, i don't think we have been taught that or the r vector?

Torque is another word for moment. What were you taught about calculating moments? What do you need to know in order to figure out their magnitude and direction?

Ah okay thanks, so for moments its = force x perpendicular distance from line of action to the pivot. For direction, here's an example I know:

And for the principle of moments Sum of CWM= Sum of ACWM for a body in equilibrium

ols500 said:
Ah okay thanks, so for moments its = force x perpendicular distance from line of action to the pivot. For direction, here's an example I know:

View attachment 300856
Great. You have two forces acting on the sliding block, the pushing force ##P## and the force of friction ##f##. You are not asked anything about the moment of ##P## so you don't care where it is applied. To find the magnitude moment of the frictional force ##f##, you need
1. the size (magnitude) of ##f##
2. the perpendicular distance from the CoG to where ##f## is applied.
Do you know these? What are they and why?

To find the direction of this moment, ask yourself this question, "if ##f## were the only force acting on the block, would the block rotate clockwise or counterclockwise?" That is always a good question to ask when trying to figure out the direction of moments.

Lnewqban
ols500 said:
...
Hi so with this question, I am really confused. Even from the start. Like it doesn't tell us where the force is acting on. From top of block, middle of block etc? And I know how to determine the direction of moment for something on a pivot, not in this scenario though. The answer is A. They did 200N x 0.75M = 150NM and somehow this is clockwise.
Welcome!
What else is still confusing you?

Where the force of 200 N is acting on is not important in this particular case.
What is important is the result of the application of that force: horizontal movement toward the right, with constant velocity.

Therefore, we can now that the friction force is of equal magnitude and opposite direction, which of course is located on the bottom surface of our block.

You have learned how to determine the direction of moment for something on a pivot; nevertheless, moment can still exist for things that do not have a pivot, or that are restricted from rotating.

This is the case of our block, it is not rotating or falling forward.
If it would, it would rotate about its center of gravity.
We still have a "latent" moment that growing big enough, could induce that rotation.
That moment is the friction force x perpendicular distance from line of action (rough surface) to the pivot (center of gravity).

ols500 said:
Homework Statement:: Moments
Relevant Equations:: Moment= Force x Distance

it doesn't tell us where the force is acting on. From top of block, middle of block etc?
As others have advised, you do not need to know where on the block the force is applied since you do know where on the block the frictional force acts. However, you will need to assume the applied force is horizontal: that should have been stated.

jbriggs444
Lnewqban said:
This is the case of our block, it is not rotating or falling forward.
If it would, it would rotate about its center of gravity.
Thank you this made sense now, like if the force pushing it forward suddenly stopped, the frictional force would "rotate" it forward, so from the questions point of view this is clockwise.

Lnewqban
Hi so would this be a correct diagram?

Not quite. Does it look like the sum of the moments about the CoG is zero? You are doing more than the problem asked, so this has become a learning experience.

Ah okay. Yes sum of moments about COG is 0

ols500 said:
Ah okay. Yes sum of moments about COG is 0
How do you figure? It looks like both of them are clockwise so they reinforce each other, not cancel. Check it out. You need a counterclockwise moment about the CoG and a third force and a third force to generate it. What could that be? Clearly it's not the weight because that's at the CoG by definition.

kuruman said:
a third force and a third force to generate it.
You meant a third and a fourth force, yes?

Yes.

im so confused now

ols500 said:
im so confused now
Sorry about the confusion. We know that the block is moving at constant velocity. This means that the sum of all the forces acting on it must be zero and the sum of all the moments relative to the CoG (or any other point) must be zero. In the horizontal direction, you have (a) the pushing force ##P## and (b) the force of friction ##f##. In the vertical direction, you have (c) the weight ##W## and (d) the normal force ##N##.

Can you produce a drawing, roughly to scale, that shows all these forces such the sum of all the forces and the sum off all the moments is zero?

no its my fault not understanding!

The forces balance and that's easy to figure out. The hard part is to balance the moments about the CoG.
Note that
The moments from "W" and "Normal" and "P" are zero because their line of action goes through the CoG.
The moment from "friction" is clockwise and non-zero.
It follows that he block should tip about the CoG if you have drawn the picture correctly. The block is just moving along without tipping therefore something in your picture is incorrectly drawn. What could that be? I could tell you but I think this could be an "aha moment" of discovery for you so I will guide your thinking.
1. The magnitudes of the horizontal forces to the right and to the left are drawn equal. Check.
2. The magnitudes of the vertical forces to the right and to the left are drawn equal. Check.
You want to balance the moments and you have already established that the magnitudes are drawn correctly to balance the forces. This leaves the points of application of each force.
3. The point of application of "W" is at the CoG by definition.
What about the points of application of the remaining 3 forces? Can you justify why you drew them where they are? Hint: Start by thinking why it is easy to tip over a book by pushing on it when it's standing upright and doesn't tip at all but just moves when it's lying on its side.

kuruman said:
Start by thinking why it is easy to tip over a book by pushing on it when it's standing upright and doesn't tip at all but just moves when it's lying on its side
Is it because the weight force lies outside the base area of the object so it tips over. With a book vertical the base area is much smaller so much less angle can be applied before it tips over.

ols500 said:
Is it because the weight force lies outside the base area of the object so it tips over. With a book vertical the base area is much smaller so much less angle can be applied before it tips over.
Angles are not applied, forces are. Let me try a different approach. When force P is pushing on the book so that it is tipped at an angle but still in equilibrium on one of its edges, how and where would you draw the four forces? Please show a new drawing like the one you have above but with the block tipped.

Here:

Very good. Now please look at the forces that you have drawn and tell me the direction of their moments about the CoG. For each force, choose one of
(a) clockwise
(b) counterclockwise
(c) no direction - the moment is zero.

ols500 said:
Is it because the weight force lies outside the base area of the object so it tips over.
Not so fast; there's a stage before that. If a small horizontal force is applied at the middle it won't be enough to tip it over, but the torques need to balance.
In your diagram you made an unjustified assumption when you drew the normal force. Can you think of a small change to it which would counter the torque from the horizontal forces?

haruspex said:
Which diagram? If you mean #19, in my intended incremental approach, correcting it would have been the step after understanding the adjustment to the normal force that was made in #23 to balance the moments. The lesson to be learned from #23 is that yes, it is OK to move the line of action of the normal force in order to achieve equilibrium. The difference between #23 and #19 is that in the former one is forced to draw the normal force at the only point that is in contact with the surface. Having made this realization, one might start wondering how to fix #19 in which the block cannot be in equilibrium as drawn and there are infinitely many points of contact where the normal force can be applied.

kuruman said:
it would have been the step after understanding the adjustment to the normal force that was made in #23 to balance the moments.
Ok, but I would have taken it in the other order. First, what happens when a small horizontal force is applied, then what critical point is reached as it is increased.

haruspex said:
Ok, but I would have taken it in the other order. First, what happens when a small horizontal force is applied, then what critical point is reached as it is increased.
I see your point and here is mine. I think there is an understandable beginners' preconception at work here. In the standard curriculum, torques come after forces and for good reason. Until torques are introduced, blocks in FBDs are drawn as extended masses, even though they are treated as point masses. For that reason, no effort is made to draw the normal force where it belongs. Diagrams such as in #19 are routinely shown in textbooks with a wink and a nod and nobody bats an eyelash (the pun is intentional), because why should one worry about drawing them correctly when it makes no difference to the linear application of the second law?

The underside of this is that beginners acquire the preconception that when an extended block is pushed horizontally across a rough surface, the normal force acts at a point directly underneath the CoG. I agree that drawing it where it belongs is an unnecessary and often confusing complication in a linear dynamics FBD. However, every time a beginner sees incorrectly drawn normal forces in linear dynamics FBDs, the preconception is reinforced. By the time one reaches rotational dynamics and has to worry about the point of application of a force on an extended body, the location of the normal force has found a permanent spot under the CoG.

That is why I think it is important to first bust the belief that the normal force is immovable and then see how it can be moved to account for the observed acceleration. As educators, we owe that much to the students after leading them down the garden path with our acquiescence.

Last edited:

## 1. What is the relationship between moments and friction force for a sliding block?

The relationship between moments and friction force for a sliding block can be described by the principle of moments, which states that the sum of the clockwise moments is equal to the sum of the counterclockwise moments. In the case of a sliding block, the friction force acts in the opposite direction of the motion and creates a moment that must be balanced by an equal and opposite moment.

## 2. How does the coefficient of friction affect the moments and friction force for a sliding block?

The coefficient of friction, which is a measure of the frictional force between two surfaces, directly affects the magnitude of the friction force for a sliding block. A higher coefficient of friction will result in a larger friction force and therefore a larger moment that must be balanced by the applied force.

## 3. Can moments and friction force be calculated for a sliding block on an inclined plane?

Yes, the principles of moments and friction force can be applied to a sliding block on an inclined plane. The weight of the block and the angle of the incline will affect the magnitude and direction of the friction force, which in turn affects the moments acting on the block.

## 4. How do external forces impact the moments and friction force for a sliding block?

External forces, such as applied forces or weight, can create additional moments that must be balanced by the friction force for a sliding block. These external forces can also affect the direction and magnitude of the friction force, and therefore the overall moments acting on the block.

## 5. What are some real-world applications of understanding the relationship between moments and friction force for a sliding block?

Understanding the relationship between moments and friction force for a sliding block is important in many engineering and physics applications. For example, it is crucial in designing and analyzing structures and machines that involve sliding or rolling components, such as conveyor belts, gears, and brakes. It is also important in understanding the stability and equilibrium of objects on inclined planes or surfaces with friction, such as a car driving on a curved road or a book sliding down a ramp.

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