Moment at a point when the object is sliding

[goldfish9776]

Homework Statement

i can't understand why when the total moment at B is positive , the object is not sliding ? can someone help to explain ?

Homework Equations

The Attempt at a Solution

 

[CWatters]

I agree there is something wrong with the note.

 

[stockzahn]

Dependent on h, a and μ, if you increase the force P, the block will either start to slide or overturn, depending on which of the following cases will be reached first:

Case 1: P > μ ⋅ W (horizontal forces / sliding)

Case 2: P ⋅ h > a ⋅ W → P > a/h ⋅ W (momentum around B / overturning)

Imagine Case 2 is reached first: If the block starts turning what happens with the distances a and h?

Of course there is also the indifferent case μ = a/h, I didn't take into account, but in my opinion that's not scope of the note.

 

[goldfish9776]

Dependent on h, a and μ, if you increase the force P, the block will either start to slide or overturn, depending on which of the following cases will be reached first:

Case 1: P > μ ⋅ W (horizontal forces / sliding)

Case 2: P ⋅ h > a ⋅ W → P > a/h ⋅ W (momentum around B / overturning)

Imagine Case 2 is reached first: If the block starts turning what happens with the distances a and h?

Of course there is also the indifferent case μ = a/h, I didn't take into account, but in my opinion that's not scope of the note.

P ⋅ h > a ⋅ W → P > a/h ⋅ W

why overturning will occur ?

 

[stockzahn]

P ⋅ h > a ⋅ W → P > a/h ⋅ W

why overturning will occur ?

ΣM_B = P ⋅ h - W ⋅ a > 0

 

[haruspex]

Dependent on h, a and μ, if you increase the force P, the block will either start to slide or overturn, depending on which of the following cases will be reached first:

Case 1: P > μ ⋅ W (horizontal forces / sliding)

Case 2: P ⋅ h > a ⋅ W → P > a/h ⋅ W (momentum around B / overturning)

Imagine Case 2 is reached first: If the block starts turning what happens with the distances a and h?

Of course there is also the indifferent case μ = a/h, I didn't take into account, but in my opinion that's not scope of the note.

Sure, but goldfish' question is why the book says that if the net moment is positive then sliding will not occur. As you say, it can occur.

 

[stockzahn]

Sure, but goldfish' question is why the book says that if the net moment is positive then sliding will not occur. As you say, it can occur.

I tried to clarify the point, I supposed the author of the book was trying to point out, but maybe I missed it too.

 

[goldfish9776]

Dependent on h, a and μ, if you increase the force P, the block will either start to slide or overturn, depending on which of the following cases will be reached first:

Case 1: P > μ ⋅ W (horizontal forces / sliding)

Case 2: P ⋅ h > a ⋅ W → P > a/h ⋅ W (momentum around B / overturning)

Imagine Case 2 is reached first: If the block starts turning what happens with the distances a and h?

Of course there is also the indifferent case μ = a/h, I didn't take into account, but in my opinion that's not scope of the note.

the moment of P.h and a.W are counterclockwise, right? so , when let say when P.h = -8 , a.W = -5 and -aa respectively .
case 1, -8-(-5) = -3
case 2 , -8-(-11) = +3
in case 1 , since the total moment at B is negative , it will start sliding , right ? P ⋅ h > a ⋅ W is also satisfied here.
in case 2 , since the total moment at B is positive , it will not start to slide, right ? P ⋅ h <a ⋅ W here.

 

[stockzahn]

the moment of P.h and a.W are counterclockwise, right? so , when let say when P.h = -8 , a.W = -5 and -aa respectively .

Considering the moments around point B

P ⋅ h → clockwise
W ⋅ a → counterclockwise

case 1, -8-(-5) = -3
case 2 , -8-(-11) = +3
in case 1 , since the total moment at B is negative , it will start sliding , right ? P ⋅ h > a ⋅ W is also satisfied here.
in case 2 , since the total moment at B is positive , it will not start to slide, right ? P ⋅ h <a ⋅ W here.

Case 1:

ΣM_B = P ⋅ h - W ⋅ a + N ⋅ a = 0 (as the block still touches the ground with the whole bottom, N is located exactley beneath W → the block doesn't turn as the net moment is 0)
ΣF_x = P - F_kmax = P - N ⋅ μ₀ = P - W ⋅ μ₀ > 0 (the force P exceeds the maximal possible friction force → the block starts to move in horizontal direction (sliding))

The sum of the moments around B must be zero, else it would turn (and if the net moment would be negative it would turn counterclockwise, so "into" the ground)

Case 2:

ΣM_B = P ⋅ h - W ⋅ a > 0 (in this situation the block starts turning → N (and also F_k) will positioned at the right bottom edge (point B), as the bottom starts to turn around B)
ΣF_x = P - F_k = P - N ⋅ μ = P - W ⋅ μ = 0 (the force F_k < F_kmax, which means as reaction force it will have the same value as P → the block doesn't slide)

The sum of the moments is larger than 0, this means turning, but as the force P doesn't exceed the maximal possible friction force F_kmax (= N ⋅ μ₀ = P - W ⋅ μ₀), the block doesn't start sliding as the sum of the forces in horizontal direction is zero.

 

[goldfish9776]

Considering the moments around point B

P ⋅ h → clockwise
W ⋅ a → counterclockwise
Case 1:

ΣM_B = P ⋅ h - W ⋅ a + N ⋅ a = 0 (as the block still touches the ground with the whole bottom, N is located exactley beneath W → the block doesn't turn as the net moment is 0)
ΣF_x = P - F_kmax = P - N ⋅ μ₀ = P - W ⋅ μ₀ > 0 (the force P exceeds the maximal possible friction force → the block starts to move in horizontal direction (sliding))

The sum of the moments around B must be zero, else it would turn (and if the net moment would be negative it would turn counterclockwise, so "into" the ground)

Case 2:

ΣM_B = P ⋅ h - W ⋅ a > 0 (in this situation the block starts turning → N (and also F_k) will positioned at the right bottom edge (point B), as the bottom starts to turn around B)
ΣF_x = P - F_k = P - N ⋅ μ = P - W ⋅ μ = 0 (the force F_k < F_kmax, which means as reaction force it will have the same value as P → the block doesn't slide)

The sum of the moments is larger than 0, this means turning, but as the force P doesn't exceed the maximal possible friction force F_kmax (= N ⋅ μ₀ = P - W ⋅ μ₀), the block doesn't start sliding as the sum of the forces in horizontal direction is zero.

1.)why the net moment is negative , it would go 'into' the ground?

2.)is it possible that the object start to turn/overturn while it's not sliding ?

3.)is it wrong that the author doesn't take the N.a into calculation ?

 

[stockzahn]

1.)why the net moment is negative , it would go 'into' the ground?

According to the picture you posted, clockwise moments are defined as positive. If the moment around point B is negative, the block would turn counterclockwise around point B.

2.)is it possible that the object start to turn/overturn while it's not sliding ?

Yes it can and I think that is the point of the whole text you posted. If μ₀ > a/h the net moment around B will be > 0, before the force P > F_kmax → no sliding, but turning

3.)is it wrong that the author doesn't take the N.a into calculation ?

I don't know what the author wrote before the paragraph you posted, but the picture shows the instant, when the block starts to turn. The only spot the block is touching the ground with is the right bottom edge (B). That means all the reaction forces have to affect this edge, also N. But if the sum of all moments is calculated around this point, the lever of N is zero. So based on the situation in the picture the moment produced by N is zero, so it's not wrong.

 

[goldfish9776]

According to the picture you posted, clockwise moments are defined as positive. If the moment around point B is negative, the block would turn counterclockwise around point B.
Yes it can and I think that is the point of the whole text you posted. If μ₀ > a/h the net moment around B will be > 0, before the force P > F_kmax → no sliding, but turning
I don't know what the author wrote before the paragraph you posted, but the picture shows the instant, when the block starts to turn. The only spot the block is touching the ground with is the right bottom edge (B). That means all the reaction forces have to affect this edge, also N. But if the sum of all moments is calculated around this point, the lever of N is zero. So based on the situation in the picture the moment produced by N is zero, so it's not wrong.

why when If μ0 > a/h the net moment around B will be > 0?

 

[goldfish9776]

Considering the moments around point B

P ⋅ h → clockwise
W ⋅ a → counterclockwise
Case 1:

ΣM_B = P ⋅ h - W ⋅ a + N ⋅ a = 0 (as the block still touches the ground with the whole bottom, N is located exactley beneath W → the block doesn't turn as the net moment is 0)
ΣF_x = P - F_kmax = P - N ⋅ μ₀ = P - W ⋅ μ₀ > 0 (the force P exceeds the maximal possible friction force → the block starts to move in horizontal direction (sliding))

The sum of the moments around B must be zero, else it would turn (and if the net moment would be negative it would turn counterclockwise, so "into" the ground)

Case 2:

ΣM_B = P ⋅ h - W ⋅ a > 0 (in this situation the block starts turning → N (and also F_k) will positioned at the right bottom edge (point B), as the bottom starts to turn around B)
ΣF_x = P - F_k = P - N ⋅ μ = P - W ⋅ μ = 0 (the force F_k < F_kmax, which means as reaction force it will have the same value as P → the block doesn't slide)

The sum of the moments is larger than 0, this means turning, but as the force P doesn't exceed the maximal possible friction force F_kmax (= N ⋅ μ₀ = P - W ⋅ μ₀), the block doesn't start sliding as the sum of the forces in horizontal direction is zero.

why μ0 is used in case 1? and μ is used in case 2? the value of μ remain constant all the times , right ?

why when the object is turning , the N will shift to point B ? but not beneath W ?

 

[goldfish9776]

when P.h greater than W.a , it will start to turn , right? in this condition , the total moment at B= positive ... why the book gave when it is positive , the object is not sliding / overturning ? sliding and overturning are different things , right ? why the author mix them together ?

 

[stockzahn]

why μ0 is used in case 1? and μ is used in case 2? the value of μ remain constant all the times , right ?

μ₀ shall be the static friction factor which remains constant all the time, using it it's possible to calculate the maximal friction force F_kmax. If P < F_kmax the sum of all forces in horizintal direction is ΣF_x = 0 = P - F_k → P = F_k → P = N ⋅ μ = W ⋅ μ. If instead μ₀ would be used, then the block would accelerated to the left (because P < F_kmax). So the μ without the index 0 indicates, that P can still be increased without starting to move the block.

why when the object is turning , the N will shift to point B ? but not beneath W ?

Because at point B will be the only point the block is touching the ground. But you can try it yourself, take for example a tall bottle and push it close to the bottom - it will slide. If you increase the distance of your finger to the ground/desk, there will be a point, when the bottle doesn't start to slide, but turn (and fall, if you don't stop).

 

[stockzahn]

when P.h greater than W.a , it will start to turn , right? in this condition , the total moment at B= positive ... why the book gave when it is positive , the object is not sliding / overturning ? sliding and overturning are different things , right ? why the author mix them together ?

I'd say "not sliding / overturning" means "overturning and not sliding". As already mentioned I think the point of this paragraph should be to show that dependent on the height h (and the breadth a and the friction represented by μ₀), the block will either turn OR slide (except at the indifferent case μ₀ = a/h). But honestly, reading what you've posted, I'm not sure about the quality of this notice (but I only saw one paragraph, so I don't want to judge it to soon). Maybe of course, I don't get the point of the author either.

 

[goldfish9776]

μ₀ shall be the static friction factor which remains constant all the time, using it it's possible to calculate the maximal friction force F_kmax. If P < F_kmax the sum of all forces in horizintal direction is ΣF_x = 0 = P - F_k → P = F_k → P = N ⋅ μ = W ⋅ μ. If instead μ₀ would be used, then the block would accelerated to the left (because P < F_kmax). So the μ without the index 0 indicates, that P can still be increased without starting to move the block.
Because at point B will be the only point the block is touching the ground. But you can try it yourself, take for example a tall bottle and push it close to the bottom - it will slide. If you increase the distance of your finger to the ground/desk, there will be a point, when the bottle doesn't start to slide, but turn (and fall, if you don't stop).

in order to prevent the block sliding to the left before the P is large enuf to 'counter' the frictional force , so we use μ instead of μ₀ ? if we use μ₀ , the object will slide to the left , it will looks weird?

 

[stockzahn]

in order to prevent the block sliding to the left before the P is large enuf to 'counter' the frictional force , so we use μ instead of μ₀ ? if we use μ₀ , the object will slide to the left , it will looks weird?

Of course it wouldn't slide to the left, because the friction force is a reaction force. It's just the formalism for the paper, because the equations are not correct using μ₀ (as then the calculated net force would point to the left).

 

[goldfish9776]

Yes it can and I think that is the point of the whole text you posted. If μ₀ > a/h the net moment around B will be > 0, before the force P > F_kmax → no sliding, but turning

how did u come to the equation μ₀ = a/h ?

 

[stockzahn]

how did u come to the equation μ₀ = a/h ?

ΣF_x = P - F_kmax = P - N ⋅μ = P - W ⋅μ → P = W ⋅μ
ΣM_B = P ⋅ h - W ⋅ a → P ⋅ h = W ⋅ a
→ W ⋅μ ⋅ h = W ⋅ a → μ ⋅ h = a → μ = a / h

If μ < μ₀ → μ₀ > a / h ... turning
If P > F_kmax → μ₀ < a / h ... sliding

 

[goldfish9776]

ΣF_x = P - F_kmax = P - N ⋅μ = P - W ⋅μ → P = W ⋅μ
ΣM_B = P ⋅ h - W ⋅ a → P ⋅ h = W ⋅ a
→ W ⋅μ ⋅ h = W ⋅ a → μ ⋅ h = a → μ = a / h

If μ < μ₀ → μ₀ > a / h ... turning
If P > F_kmax → μ₀ < a / h ... sliding

how do u come up with the equation W ⋅μ ⋅ h ?
why
If μ < μ0 → μ0 > a / h ... turning
If P > Fkmax → μ0 < a / h ... sliding

 

[stockzahn]

how do u come up with the equation W ⋅μ ⋅ h ?

P = F_k = N ⋅ μ = W ⋅ μ

If μ < μ0 → μ0 > a / h ... turning

Case no sliding → μ ≤ μ₀ → If μ = a / h and μ ≤ μ₀ ⇒μ₀ ≥ a / h

If P > Fkmax → μ0 < a / h ... sliding

Case sliding (and no turning) → P > F_kmax = N ⋅ μ₀ = W ⋅ μ₀ and ΣM_B = 0 = P ⋅ h - W ⋅ a → P ⋅ h = W ⋅ a ⇒
W ⋅ a / h > W ⋅ μ₀ → μ₀ < a / h

 

[goldfish9776]

P = F_k = N ⋅ μ = W ⋅ μ
Case no sliding → μ ≤ μ₀ → If μ = a / h and μ ≤ μ₀ ⇒μ₀ ≥ a / h
Case sliding (and no turning) → P > F_kmax = N ⋅ μ₀ = W ⋅ μ₀ and ΣM_B = 0 = P ⋅ h - W ⋅ a → P ⋅ h = W ⋅ a ⇒
W ⋅ a / h > W ⋅ μ₀ → μ₀ < a / h

ok, understand all now . Thanks for your effort.
can you take time to explain why when total moment at B = negative , the sliding and overturning will occur at the same time by mathematical proof? if the P applied doesn't exceed W.a , how can the turning occur ?

i tried to figure out how can i prove when total moment at B = negative , the obejct is sliding and turning ...
here's what i gt :
P-Fk max >0
P> Fk max

total moment about B <0
total moment about B = P(h) -W(a)
= P(h) < W(a)
= P <W(a/h)

 

[stockzahn]

can you take time to explain why when total moment at B = negative , the sliding and overturning will occur at the same time by mathematical proof? if the P applied doesn't exceed W.a , how can the turning occur ?

i tried to figure out how can i prove when total moment at B = negative , the obejct is sliding and turning ...
here's what i gt :
P-Fk max >0
P> Fk max

total moment about B <0
total moment about B = P(h) -W(a)
= P(h) < W(a)
= P <W(a/h)

To be honest I can't for one of the follwing two reasons:

1) I don't understand what the author wants to proof/explain
2) The text is wrong/badly written

A negative net moment around B means, that the block will turn around point B counterclockwise. As long as it is on the floor, turning can't occur (except the ground breaks). So assuming the block already has started to turn affected by the force P at height h. If the block turns, h increases and a must be substituted with a ⋅ cos(θ). With increasing angle θ (and constant force P) the block will turn even faster. The only possibility to generate a negative net moment is to decrease the force P, so the block can turn back to its initial position. But if P is decreased, there won't be any sliding anyway.

Maybe the author wants to point out the case μ₀ = a / h (indifferent). That would mean sliding and turning could occur at the same time. If force P exceeds F_kmax only a little bit, both movements could be possible (even at the same time) - that also works as experiment with the tall bottle. You will find the height h to push at to see tendencies of the bottle for both movements. However, as soon as the block/bottle turned only for a little bit, it will turn faster and faster, because the lever of P increases, while the lever of W decreases. The only possibility to prevent it from falling is to decrease P. Now there is a difference between the static friction factor and the dynamic friction factor - generally the maximal friction force is higher, if the relative velocity between to bodies is zero. So this is the only way i can think about this problem:

1) The force P is positioned at h to fulfil the condition μ₀ = a / h
2) The block starts to slide and to turn at the same time - as it starts to slide the friction force decreases (because μ₀ > μ_dyn)
3) To prevent the block from falling force P is decreased (at the turning angle θ₁), the net moment becomes negative and the block turns back in his initial position
4) The decrease of P small enough, so that P > F_kdyn = W ⋅ μ_dyn

Now, for a short amount of time, we have sliding and counterclockwise turning of the block with a negative net moment (with μ_dyn < (a ⋅ cos(θ₁) / h(θ₁)) and P > W ⋅ μ_dyn ⇒μ_dyn < μ₀ ⋅ h(θ = 0) ⋅ cos(θ₁) / h(θ₁)).

But as you can see, it's a quite special situation - so probably one of the reasens mentioned in the beginning.

 

[goldfish9776]

that the block will turn around point B counterclockwise

how do u knw that when the block is turned counterclockwise, it's falling downwards?

 

[stockzahn]

how do u knw that when the block is turned counterclockwise, it's falling downwards?

Counterclockwise rotation around point B: If the bottom right edge of the block touches the floor (and that's the condition the calculations are based on), the only possibility to perform the rotation is "through" the ground.

 

[goldfish9776]

Counterclockwise rotation around point B: If the bottom right edge of the block touches the floor (and that's the condition the calculations are based on), the only possibility to perform the rotation is "through" the ground.

when the block is rotated anticlockwise, the left edge of the block touches the ground , right ? why you said it's right edge that's touches the ground ?

 

[stockzahn]

when the block is rotated anticlockwise, the left edge of the block touches the ground , right ? why you said it's right edge that's touches the ground ?

As indicated in the drawing, N and F_k affect the right bottom edge. To rotate counterclockwise, the force P has to push from the other side, otherwise there is no reason to turn around the bottom left edge. Additionally, when the force P doesn't push from the left side anymore, N and F_k would change their positions towards the middle of the block and the ΣM_B would look completely different.