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P ⋅ h > a ⋅ W → P > a/h ⋅ Wstockzahn said: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 said:P ⋅ h > a ⋅ W → P > a/h ⋅ W
why overturning will occur ?
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 said: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.
haruspex said: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.
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 .stockzahn said: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 said: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 .
goldfish9776 said: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.
1.)why the net moment is negative , it would go 'into' the ground?stockzahn said:Considering the moments around point B
P ⋅ h → clockwise
W ⋅ a → counterclockwise
Case 1:
ΣMB = 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)
ΣFx = P - Fkmax = 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:
ΣMB = P ⋅ h - W ⋅ a > 0 (in this situation the block starts turning → N (and also Fk) will positioned at the right bottom edge (point B), as the bottom starts to turn around B)
ΣFx = P - Fk = P - N ⋅ μ = P - W ⋅ μ = 0 (the force Fk < Fkmax, 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 Fkmax (= N ⋅ μ0 = P - W ⋅ μ0), the block doesn't start sliding as the sum of the forces in horizontal direction is zero.
goldfish9776 said:1.)why the net moment is negative , it would go 'into' the ground?
goldfish9776 said:2.)is it possible that the object start to turn/overturn while it's not sliding ?
goldfish9776 said:3.)is it wrong that the author doesn't take the N.a into calculation ?
why when If μ0 > a/h the net moment around B will be > 0?stockzahn said: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 > Fkmax → 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.
stockzahn said:Considering the moments around point B
P ⋅ h → clockwise
W ⋅ a → counterclockwise
Case 1:
ΣMB = 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)
ΣFx = P - Fkmax = 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:
ΣMB = P ⋅ h - W ⋅ a > 0 (in this situation the block starts turning → N (and also Fk) will positioned at the right bottom edge (point B), as the bottom starts to turn around B)
ΣFx = P - Fk = P - N ⋅ μ = P - W ⋅ μ = 0 (the force Fk < Fkmax, 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 Fkmax (= N ⋅ μ0 = P - W ⋅ μ0), the block doesn't start sliding as the sum of the forces in horizontal direction is zero.
goldfish9776 said:why μ0 is used in case 1? and μ is used in case 2? the value of μ remain constant all the times , right ?
goldfish9776 said:why when the object is turning , the N will shift to point B ? but not beneath W ?
goldfish9776 said: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 said:μ0 shall be the static friction factor which remains constant all the time, using it it's possible to calculate the maximal friction force Fkmax. If P < Fkmax the sum of all forces in horizintal direction is ΣFx = 0 = P - Fk → P = Fk → P = N ⋅ μ = W ⋅ μ. If instead μ0 would be used, then the block would accelerated to the left (because P < Fkmax). 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).
goldfish9776 said: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?
how did u come to the equation μ0 = a/h ?stockzahn said: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 > Fkmax → no sliding, but turning
goldfish9776 said:how did u come to the equation μ0 = a/h ?
how do u come up with the equation W ⋅μ ⋅ h ?stockzahn said:ΣFx = P - Fkmax = P - N ⋅μ = P - W ⋅μ → P = W ⋅μ
ΣMB = 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 > Fkmax → μ0 < a / h ... sliding
P = Fk = N ⋅ μ = W ⋅ μgoldfish9776 said:how do u come up with the equation W ⋅μ ⋅ h ?
goldfish9776 said:If μ < μ0 → μ0 > a / h ... turning
goldfish9776 said:If P > Fkmax → μ0 < a / h ... sliding
ok, understand all now . Thanks for your effort.stockzahn said:P = Fk = 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 > Fkmax = N ⋅ μ0 = W ⋅ μ0 and ΣMB = 0 = P ⋅ h - W ⋅ a → P ⋅ h = W ⋅ a ⇒
W ⋅ a / h > W ⋅ μ0 → μ0 < a / h
goldfish9776 said: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)
how do u knw that when the block is turned counterclockwise, it's falling downwards?stockzahn said:that the block will turn around point B counterclockwise
goldfish9776 said:how do u knw that when the block is turned counterclockwise, it's falling downwards?
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 said: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 said: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 ?
A moment at a point is a measure of the tendency of a force to cause rotation around a specific point. It is calculated by multiplying the magnitude of the force by the perpendicular distance from the point to the line of action of the force.
The moment at a point is related to an object sliding by the force of friction. As an object slides, frictional forces act on it at a point of contact. These forces create a moment at that point, causing the object to rotate.
The factors that affect the moment at a point when an object is sliding include the magnitude of the force, the distance from the point to the line of action of the force, and the coefficient of friction between the object and the surface it is sliding on.
The moment at a point can be calculated by multiplying the magnitude of the force by the perpendicular distance from the point to the line of action of the force. This can be represented by the formula M = F x d, where M is the moment, F is the force, and d is the distance.
The moment at a point when an object is sliding is significant because it determines the rotational motion of the object. It also affects the stability and equilibrium of the object, as well as the amount of energy required to keep the object in motion.