In the formula of sliding friction f = μ n, can μ be greater than 1?

In the formula of sliding friction f = μ n, can μ be greater than 1?

Yes, see the following Citation:
Is the coefficient of sliding friction always less than 1-
Friction is divided into two categories: dry friction and wet friction. The friction between solid surfaces is called dry friction (the friction discussed in middle school mostly belongs to this). The friction coefficient μ of dry friction is generally less than 1, but there are also more than 1. Here are the static friction coefficients between several materials:
Wood and wood: 0.30-0.70
Leather and metal: 0.30-0.60
Aluminum and aluminum: 1.10-1.70
The sliding friction coefficient is only slightly smaller than the static friction coefficient, so it can also be greater than 1. Because of the nature of friction, there are various theories: concave convex collision theory, adhesion theory, electromagnetic force theory. All kinds of reasons for friction should be both, not a single reason. It is found that the friction between two very rough surfaces will be very large, which may be the main reason for the collision, grinding the surface smooth, The friction force will decrease, but the friction force between the two surfaces which are polished to be quite smooth (so-called "super smooth") will increase greatly. This may be because the atoms on the surface of the two surfaces are quite close to each other, which causes the electromagnetic force between the atoms to act. It has also been reported that the dry friction coefficient between the two surfaces can reach as much as 50, In middle school, the friction coefficient is less than 1
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For static friction and sliding friction, can the formula be FF = UFN (pressure), don't mislead people!

The static friction force is not calculated in this way

Is static friction related to pressure?

Static friction is the friction when the object is still. If the static friction is related to the pressure on the contact surface, the greater the pressure is, the greater the horizontal static friction is. Then physics does not move under the action of static friction (obviously impossible). The size of static friction has nothing to do with the pressure, so it is important to look at the force analysis

Is static friction related to positive pressure?

The maximum static friction is related to the positive pressure

The point m (4,2) f is the focus of the parabola y = 4x. Find a point P on the parabola, which is the minimum of | PM | + | PF | and find the coordinates of the point P at this time, and find the minimum

M is in the parabola
Then make Mn vertical guide line through M, x = - 1
Defined by parabola
The distance from P to the guide line is equal to the distance to the focus
So | PM | + | PF | = distance from P to guide line + PM
Obviously, when p is the intersection of Mn and parabola, it is minimum
In this case, the ordinate of P and m are equal,
Y = 2, so x = 1
So p (1,2)
Minimum distance = Mn = 4 - (- 1) = 5

When x satisfies, the value of algebraic formula 3 (1 + 3x) is greater than - 1 and not greater than 6?

-1＜3（1+x)≤6
-1/3＜1+x≤2
-4/3＜x≤1

When the line L passes through the point (0, - 4), make the tangent PA, Pb of the circle C: X & # 178; + Y & # 178; - 2Y = 0 at a point P on the line, (a, B are on the circle C), if the minimum area of the quadrilateral PACB is 2, find the slope of the line L

Spacb=cb*pb=2
Because CB = r = 1
So Pb = 2
pc=√1^+2^2=√5
When the area is minimum, because the length of CB is radius, Pb must be minimum. At this time, PC is perpendicular to the line L
So PC is the distance between the center of the circle C and the line L
Let the slope of the line l be K, and write the equation kx-y-4 = 0
pc=5/√1+k^2=√5
We get k = 2 or - 2

The straight line passing through point P (3,2m) and point Q (m, 2) is parallel to the straight line passing through point m (2, - 1) and point n (- 3,4)

PQ slope = (2-2m) / (M-3)
Mn slope = (4 + 1) / (- 3-2) = - 1
The two lines are parallel and have the same slope
∴(2-2m)/(m-3)=-1
2-2m=3-m
m=-1

In rectangular paper ABCD, ab = 3cm, BC = 4cm. Now fold and flatten the paper so that a and C coincide. If the crease is EF, the area of the overlapping part △ AEF is equal to______ ．

Let AE = x, from the folding, EC = x, be = 4-x, in RT △ Abe, AB2 + be2 = AE2, that is 32 + (4-x) 2 = X2, the solution is: x = 258, from the folding, we can know ∠ AEF = ∠ CEF, ∵ ad ‖ BC, ∵ CEF = ∠ AFE, ∵ AEF = ∠ AFE, that is AE = AF = 258, ∵ s △ AEF = 12 × AF × AB = 12 × 258 × 3 = 7516

(the third power of 12ax - 27ax) divided by 3ax

(the third power of 12ax - 27ax) divided by 3ax
=The third power of 12ax divided by 3ax-27ax divided by 3ax
=4x²-9