How to determine the direction of friction in physics?

How to determine the direction of friction in physics?

The action effect of friction is to hinder the relative movement or relative movement trend between objects, not the movement of objects. The direction of sliding friction is not necessarily opposite to the movement direction of objects. When the relative movement trend direction is perpendicular to the movement direction, the direction of static friction is perpendicular to the movement direction. Friction is a tangential force, so it is parallel to the contact surface

How to determine the friction direction of bicycle wheels/

Under normal circumstances, the friction is opposite to the forward direction. If the bicycle wheel is forward, the friction with the surface is backward

How to determine the magnitude and direction of friction?

Direction:
Contrary to the relative trend of motion, note that not all are in the opposite direction of motion
example:
In the opposite direction: the slope slides
Same direction: walking
size:
The sliding friction is related to the roughness of the contact surface and the pressure
Rolling friction is smaller than sliding friction
The static friction force is the same as the force horizontal to the contact surface

The conditions for friction are ___, __

The conditions of friction are: ① two objects are in contact with each other; ② Having or tending to move relatively
So the answer is: two objects are in contact with each other; Having or tending to move relatively

Formula of inscribed circle radius of right triangle: r = (a + B-C) / 2 how is this formula derived?

Let RT △ ABC, ∠ C = 90 degrees, BC = a, AC = B, ab = C
The conclusion is that the radius of inscribed circle r = (a + B-C) / 2
There are generally two methods of proof:
Method 1:
As shown in the figure, the center of the inscribed circle is O, and the three tangent points are D, e and F, connecting od and OE
Obviously, there is OD ⊥ AC, OE ⊥ BC, OD = OE
So the quadrilateral CDOE is a square
So CD = CE = R
So ad = B-R, be = A-R,
Because ad = AF, CE = CF
So AF = B-R, CF = A-R
Because AF + CF = AB = R
So B-R + A-R = R
Radius of inscribed circle r = (a + B-C) / 2
That is, the diameter of inscribed circle L = a + b-c
Method 2:
As shown in the figure, the center of the inscribed circle is O, and the three tangent points are D, e and F, connecting OD, OE, of, OA, OB and OC
Obviously, there are OD ⊥ AC, OE ⊥ BC, of ⊥ ab
So s △ ABC = s △ OAC + s △ OBC + s △ OAB
So AB / 2 = br / 2 + Ar / 2 + Cr / 2
So r = AB / (a + B + C)
＝ab(a＋b－c)/(a＋b＋c)(a＋b－c)
＝ab(a＋b－c)/[(a＋b)^2－c^2]
Because a ^ 2 + B ^ 2 = C ^ 2
Therefore, the radius of inscribed circle r = (a + B-C) / 2
That is, the diameter of inscribed circle L = a + b-c

Why is the radius of inscribed circle of right triangle (a + B-C) / 2? 2S / (a + B + C) I understand Shouldn't it be (AB) / (a + B + C) instead of a right triangle? How is (AB) / (a + B + C) equal to (a + B-C) / 2? Please give me your advice

(ab)/(a+b+c)
=[(a+b)^2-a^2-b^2]/2(a+b+c)
=[(a+b)^2-c^2]/2(a+b+c)
=(a+b+c)(a+b-c)/2(a+b+c)
=(a+b-c)/2