The parallelogram rule and triangle rule of proving force in theory

The parallelogram rule and triangle rule of proving force in theory

Vector has size and direction. Parallelogram rule and triangle rule are part of vector rule. Vector can be represented by line segment with arrow, which is similar to free vector. If the object is balanced by force, it can form a closed figure. Parallelogram rule and triangle rule are the embodiment of balance of three forces

In the experiment to verify the parallelogram rule of force, the angle between two springs should be 90 degrees,

No, since it is the parallelogram rule to verify the force, the two springs should be the two sides of the parallelogram

The parallelogram rule requires two inner triangles or something

Two equal triangles

The velocity composition of two uniform motions with different directions follows the parallelogram rule. What is the parallelogram rule? How is it applied?

The parallelogram rule of force is selected from the composition rule of common point force in practical complete book of middle school teaching. This rule is usually expressed as: make a parallelogram with the directed line segment representing two common point forces as the adjacent side, and the diagonal line between the two adjacent sides represents the size and direction of the resultant force of two forces

Finding definite integral DX / (1 + (cosx) ^ 2) Pai / 2

∫(0->π/2) dx/(1+cos²x)= ∫(0->π/2) dx/(sin²x+cos²x+cos²x)= ∫(0->π/2) dx/(sin²x+2cos²x)= ∫(0->π/2) sec²x/(tan²x+2)= ∫(0->π/2) d(tanx)/(2+tan²x)= (1...

It is proved that definite integral (0 to 2 π) f (| cosx |) DX = 4 definite integral (0 to π / 2) f (cosx) DX

Indefinite integral of ∫ (3sinx + 2 / cos square x) DX

∫ 3sinx dx = －3 cosx + C
∫ 2sec²x dx = 2tanx + C
The original formula = - 3cosx + 2tanx + C

Solving definite integral ∫ (2,0) 1 / (1-x ^ 2) DX

=∫(0,1)1/(1-x^2) dx+∫(1,2)1/(1-x^2) dx
Because the integral ∫ (0,1) 1 / (1-x ^ 2) DX diverges, the original integral diverges

Definite integral ∫ [- 1,1] DX / [1 + 2 ^ (1 / x)],

Let x = - y、dx = - dy
N = ∫(- 1→1) 1/[1 + 2^(1/x)] dx
= ∫(1→- 1) 1/[1 + 2^(- 1/y)] * (- dy)
= ∫(- 1→1) 1/[1 + 2^(- 1/x)] dx
= ∫(- 1→1) 1/[1 + 1/2^(1/x)] dx
= ∫(- 1→1) 2^(1/x)/[2^(1/x) + 1] dx = N
∵
N + N = ∫(- 1→1) 1/[1 + 2^(1/x)] dx + ∫(- 1→1) 2^(1/x)/[2^(1/x) + 1] dx
2N = ∫(- 1→1) [1 + 2^(1/x)]/[1 + 2^(1/x)] dx = ∫(- 1→1) dx = 2
→ N = 1
∴∫(- 1→1) 1/[1 + 2^(1/x)] dx = 1

Finding the definite integral | 2-x | DX from 0 to 4

Definite integral from 0 to 4 | 2-x | DX = definite integral from 0 to 2 (- 2 + x) DX + definite integral from 2 to 4 (2-x) DX = - 2 + 2 = 0