Opruge u seriji i paralelno

Article about the Opruge u seriji i paralelno

1. Springs in series

If the spring is connected in series, as in the figure on the side, then:

1. The increase in the length of spring = the increase in length 1 + the increase in length 2

Δy = Δy₁ + Δy₁

2. The force experienced by equivalent spring = the force experienced by spring 1 = the force experienced by spring 2

F_s =F₁ =F₂

3. The equivalent spring’s constant (k_s)

1/k_s = 1/k₁ + 1/k₂

Primjer problema 1:

Two identical springs each have a constant of 100 N / m connected in series. If the spring’s arrangement is given a load so that it increases 4 cm in length, then the increase in the length of each spring is …

Rješenje:

The total increase in the length of the two springs is 4 cm, therefore the increase in the length of each spring is 2 cm.

2. Springs in parallel

If the spring is connected in parallel, as in the figure on the side, then:

1. The increase in the length of the equivalent spring = the increase in the length of spring 1 = the increase in the length of spring 2

Δy = Δy₁ + Δy₁

2. The force experienced by the equivalent spring = the force that is experienced by spring 1 + the force experienced by spring 2

F_s =F₁ + F₂

3. The equivalent spring’s constant (k_p)

kp = k₁ +k₂

Primjer problema 2:

Two springs each with constant c arranged in parallel. The konstanta opruge of this arrangement becomes …

Rješenje:

The equivalent spring’s constant (k_p) = c + c = 2c