Mechanics

Contents

Applications

# Static Equilibrium

### 1. Definition

For Static system, to be at equilibrium, The net force and the total moment of a system are zero.

ΣF = Fnet = 0
Σ M(F)/any-axis = 0

The total moment is defined as:

Σ M(F)/axis = Σ Mx(F)/axis i + Σ My(F)/axis j + Σ Mz(F)/axis k

Therefore:

Σ Mx(F)/axis = 0
Σ My(F)/axis = 0
Σ Mz(F)/axis = 0

The following relationships are true:

ΣF = Fnet = 0
Fnet = Fnet-x + Fnet-y = 0
Fnet-x = ΣF (over x) = 0
Fnet-y = ΣF (over y) = 0

Σ M(F)(over x)(about z) = 0
Σ M(F)(over y)(about z) = 0

### 2. Example

Two forces F1 and F2 are exerted on the rod at the distance d1 and d2 from the fulcrum respectively.
We have the following formula:

ΣF = Fnet = F1 + F2 = 0

The rod is at equilibrium on a fulcrum. Its mass is of 0.5 kg, d = 20 cm, d1 = 30 cm. Where to exert a force F2 to have F1 = F2 = 10.0 N, and stay at equilibrium?

Fnet-x = ΣF (over x) = 0 + 0 + 0 + 0 = 0

Fnet-y = ΣF (over y) = 0
- F2 - W - F1 + R = 0
Then
R = F2 + W + F1
With F2 = 2F1, we have
R = W + 2F1
= 0.5 9.8 + 2 10.0 = 24.9 N

R= 24.9 N

Σ M(F)(z-axis-fulcrum) = 0

M(- F2) + M(- W) + M(- F1) + M(R) = 0 d2 F2 - d W - d1 F1 + 0 = 0
Then:
d2 F2 - d1 F1 = d W
With F2 = 2 F1, we have:
2 F1 d2 = d W + F1 d1
d2 = (d W / 2F1)+ d1/2

d2 = (d W /2 F1) + d1/2

0.20 0.5 9.8 / 2 10.0) + 0.30/2 = 0.20
d2 = 20 cm

d2 = 20 cm

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