MACE 11010: Engineering Mechanics
Lecture 2, Equilibrium of Many-Particle System: Force
Lecture 3, Equilibrium of Many-Particle System: Moment
5/10, Friday
Some mathematics for N-particle system
For a system of N particles at equilibrium, applying Newton’s 1st law on the i-th particle
Fi + ∑ fij = 0 ,
(1)
j ≠i
where i = 1, 2, , N , Fi is the external force acted on the i-th particle, f ij is the internal force
acted on particle i from particle j,
∑f
ij
means the summation for all the particle j except when
j ≠i
j = i.
Summation of the above equation over all particles, i = 1, 2, , N , gives
∑ F + ∑∑ f
i
i
i
ij
=0,
(2)
j ≠i
From Newton’s 3rd law, it can be proved (can you prove?) that
∑∑ f
i
ij
=0,
(3)
j ≠i
Therefore, from Eq. (2), one has
∑F =0,
(4)
i
i
which means for a system of N particles at equilibrium, the resultant force
∑F
i
(a vector)
i
equals 0.
Further, choose a fixed point O in the space, the position of particle i to point O is denoted
as ri . The cross product of ri to equation (1) for each particle i gives
1
ri × Fi + ∑ ri × f ij = 0 .
(5)
j ≠i
Summation of the above equation over all particles, i = 1, 2, , N , gives
∑ r × F + ∑∑ r × f
i
i
i
i
i
ij
=0.
(6)
j ≠i
From Newton’s 3rd law, it can be proved (can you prove?) that
∑∑ r × f
i
i
ij
=0.
(7)
j ≠i
Therefore, from Eq. (6), one has
∑r ×F = 0 ,
i
(8)
i
i
where ri × Fi , defined as M i , is the moment of the external force on particle i to the fixed point
O. For a system of N particles at equilibrium, the resultant moment
∑M
i
(a vector) equals 0,
i
i.e.,
∑M
i
=0,
(9)
i
To summary, for a system of N particles at equilibrium,
∑F = 0,
(10)
i
i
∑r ×F = 0 ,
i
(11)
i
i
2