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Equations
As the relationship with the action force ($F_A$) and the reaction force ($F_R$) in one dimension is
| $ F_R =- F_A $ |
it can be applied to each component of the action force (vector) ($\vec{F}_A$) and the reaction force (vector) ($\vec{F}_R$), resulting in
$\vec{F}R=(F{Rx},F_{Ry},F_{Rz})=(-F_{Ax},-F_{Ay},-F_{Az})=-\vec{F}_A$
hence
| $ \vec{F}_R = - \vec{F}_A $ |
(ID 3240)
(ID 4644)
(ID 4645)
(ID 7334)
Examples
(ID 2695)
(ID 2696)
G_r=-F_r
(ID 4642)
G_x=-F_x
(ID 4644)
G_y=-F_y
(ID 4645)
(ID 1346)
F_x=F_l\sin\theta+F_r\cos\theta
(ID 4637)
F_y=-F_l\cos\theta+F_r\sin\theta
(ID 4638)
F_l=F_x\sin\theta-F_y\cos\theta
(ID 4639)
F_r=F_x\cos\theta+F_y\sin\theta
(ID 4640)
(ID 1347)
T=\displaystyle\frac{l}{r}F_r
(ID 4641)
Para que el centro de masa no se desplace es necesario que las fuerzas radiales se compensen, es decir
$G_r+F_r=0$
(ID 7334)
G_l=\displaystyle\frac{lF_r-rF_l}{r}
(ID 4643)
Cuando un cuerpo esta en equilibrio su centro de masa no se esta desplazando. Para ello la suma de las fuerzas sobre este deben ser nula o sea con debe ser
| $\displaystyle\sum_i \vec{F}_i=0$ |
(ID 10753)
When a body is in rotational equilibrium, it does not rotate around its center of mass. To achieve this, the sum of the torques acting on it must be zero. This implies that:
| $\displaystyle\sum_i \vec{T}_i=0$ |
(ID 10754)
The relationship between the action force ($F_A$) and the reaction force ($F_R$) in one dimension:
| $ F_R =- F_A $ |
can be generalized to more dimensions with the action force (vector) ($\vec{F}_A$) and the reaction force (vector) ($\vec{F}_R$), as follows:
| $ \vec{F}_R = - \vec{F}_A $ |
(ID 3240)
ID:(65, 0)