If a body is in equilibrium under a set of non-collinear forces,then the minimum number of forces has to be

Four forces are acting at a point $P$ in equilibrium as shown in the figure. The ratio of force $F_{1}$ to $F_{2}$ is $1: x$,where $x = ....$

$A$ uniform sphere of weight $W$ and radius $R = 3\,m$ is held by a string of length $L = 2\,m$ attached to a frictionless wall as shown in the figure. The tension in the string is:

Three forces $\vec{P}, \vec{Q},$ and $\vec{R}$ act at a point in a plane. The angles between $\vec{P}$ and $\vec{Q}$,and $\vec{Q}$ and $\vec{R}$ are $150^{\circ}$ and $120^{\circ}$ respectively. What is the ratio of the magnitudes of the forces $\vec{P}, \vec{Q},$ and $\vec{R}$ if the system is in equilibrium?

Describe the different types of common forces.

As shown in the figure,the two sides of a step ladder $BA$ and $CA$ are $1.6 \; m$ long and hinged at $A$. $A$ rope $DE$ of length $0.5 \; m$ is tied halfway up. $A$ weight of $40 \; kg$ is suspended from a point $F$,$1.2 \; m$ from $B$ along the ladder $BA$. Assuming the floor to be frictionless and neglecting the weight of the ladder,find the tension in the rope and the forces exerted by the floor on the ladder. (Take $g = 9.8 \; m/s^2$)