STATIC EQUILIBRIUM: Concurrent Force Systems

A system that is acted upon by concurrent forces is in equilibrium when it remains motionless relative to an observer within the system’s frame of reference:

In the diagram above, the system is a 200 N object that is suspended from a pulley. Since the system is not accelerating, we must search for forces that are equal and opposite to the object’s weight. In such circumstances, the sum of the horizontal and vertical forces acting upon the system must be zero:

∑F_x = 0

∑F_y = 0

Regardless of the relative simplicity or complexity of a system at hand, the same principle applies. In the diagram below, four pulleys “ share “ the task of stabilizing the weight of the suspended object:

A slight bit of complexity arises when we evaluate systems that are stabilized by forces that lie outside of the horizontal and vertical axis. No need to fear, however; many such circumstances are handled by breaking diagonal vectors into horizontal and vertical components. Once this has been done, simple trigonometric functions are used to solve for any unknown values of interest:

Q: What is the value of the force vectors that stabilize this system?

A: It will be simplest to follow tradition by designating the upward and rightward directions as being positive. Likewise, each vector is equal in value:

F₁ = F₂

We will arbitrarily allow the rightmost vector to be F₂. Let’s first evaluate the sum of forces acting horizontally:

∑F_x = 0

F₂ sin θ – F₁ sin θ = 0

F₂ sin θ = F₁ sin θ

Well, we’ve mathematically proven that the horizontal forces are equal, but this didn’t help us solve the problem at hand. If we’d taken into account that the downward force of weight has a known value, we could have more hastily begun to search for our answer:

∑F_y = 0

2F₂ sin θ – F_w = 0

2F₂ sin θ = F_w 

F₂ = ( F_w / 2sin θ )

sin 20^o = 0.34

F₂ = ( 400 N  / 0.68 )

F₂ = 588 N