#头条文章养成计划#
The actual middle pull F direction is not vertically upward, and the remaining two pull F directions are vertically upward. Forces with different directions cannot be directly added or subtracted numerically (because forces are vectors and follow the parallelogram rule). Therefore, the F = 1/3 (G + G motion) we get is actually an approximation, not a strict-like relationship
Pulley set is almost a mandatory knowledge point for the middle school entrance examination, many students in the second and third years of junior high school always memorize a bunch of formulas for the pulley group, and they can't remember it, and they won't use it if they remember. Here is to show you the derivation process, easy to understand how these formulas come about, what problems can be solved ··
The pulley set is a simple machine composed of fixed pulley and moving pulley, and the formulas involved are:
II. [Character Explanation]:
F : The pull force of the free end of the rope
n: The number of rope segments wound on all moving pulleys
G : The gravity of the object being pulled up
G motion : the gravity of all moving pulleys
s : The distance pulled by the free end of the rope
h : The height at which the object is pulled up
P: The power of the rope free end pull force F
vF : The speed at which the free end of the rope moves
vG : The speed at which the object moves
η: Mechanical efficiency
W has : useful work of the pulley set
W total : The total work of the pulley set
There are 3 sections of rope on the moving pulley, and it is these three sections that save effort to lift the moving pulley and the heavy objects attached to the moving pulley, so you can get:
F + F + F = G动 + G
3F = G + G动
F = 1/3（G + G动）
[Extension]: When there is n segment rope on the common pulley set, F = 1/n (G + G movement)
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The pulley set can save effort, but it certainly does not save effort. When the weight of the rope and the friction between the pulley and the rope are not considered, and the tension is vertical and uniform to pull the weight and the moving pulley rise for a distance h and then remain stationary, the work done by the pulling force achieves the effect of pulling the moving pulley and the heavy object up H. In other words, the work done by the pull increases the gravitational potential energy of the moving pulley and the heavy object.
F · s = (G + G动)· h
And because F = 1/n(G + G movement)
所以 1/n(G + G动)· s = (G + G动)· h
You can get :s = n · h
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At the same time, we can get: s / t = (n · h）/ t
So there is :vF = n · vG
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Work power P of pull force F :
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Useful work W has :
During uniform pulling, the height s0 = h and F0 = G on the object
W有 = F0 · s0 = G · h
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Total work W total = F · s
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Mechanical efficiency:
In fact, in the derivation of mechanical efficiency, in addition to the original Gh/Fs, the other two in the derivation process, one only replaced s, and the other was that F and s were replaced. It's that simple.
Can you write it yourself?