# Linear programming problem and Newtonian mechanics

I am trying to learn Linear Programming. However, I don’t know how to solve the following problem. Maybe you can help, because I am curious to the right approach and solution for this problem. It involves Newtonian mechanics.

A company uses a crane for lifting (see picture). The crane has $$2$$ hoists. They can work at the same time. Both ends can take $$10000 \,\rm N$$. If the two hoist both pick 1 artefact at the same time, what is the maximum force that the first hoist can take?

• The force on hoist $$1$$ is represented by $$x$$.

• The force on hoist $$2$$ is $$9000 \,\rm N$$.

• The position of hoist $$1$$ is fixed.

• The position of hoist $$2$$ is not fixed, as long as it is to the right of the first hoist.

The distance between both hoist is represented by $$y$$.

A force $$z$$ down at position $$a$$ from the left will exert a force of $$\frac{40-a}{40}z$$ on the left support and $$\frac{a}{40}z$$ on the right.

The force distribution of hoist 2 does not change that the force acts on the crane effectively at the same center position. Thus you get inequalities $$10\,000 \ge \frac{35}{40}⋅x+\frac{32-y}{40}⋅9000\\ 10\,000 \ge \frac{5}{40}⋅x+\frac{8+y}{40}⋅9000$$ you get the geometry constraint $$0\le y\le 29$$ and you want of course maximize $$x$$.

Graphically the maximum is at $$y=29$$ with $$x=\frac{40⋅10\,000-3⋅9000}{35}=10657.142857$$

• From an Integer Linear Programming perspective, how do I formulate the objective function for this problem? Such as: Max z 5(x1 + y1 + x2) - 2(y1). – Freddy1996 Sep 17 '19 at 18:12

Calling $$V_1, V_2$$ the vertical reactions al left and right, and considering that the supporting beam length is $$40$$ we have

$$\max x \ \ \text{s. t. } \cases{ 0 \le y \le 29\\ 0\le V_1\le 10000\\ 0\le V_2\le 10000\\ V_1+V_2 = x + 9000\\ 5x + (y+8)9000 -40 V_2 = 0 }$$

The solution is

$$\cases{ x = 10657.14\\ y = 29\\ V_1 = 10000\\ V_2 = 9657.14 }$$