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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.

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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 $y$.

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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 $x=\frac{40⋅10\,000-3⋅9000}{35}=10657.142857$

plot of the constraints

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  • $\begingroup$ 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). $\endgroup$ – Freddy1996 Sep 17 '19 at 18:12
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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

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

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