Beam force distribution

A body weighing 400 kg is suspended at a distance of 1 m from one end of the beam on a beam weighing 100 kg and 5 m. How large do forces act at the ends of the beam?

Final Answer:

F₁ =  3700 N
F₂ =  1300 N

Step-by-step explanation:

$$\begin{gathered} g=10{\text{m/s}}^2 \\ {m}_1=100\text{kg} \\ l=5\text{m} \\ x=1\text{m} \\ {m}_2=400\text{kg} \\ {G}_1=m_1\cdot g=100\cdot 10=1000\text{N} \\ {G}_2=g\cdot m_2=10\cdot 400=4000\text{N} \\ a+b=G_2 \\ a\cdot x=b\cdot (l-x) \\ a+b=4000 \\ a\cdot 1=b\cdot (5-1) \\ a+b=4000 \\ a-4b=0 \\ Row2-Row1\to Row2 \\ -5b=-4000 \\ b=\frac{-4000}{-5}=800 \\ a=4000-b=4000-800=3200 \\ a=3200 \\ b=800 \\ {F}_1=a+G_1/2=3200+1000/2=3700\text{N} \end{gathered}$$

$F_2=b+G_1/2=800+1000/2=1300\text{N}$

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