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A person carries a plank of wood 1.9 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 56 cm from the end of the plank with force F2. If the plank has a mass of 15 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces F1 and F2? (The distance of 56 cm is measured from the location of F1.)F1 = F2 =



Answer :

We will have the following:

[tex]\sum\tau=0[/tex]

Thus, assuming the axis of rotation is at F1, we will have that:

[tex]\tau_2-\tau_3\Rightarrow\tau_2=\tau_3[/tex]

Then:

[tex]\begin{gathered} r_2F_2=r_3F_3\Rightarrow F_2=\frac{r_3F_3}{r_2} \\ \\ \Rightarrow F_2=\frac{(0.95m)(15kg\ast9.8m/s^2)}{0.56m}\Rightarrow F_2=249.375N \end{gathered}[/tex]

So, the magnitude of F2 is 249.375 N.

Now, since the system is in equilibrium we will have that:

[tex]\begin{gathered} \sum F=0=-F_1+F_2-F_3=0 \\ \\ \Rightarrow F_1=F_2-F_3\Rightarrow F_1=(249.375N)-(15kg\ast9.8m/s^2) \\ \\ \Rightarrow F_1=102.375N \end{gathered}[/tex]

So, the magnitude of F1 is 102.375 N.

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