Question

Figure (5−E8) shows a uniform rod of length 30 cm and mass 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light.
Figure

Answer 1

The force exerted by the 20 cm part of the rod on the 10 cm part is 24 N.

Explanation:

Step 1:

Mass per length of unit = $\frac{3}{30}$ kg/cm=330 kg/cm = 0.10 kg/cm

10 cm  Mass part, $m_1$ = 1 kg

20 cm  Mass part part, $m_2$ = 2 kg

Let F = The force of interaction between them.

Step 2:

From diagram of the  free-body,

$\begin{array}{l}{m_{1} a=F-20 \ldots e q^{n}(i)} \\{m_{2} a=32-F \quad \ldots e q^{n}(i i)}\end{array}$

Adding the two equations, we obtain:

$a=\frac{12}{m_{1}+m_{2}}=\frac{12}{3}=4 m / s^{2}$

So,linked force,  

F = 20 + 1 a

F = 20 + 4 = 24 N

Answer 2

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