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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.

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प्रश्न

In the following figure 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.

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उत्तर

Mass per unit length \[= \frac{3}{30} kg/cm\] = 0.10 kg/cm
Mass of the 10 cm part, m1 = 1 kg
Mass of the 20 cm part, m2 = 2 kg

Let F = contact force between them.

From the free-body diagram, 
\[m_1 a = F - 20 . . . \left( i \right)\]
\[ m_2 a = 32 - F . . . \left( ii \right) \]
\[\text{ Adding both the equations, we get: }\]
\[a = \frac{12}{m_1 + m_2} = \frac{12}{3} = 4 m/ s^2\]
So, contact force,
F = 20 + 1a
F = 20 + 4 = 24 N

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पाठ 5: Newton's Laws of Motion - Exercise [पृष्ठ ८०]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 5 Newton's Laws of Motion
Exercise | Q 24 | पृष्ठ ८०

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