Figure shows a uniform rod of length 30 cm having a mass of 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 20 cm part of the rod on the 10 cm pan. All the surfaces are smooth and the strings and pulleys are light

Question

Figure shows a uniform rod of length 30 cm having a mass of 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 20 cm part of the rod on the 10 cm pan. All the surfaces are smooth and the strings and pulleys are light (A) 36 N (B) 12 N (C) 64 N (D) 24 N

Tardigrade Admin · Accepted Answer

Net force on the rod, f=32-20=12 N Acceleration of rod =(f/m)=(12/3)=4 ms -2 Equation of motion of 10 cm part is F-20 =m × a=1 × 4 F =4+20=24 N Similarly, Equation of motion 20 cm part is 32-F =m' a=2 × 4 F =32-8=24 N

Q. Figure shows a uniform rod of length $30 c m$ having a mass of $3.0 k g$ . The strings shown in the figure are pulled by constant forces of $20 N$ and $32 N$ . Find the force exerted by $20 c m$ part of the rod on the $10 c m$ pan. All the surfaces are smooth and the strings and pulleys are light

36 N

12 N

64 N

24 N

Net force on the rod, $f = 32 - 20 = 12 N$
Acceleration of rod $= \frac{f}{m} = \frac{12}{3} = 4 m s^{- 2}$
Equation of motion of $10 c m$ part is
$F - 20 = m \times a = 1 \times 4$
$F = 4 + 20 = 24 N$
Similarly, Equation of motion $20 c m$ part is
$32 - F = m^{'} a = 2 \times 4$
$F = 32 - 8 = 24 N$

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

36 N

12 N

64 N

24 N

Net force on the rod, f=32−20=12N Acceleration of rod =mf​=312​=4ms−2 Equation of motion of 10cm part is F−20=m×a=1×4 F=4+20=24N Similarly, Equation of motion 20cm part is 32−F=m′a=2×4 F=32−8=24N