Question

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

Answer 1

According to Newton's law of motion,

Body moves where force exists . here in right side force $32N$ is greater than left side side {20N}. so, body of mass 3kg moves in rightward.

Let the force exerted by $20cm$ part of the rod on the $10cm$ part is $F$.

first we have to find mass of $20cm$ part and $10cm$ part of body .

mass of $20cm$ part , $M=3kg\times 20cm/(10cm+20cm)=2kg$

mass of $10cm$ part , $m=3kg-2kg=1kg$

now, use Newton's 2nd law ,

for $20cm$ part ,

forward force $-$ backward force $=Ma$

$32-F=2a$ -------(i) [ here a is acceleration]

similarly for $10cm$ part,

$F-20=a$ -------(ii)

from eqs. (i) and (ii),

$12=3a\Rightarrow a=4m/s^2$

now, $F=32-2a=32-2\times 4=24$

hence, required force $=24N$