Question

Figure shows a uniform rod of length 30 cm having a mass 3.0 kg. The rod is pulled by constant forces of 20 N and 32 N as shown. Find the force exerted by 20 cm part of the rod on the 10 cm part (all surface are smooth) is:

• A. 36 N
• B. 12 N
• C. 64 N
• D. 24 N

Answer 1

The correct option is D 24 N

According to the newton's law of motions, body mares whore force exists, here in right side force $\left(32N\right)$ is greater than left side $\left(20N\right)$, so body of mass $3kg$moves in rightwards. Let the force exerted by $20cm$ of the rod on the lo $cm$ part 10 $N$.

$\text{mass of}20cm\text{part,}M=3\times \frac{20}{(10+20)}$

mass of $10cm$ post , $m=3-2$$

$m=1\mathrm{kg}.$

From newtons 2nd law for $20cm$ part $F_{\text{forward}}-F_{\text{backward}}=Ma$

$32-N=2a$

Similarly for $10$ cm part

$N-20=a$

from (1) and (2)

$12=3a$

$a=4m/s^2$

$N=32-2a$

$N=24N$

[D] $24N$