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 part. All the surfaces are smooth and the strings and pulleys are light





a) 36 N

(b) 12 N

(c) 64 N

(d) 24 N

Answer 1

Mass of 10cm= M(L-x)/L=M*10/30=M/3
Similarly, mass of 20cm= M*20/30=2M/3
Draw the FBD for both the masses with their reaction force on each block.
From mass M/3 we get:
R-20=Ma/3....1)
R=Ma/3 + 20
From mass 2M/3:
32-R= 2M/3
Substitute the value of R here.. we get Ma=12
Substitute this value of "Ma" in equation 1)
Hence we get the value of reaction force R as 24N

Answer 2

you forgot to attach figure. well it's okay , I attached figure.

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 N.
first we have to find mass of 20cm part and 10cm part of body .
mass of 20cm part , M = 3kg × 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 - N = 2a -------(i) [ here a is acceleration]

similarly for 10cm part,
N - 20 = a -------(ii)

from eqs. (i) and (ii),
12 = 3a => a = 4 m/s²
now, N = 32 - 2a = 32 - 2 × 4 = 24

hence, required force = 24N