Question

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. 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.

Answer 1

Given :

Mass per unit length

=330 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,

m₁a=F-20    …i

m₂a=32-F    …ii

Adding both the equations, we get:

a=12/m1+m2

=12/3=4 m/s2²

So, contact force,

F = 20 + 1a

F = 20 + 4 = 24 N

Answer 2

Answer:

Mass per unit length=

30

3

kg/cm=0.1kg/cm

Mass of 10cm part=m

1

=1kg

Mass of 20cm part=m

2

=2kg

Let F be the force of contact between them

From free body diagram,

F−20−a=0 ......(1)

32−F−2a=0 ......(2)

⇒32−20−2a−a=0

⇒12−3a=0

⇒a=

3

12

=4ms

−2

Contact forceF=20+a=20+4=24N

∴ force exerted=24N