Question

Two smooth cylindrical bars weighing W each lie next to each other in contact. A similar third bar is placed over the two bars as shown in figure. Neglecting friction, the minimum horizontal force on each lower bar necessary to keep them together is

1) W/2
2) W
3) W/√3
4) W/2√3​

Answer 1

Answer:

For the system to be in equilibrium, the normal force between the lower cylinders should be zero which can be achieved by applying and external force F as mentioned in the figure.

Need to find:

F=?

On the basis of free body diagram of left lower cylindrical bar, we can write:

F=Nsin30

∘

----------[1]

and, On the basis of free body diagram of left lower cylindrical bar, we can write:

W=2Ncos30

∘

----------[2]

From eq [1] and [2]: [1]/[2]

⇒

W

F

=

2Ncos30

∘

Nsin30

∘

⇒F=

2cos30

∘

Wsin30

∘

=

2

Wtan30

∘

On putting, tan30

∘

=

3

1

and ,W=20

3

⇒F=

2

(20

3

)(

3

1

)

⇒F=

2

3

20

3

=

2

20

=10N

Hence, the minimum horizontal force on each lower bar necessary to keep them together is, F=10N

Answer 2

The minimum horizontal force on each lower bar necessary to keep them together is $F=\frac{W}{2\sqrt{3} }$.

Explanation:

On the basic of free body diagram of left lower cylindrical bar,

$F=Nsin{30}^{0}$

On this basic of free body diagram of left lower cylindrical bar,

$W=2Ncos30^{0}$

$F=\frac{Wsin30^{0}}{2cos30^{0}}$

$F=\frac{Wtan30^{0}}{2}$

on putting,$tan30^{0}=\frac{1}{\sqrt{3} }$

$F=\frac{W(\frac{1}{\sqrt{3} }) }{2}$

$F=\frac{W}{2\sqrt{3} }$.