Question

In a smooth hemispherical shell of radius R, a rod of mass 2kg is placed horizontally and is in equilibrium. The length of rod is . Find the normal reaction at any end of the rod (in N). ​

Answer 1

In a smooth hemispherical shell of radius R, a rod of mass 2 kg is placed horizontally and to equilibrium. The length of rod is R . Find the normal reaction at any end of the rod (in N ). [ Take g=10m/s ^2  ]

The normal reaction at any end of the rod is $\frac{20}{\sqrt{3} }$ N.

Given:

Mass=2kg

g=10 m/s^2

To Find:

The normal reaction at any end of the rod.

Solution:

The Force acting at the end of the rod is given by

F=mg/2

F=2*10/2

F=10 N

The normal reaction is

F=NSin60

N=F/Sin60

$N=\frac{10}{\frac{\sqrt{3} }{2} } \\N=\frac{20}{\sqrt{3} }$

Therefore,the normal reaction at any end of the rod is $\frac{20}{\sqrt{3} }$ N.

Answer 2

In a smooth hemispherical shell of radius R, a rod of mass 2 kg is placed horizontally and to equilibrium. The length of the rod is R. Find the normal reaction at any end of the rod (in N). [ Take g=10m/s^2 ].

The normal reaction at the end of the rod is 20/√3 N.

Given:

A smooth hemispherical shell of radius R, a rod of mass 2 kg is placed horizontally and to equilibrium. The length of the rod is R.

To Find:

The normal reaction at any end of the rod (in N).

Solution:

To find the normal reaction at any end of the rod we will follow the following steps:

As we know,

The force at the end of the rod =

$\frac{mg}{2} = \frac{2 \times 10}{2} = 10N$

Here, m is the mass of road.

N×sin60° is the vertical component of normal reaction.

The normal reaction produced at end of the rod = N×sin60° =

$N \frac{ \sqrt{3} }{2} = \frac{ \sqrt{3} }{2} N$

Here, N is the normal reaction produced by the road and sin60° is

$\frac{ \sqrt{3} }{2}$

As the forces are equal and opposite,

Normal reaction = force at the end of the road

$\frac{ \sqrt{3} }{2} N = 10$

$N = \frac{20}{ \sqrt{3} }$

Henceforth, the normal reaction at the end of the rod is 20/√3 N.

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