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3 years ago

What are the components of friction?

Physics

1 answer:

Tresset [83]3 years ago

4 0

<h2>Answer:</h2>

Friction:

When an object slips on a surface, an opposing force acts between the tangent planes which acts in the opposite direction of motion. This opposing force is called Friction. Or in other words, Friction is the opposing force that opposes the motion between two surfaces.

The main component of friction are:

Normal Reaction (R):

Suppose a block is placed on a table in the above picture, which is in resting state, then two forces are acting on it at that time.

The first is due to its weight mg which is working from its center of gravity towards the vertical bottom.

The second one is superimposed vertically upwards by the table on the block, called the reaction force (P). This force passes through the center of gravity of the block.

Due to P = mg, the box is in equilibrium position on the table.

Coefficient of friction ( μ ):

The ratio of the force of friction and the reaction force is called the coefficient of friction.

Coefficient of friction, µ = force of friction / reaction force

μ = F / R

The coefficient of friction is volume less and dimensionless.

Its value is between 0 to 1.

Advantage and disadvantage from friction force:

The advantage of the force of friction is that due to friction, we can walk on the earth without slipping.
Brakes in all vehicles are due to the force of friction.
We can write on the board only because of the force of friction.
The disadvantage of this force is that due to friction, some parts of energy are lost in the machines and there is wear and tear on the machines.

How to reduce friction:

Using lubricants (oil or grease) in machines.
Friction can be reduced by using ball bearings etc.
Using a soap solution and powder.

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$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

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(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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