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3 years ago

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A 25 - kg television set rests on four rubber pads , each having a height of 1 . 0 cm and a radius of 0 . 60 cm . A 200 - N hori

zontal force is applied to the television set . How far does it move sideways ? The shear modulus of rubber is 2 . 6 x 106 N / m2

1 answer:

Svetlanka [38]3 years ago

6 0

Answer:

Thus it moves 1.7 mm sideways

Solution:

As per the question:

Mass of the television, m = 25 kg

Height of the rubber pads, h = 1.0 cm = 0.01 m

Radius, R = 0.60 cm = 0.006 m

Horizontal force, F = 200 N

Shear Modulus of Rubber, G = $2.6\times 10^{6}\ N/m^{2}$

Now,

To calculate the distance moved:

G = $\frac{shear\ stress}{shear\ strain}$

G = $\frac{\frac{Force}{Area}}{\frac{\Delta l}{L}} = \frac{Fl}{A\Delta L}$

where

$\Delta L$ = distance moved

Area of 4 rubber pads, A = $4\times \pi R^{2}$

Now,

$\Delta L = \frac{200\times 0.01}{4\times \pi 0.006^{2}\times 2.6\times 10^{6}} = 1.7\times 10^{- 3}\ m = 1.7\ mm$

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<em>consider the attached diagram below</em>

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$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

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likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

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Answer

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OR

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

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