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2 months ago

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A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 cm and a cross-sectional area of 15 cm2. T

he shear modulus for this rubber is 2.0 MPa. A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways? B.) With what frequency will the motor vibrate back and forth sideways if disturbed?

2 answers:

Ostrovityanka [3.2K]2 months ago

5 0

Answer:

Part a)

$x = 3 mm$

Part b)

$f = 6.5 Hz$

Explanation:

Part a)

The shear modulus can be defined as

$\eta = \frac{F/A}{x/L}$

at this point, we find

$x = \frac{FL}{\eta A}$

here we notice

$F = 300 N$

$A = 15 cm^2$

$L = 3 cm$

$\eta = 2 \times 10^6 Pa$

now we proceed with

$x = \frac{300 \times 0.03}{(15 \times 10^{-4})(2\times 10^6)}$

$x = 3 mm$

Part b)

The frequency of oscillation is determined by

$F = \frac{\eta A}{L} x$

hence, the established frequency of oscillation is

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

we are aware that

$k = \frac{\eta A}{L}$

$k = \frac{2\times 10^6 \times (15 \times 10^{-4})}{0.03}$

$k = 10^5$

$f = \frac{1}{2\pi}\sqrt{\frac{10^5}{60}}$

$f = 6.5 Hz$

Maru [3.3K]2 months ago

3 0

<span>A.) When a lateral force of 300 N is exerted on the motor, what will the sideways displacement be?</span>

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