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. The 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?

Respuesta :

A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?

Answer:

Part a)

[tex]x = 3 mm[/tex]

Part b)

[tex]f = 6.5 Hz[/tex]

Explanation:

Part a)

As we know that shear modulus is given as

[tex]\eta = \frac{F/A}{x/L}[/tex]

now we have

[tex]x = \frac{FL}{\eta A}[/tex]

here we have

[tex]F = 300 N[/tex]

[tex]A = 15 cm^2[/tex]

[tex]L = 3 cm[/tex]

[tex]\eta = 2 \times 10^6 Pa[/tex]

now we have

[tex]x = \frac{300 \times 0.03}{(15 \times 10^{-4})(2\times 10^6)}[/tex]

[tex]x = 3 mm[/tex]

Part b)

As we know that

[tex]F = \frac{\eta A}{L} x[/tex]

now the frequency of oscillation is given as

[tex]f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

here we know

[tex]k = \frac{\eta A}{L}[/tex]

[tex]k = \frac{2\times 10^6 \times (15 \times 10^{-4})}{0.03}[/tex]

[tex]k = 10^5[/tex]

[tex]f = \frac{1}{2\pi}\sqrt{\frac{10^5}{60}}[/tex]

[tex]f = 6.5 Hz[/tex]