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?
A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?
Answer:
Part a)
\(x = 3 mm\)
Part b)
\(f = 6.5 Hz\)
Explanation:
Part a)
As we know that shear modulus is given as
\(\eta = \frac{F/A}{x/L}\)
now we have
\(x = \frac{FL}{\eta A}\)
here we have
\(F = 300 N\)
\(A = 15 cm^2\)
\(L = 3 cm\)
\(\eta = 2 \times 10^6 Pa\)
now we have
\(x = \frac{300 \times 0.03}{(15 \times 10^{-4})(2\times 10^6)}\)
\(x = 3 mm\)
Part b)
As we know that
\(F = \frac{\eta A}{L} x\)
now the frequency of oscillation is given as
\(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
here we know
\(k = \frac{\eta A}{L}\)
\(k = \frac{2\times 10^6 \times (15 \times 10^{-4})}{0.03}\)
\(k = 10^5\)
\(f = \frac{1}{2\pi}\sqrt{\frac{10^5}{60}}\)
\(f = 6.5 Hz\)