Question

Problem (1) In the event of a cable failure at D, an air spring A is used to protect the support B and prevent damage to the weight C which has a mass of 12 kg. (a) Plot (i) shows the force developed by the spring as a function of its deformation. Determine the height d from which the block needs to be suspended to cause a maximum spring deformation of 0.12 m in the event the cable fails. (b) If the cable is suspended from the same height d calculated earlier and the force developed by the spring as a function of its deformation is now given by plot (ii), find the maximum spring deformation in this case, knowing that k = 24 kN/m2 in plot (ii). Neglect the mass of the pulley and belt and use the principle of work and energy.

Answer 1

Answer:

D = 0.3387 m

S= 0.171 m

Explanation:

See attachment for detailed answer.

Answer 2

Answer:

a) d=0.3387 m

b) s=0.17 m

Explanation:

a) The force displacement relation ship B is:

$F=kx$

Clearing k:

$k=\frac{f}{x} =\frac{1500}{0.2} =7500N/m$

Spring energy B is:

$\frac{1}{2} kx^{2}$

So, at max compression:

$mg(d+x)=\frac{1}{2} kx^{2}$

Clearing d and replacing values:

$12*9.8(d+0.12)=\frac{1}{2} *7500*(0.12^{2} )\\d=0.3387 m$

b) is similar to a)

$mg(d+s)=\frac{1}{2}ks^{3} \\12*9.8(0.3387+s)=\frac{1}{2} *2400*s^{3} \\12000s^{3} 117.72s-39.87=0\\s=0.17 m$