Question

The system shown can undergo vertical motion only, and both springs are unstretched
when WD0. Determine the deflection ?in terms of W, k1,andk2

Answer 1

At the equilibrium, the net force on the mass is null.

$F_{\mathrm{net}}=0$

then,

$F_{1}+F_{2}-W=0$

where,

• $F_{1}$ is the magnitude of force exerted by the spring 1

$F_{1}=K_{1}\delta$

• $F_{2}$ is the magnitude of the force exerted by the spring 2

$F_{2}=K_{2}\delta$

then,

$F_{1}+F_{2}-W=0$

$K_{1}\delta+K_{2}\delta-W=0$

according to the above expression, the deflection must be

$\left( K_{1}+K_{2} \right)\delta=W$

$\delta=\frac{W}{K_{1}+K_{2}}$

so, when

• $W=0$.

$\delta=\frac{0}{K_{1}+K_{2}}=0$

if the weight is null there is not deflection

• $K_{1}=0$

$\delta=\frac{W}{0+K_{2}}=\frac{W}{K_{2}}$

• $K_{2}=0$

$\delta=\frac{W}{K_{1}+0}=\frac{W}{K_{1}}$

• $K_{1}\rightarrow \infty$

$K_{1}+K_{2}\rightarrow \infty$

then

$\delta=\frac{W}{K_{1}+K_{2}} \rightarrow 0$

• $K_{2}\rightarrow \infty$

$K_{1}+K_{2}\rightarrow \infty$

then

$\delta=\frac{W}{K_{1}+K_{2}} \rightarrow 0$