A common device for entertaining a toddler is a jump seat that hangs from the ho

Question

A common device for entertaining a toddler is a jump seat that hangs from the horizontal portion of a doorframe via elastic cords (the figure). Assume that only one cord is on each side in spite of the more realistic arrangement shown. When a child is placed in the seat, they both descend by a distance ds as the cords stretch (treat them as springs). Then the seat is pulled down an extra distance dm and released, so that the child oscillates vertically, like a block on the end of a spring. Suppose you are the safety engineer for the manufacturer of the seat. You do not want the magnitude of the child's acceleration to exceed 0.19g for fear of hurting the child's neck. If dm = 11 cm, what value of ds corresponds to that acceleration magnitude?

Explanation / Answer

For a child of mass M, the equilibrium position is
k x ds = Mg
where the child's weight is equal to the spring forces.
So k = Mg / ds
When the springs are extended another dm, we have
net force Ma = k(ds + dm) - Mg
given a = 0.19g and substitute the equation k we have
Mx0.19g = (Mg/ds)(ds + dm) – Mg

0.19Mg+Mg = (Mg(ds+dm))/ds

1.19 = (ds + dm)/ds
Where dm = 11 cm, given

1.19 xds=ds+dm

1.19xds=ds+11

Ds(1.19-1)=11

ds=11/0.19
therefore ds= 57.89 cm

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