Question

6 tensions and weights (15 pts)
(problem 5.6.57)
consider the baby being weighed in the figure above. the scale works using hookes law: the weight of the baby
pulls down on the box containing a spring, which stretches to show the downward force exerted. when the baby is
placed in the basket that hangs from to the bottom of the scale, the scale reads 55 n. assume the cord between the
basket and the scale box and the cord between the scale box and the ceiling both have negligible mass.
(a) what is the mass of the infant and basket? (5 pts)
(b) what is the tension ( t_1 ) in the cord attaching the baby to the scale? (5 pts)
(c) what is tension ( t_2 ) in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (5 pts)

Explanation:

Step1: Calculate the mass of the infant and basket (for part a)

Use the formula \(W = mg\), where \(W = 55\space N\) and \(g=9.8\space m/s^{2}\). Rearranging for \(m\) gives \(m=\frac{W}{g}\).

$$m=\frac{55}{9.8}\approx 5.61\space kg$$

Step2: Find the tension \(T_{1}\) (for part b)

Since the system is in equilibrium, the tension \(T_{1}\) balances the weight of the baby and basket. So \(T_{1}=W\).

$$T_{1} = 55\space N$$

Step3: Calculate the tension \(T_{2}\) (for part c)

The total weight acting on the upper - cord is the weight of the baby and basket plus the weight of the scale. The weight of the scale \(W_{scale}=m_{scale}g\), where \(m_{scale} = 0.500\space kg\) and \(g = 9.8\space m/s^{2}\), so \(W_{scale}=0.500\times9.8 = 4.9\space N\). Then \(T_{2}=W + W_{scale}\).

$$T_{2}=55+4.9=59.9\space N$$

Answer:

a. The mass of the infant and basket is approximately \(5.61\space kg\).
b. The tension \(T_{1}\) is \(55\space N\).
c. The tension \(T_{2}\) is \(59.9\space N\).