Question0. (II) As shown in Fig. 4-48, five balls (masses 2.00, 2.05, 2.10, $2.15,2.20 \mathrm{~kg}$ ) hang from a crossbar. Each mass is supported by "5-lb test" fishing line which will break when its tension force exceeds 22.2 N ( $=5.00 \mathrm{lb}$ ). When this device is placed in an elevator, which accelerates upward, only the lines attached to the 2.05 and 2.00 kg masses do not break. Within what range is the elevator's acceleration?
FIGURE 4-48 Problem 50.

Studdy Solution

STEP 1

1. The fishing line will break if the tension force exceeds 22.2 N.
2. The elevator accelerates upward.
3. The acceleration due to gravity is $g = 9.81 \, \text{m/s}^2$ .
4. We need to find the range of acceleration of the elevator such that only the lines attached to the 2.05 kg and 2.00 kg masses do not break.

STEP 2

1. Calculate the tension in the fishing line for each mass when the elevator is at rest.
2. Determine the tension in the fishing line when the elevator accelerates upward.
3. Set up inequalities for the tension forces that do not break the lines.
4. Solve for the range of acceleration.

STEP 3

Calculate the tension in the fishing line for each mass when the elevator is at rest. The tension $T$ is equal to the weight of the mass, $T = mg$ .
For the 2.00 kg mass: $T = 2.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N}$
For the 2.05 kg mass: $T = 2.05 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 20.11 \, \text{N}$
For the 2.10 kg mass: $T = 2.10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 20.60 \, \text{N}$
For the 2.15 kg mass: $T = 2.15 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 21.09 \, \text{N}$
For the 2.20 kg mass: $T = 2.20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 21.58 \, \text{N}$

STEP 4

Determine the tension in the fishing line when the elevator accelerates upward. The tension $T$ is given by $T = m(g + a)$ , where $a$ is the acceleration of the elevator.

STEP 5

Set up inequalities for the tension forces that do not break the lines. The tension must be less than or equal to 22.2 N.
For the 2.05 kg mass: $2.05(g + a) \leq 22.2$
For the 2.00 kg mass: $2.00(g + a) \leq 22.2$

STEP 6

Solve for the range of acceleration.
For the 2.05 kg mass: $2.05(9.81 + a) \leq 22.2$ $20.11 + 2.05a \leq 22.2$ $2.05a \leq 2.09$ $a \leq \frac{2.09}{2.05}$ $a \leq 1.02 \, \text{m/s}^2$
For the 2.00 kg mass: $2.00(9.81 + a) \leq 22.2$ $19.62 + 2.00a \leq 22.2$ $2.00a \leq 2.58$ $a \leq \frac{2.58}{2.00}$ $a \leq 1.29 \, \text{m/s}^2$
Thus, the range of acceleration is $0 \leq a \leq 1.02 \, \text{m/s}^2$ .

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