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Chapter 4: Problem 141

Spring scale \(A\) is attached to the ceiling. A 10.0 -kg mass is suspended from the scale. A second spring scale, \(\mathrm{B}\), is hanging from a hook at the bottom of the 10.0 -kg mass and a 4.0 -kg mass hangs from the second spring scale. (a) What are the readings of the two scales if the masses of the scales are negligible? (b) What are the readings if each scale has a mass of $1.0 \mathrm{kg} ?$

Short Answer

Expert verified
Answer: (a) When the masses of the scales are negligible, scale A reads 98.1 N and scale B reads 39.24 N. (b) When each scale has a mass of 1.0 kg, scale A reads 107.91 N and scale B reads 49.05 N.

Step by step solution

01

Analyze the forces acting on the system (Negligible masses case)

In this step, we'll analyze the forces acting on the system when the masses of the scales are negligible. In this case, we'll only consider the forces acting on the two masses (10.0 kg and 4.0 kg). For the 10.0-kg mass, there's an upward force (tension) from spring scale A and a downward gravitational force (weight): \(F_{gA}=mg_{A}\) (\(m_{A}=10.0 \mathrm{kg}\), \(g=9.81 \mathrm{m/s^2}\)). For the 4.0-kg mass, there's an upward force (tension) from spring scale B and a downward gravitational force (weight): \(F_{gB}=mg_{B}\) (\(m_{B}=4.0 \mathrm{kg}\), \(g=9.81 \mathrm{m/s^2}\)).
02

Apply Newton's second law of motion (Negligible masses case)

Since the system is in equilibrium (not accelerating), the net force acting on each mass is 0. Thus, we can write Newton's second law for each mass (using up as positive): For mass A (10.0 kg): \(T_{A} - F_{gA} = 0\) For mass B (4.0 kg): \(T_{B} - F_{gB} = 0\)
03

Calculate the tension forces and scale readings (Negligible masses case)

Now, we can solve these equations to find the tension forces and the readings of the scales (assuming the scales display the tension force). For mass A: \(T_{A} = F_{gA} = m_{A}g = 10.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 98.1 \mathrm{N}\) (reading of scale A) For mass B: \(T_{B} = F_{gB} = m_{B}g = 4.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 39.24 \mathrm{N}\) (reading of scale B) (a) Readings of scales with negligible masses: scale A: 98.1 N; scale B: 39.24 N.
04

Analyze the forces acting on the system (1.0-kg masses case)

Now, let's analyze the forces acting on the system when each scale has a mass of 1.0 kg. Both the scales and the two masses (10.0 kg and 4.0 kg) have downward gravitational forces and upward tension forces. For scale A (1.0 kg) - combined with mass A (10.0 kg): Upward force (tension): \(T_{A}\) Downward force (weight): \(F_{gA} = m_{A}g\) (\(m_{A} = 11.0 \mathrm{kg}\), \(g=9.81 \mathrm{m/s^2}\)) For scale B (1.0 kg) - combined with mass B (4.0 kg): Upward force (tension): \(T_{B}\) Downward force (weight): \(F_{gB} = m_{B}g\) (\(m_{B} = 5.0 \mathrm{kg}\), \(g=9.81 \mathrm{m/s^2}\))
05

Apply Newton's second law of motion (1.0-kg masses case)

Since the system is in equilibrium, the net force acting on each mass and scale is 0. Thus, we can write Newton's second law: For scale A (1.0 kg) - combined with mass A (10.0 kg): \(T_{A} - F_{gA} = 0\) For scale B (1.0 kg) - combined with mass B (4.0 kg): \(T_{B} - F_{gB} = 0\)
06

Calculate the tension forces and scale readings (1.0-kg masses case)

We can now solve these equations to find the tension forces and readings of the scales (as before, assuming the scales display tension force). For scale A (1.0 kg) - combined with mass A (10.0 kg): \(T_{A} = F_{gA} = m_{A}g = 11.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 107.91 \mathrm{N}\) (reading of scale A) For scale B (1.0 kg) - combined with mass B (4.0 kg): \(T_{B} = F_{gB} = m_{B}g = 5.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 49.05 \mathrm{N}\) (reading of scale B) (b) Readings of scales with 1.0-kg masses: scale A: 107.91 N; scale B: 49.05 N.

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