A light train made up of two cars is traveling at 100 km/h when the brakes are applied to both cars. Knowing that car A has a ma
ss of 35 Mg and car B a mass of 20 Mg, and that the braking force is 35 kN on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is showing down.
1 answer:
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The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.
<u>Given the following data:</u>
- Mass of cat = 2 kg
- Mass of elevator = 97 kg
- Acceleration = 2
To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:
First of all, we would calculate the total mass of the cat/elevator pair.
Total mass = 99 kilograms
Mathematically, Newton's Second Law of Motion is given by this formula;
Substituting the given parameters into the formula, we have;
Net force = 198 Newton
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Answer:
The stress is calculated as
Solution:
As per the question:
Length of the wire, l = 75.2 cm = 0.752 m
Diameter of the circular cross-section, d = 0.560 mm =
Mass of the weight attached, m = 25.2 kg
Elongation in the wire,
Now,
The stress in the wire is given by:
(1)
Now,
Force is due to the weight of the attached weight:
F = mg =
Cross sectional Area, A =
Using these values in eqn (1):