Question

problem 3
you are suspending two cows from the ceiling,
one below the other. you have really strange
hobbies. the lower cow has a mass of 2kg. the
upper cow has a mass of 1kg. what is the tension
in the rope between the top cow and the ceiling?

Explanation:

Step1: Calculate the weight of each cow

The weight formula is \(F = mg\), where \(g = 10m/s^{2}\).
For the lower cow (\(m_1=2kg\)), its weight \(F_1=m_1g = 2\times10=20N\).
For the upper cow (\(m_2 = 1kg\)), its weight \(F_2=m_2g=1\times 10 = 10N\).

Step2: Analyze the tension in the rope

The tension \(T\) in the rope between the top - cow and the ceiling must support the total weight of the two cows.
Using the equilibrium condition (since the system is at rest, \(T=\sum F\)), \(T=F_1 + F_2\).
Substitute \(F_1 = 20N\) and \(F_2=10N\) into the formula, we get \(T=20 + 10=30N\).

Answer:

30 N