Answer to Question #125867 in Mechanics | Relativity for Sammy W-S

Question

a pully system has a two boxes. box 1 is on a ledge and has a mass of 2.0×10 to the power of 3 kg, box 2 is attached by a rope at a 90° angle blow box one and has a mass of 4.4×10 to the power of 2 kg. There is a man standing on box number two, what is the apparent weight of the man if all surfaces are frictionless?

Answer 1

According to the second Newton's law

$Ma=T$

$mg-T=ma$

$m_1g-N=m_1a$

So, we get

$N=m_1g-m_1\frac{mg}{M+m}$

Assume that the man has a mass $m_1=60 kg$ . Finally

$P=N=m_1g-m_1\frac{mg}{M+m}=60\cdot9.81-60\cdot\frac{440\cdot9.81}{2000+440}=482 N$