Answer to Question #123911 in Mechanics | Relativity for Richard

Question

A mass m1=1.8kg is attached to another mass m2=3.2kg with a light inextensible string passing over a frictionless pulley.m2 is surrounded by a viscous liquid .if the acceleration of m2 downward is 0.6m/s^2 . calculate the frictional force acting as a drag on m2

Answer 1

According to the second Newton's law

$m_1g-T=-m_1a$ and $m_2g-T-F_f=m_2a$

We have

$T=m_1g+m_1a$

$m_2g-m_1g-m_1a-F_f=m_2a \to F_f=m_2g-m_1g-m_1a-m_2a=$

$=(m_2-m_1)g-(m_1+m_2)a=(3.2-1.8)\cdot 9.81-(3.2+1.8)\cdot 0.6=10.7N$