- #1

issacnewton

- 1,026

- 36

Hi

I have solved this problem but I have some doubts about the logic I have used at one point.

I have attached the problem in 1.jpg. Now I am trying to construct a free body diagram

for the mass M. First, there is force [tex]\inline{F}[/tex] in the right direction. And as usual, there is force of gravity [tex]\inline{Mg}[/tex] downward and the normal reaction

[tex]\inline{N}[/tex] upward. Also there would be a force exerted by the mass [tex]\inline{m_2}[/tex] towards the left, let's call it [tex]\inline{F_{21}}[/tex]. From top, there would be a force exerted by the mass [tex]\inline{m_1}[/tex] on [tex]\inline{M}[/tex].

Now let's say that there is a tension [tex]\inline{T}[/tex] in the string when the system is going to the right. Since pulley is frictionless, the tension in both sides would be same and since the pulley is attached to the mass [tex]\inline{M}[/tex] any force exerted on the pulley would be exerted on the mass [tex]\inline{M}[/tex]. Since force [tex]\inline{T}[/tex] is exerted

on the mass [tex]\inline{m_1}[/tex] ,by Newton's third law, the force [tex]\inline{T}[/tex] would be exerted by the m

[tex]\inline{M}[/tex] , towards left.. With similar reasoning, we can say that a downward force

[tex]\inline{T}[/tex] is exerted on the mass [tex]\inline{M}[/tex].

Do you think that was the valid reasoning. I got the right answer. The book's solution

manual does it differently but doing my way gives more insight into the forces

between the masses m

thanks

I have solved this problem but I have some doubts about the logic I have used at one point.

I have attached the problem in 1.jpg. Now I am trying to construct a free body diagram

for the mass M. First, there is force [tex]\inline{F}[/tex] in the right direction. And as usual, there is force of gravity [tex]\inline{Mg}[/tex] downward and the normal reaction

[tex]\inline{N}[/tex] upward. Also there would be a force exerted by the mass [tex]\inline{m_2}[/tex] towards the left, let's call it [tex]\inline{F_{21}}[/tex]. From top, there would be a force exerted by the mass [tex]\inline{m_1}[/tex] on [tex]\inline{M}[/tex].

Now let's say that there is a tension [tex]\inline{T}[/tex] in the string when the system is going to the right. Since pulley is frictionless, the tension in both sides would be same and since the pulley is attached to the mass [tex]\inline{M}[/tex] any force exerted on the pulley would be exerted on the mass [tex]\inline{M}[/tex]. Since force [tex]\inline{T}[/tex] is exerted

on the mass [tex]\inline{m_1}[/tex] ,by Newton's third law, the force [tex]\inline{T}[/tex] would be exerted by the m

_{1}on the pulley, which means on the mass[tex]\inline{M}[/tex] , towards left.. With similar reasoning, we can say that a downward force

[tex]\inline{T}[/tex] is exerted on the mass [tex]\inline{M}[/tex].

Do you think that was the valid reasoning. I got the right answer. The book's solution

manual does it differently but doing my way gives more insight into the forces

between the masses m

_{1},m_{2}and [tex]\inline{M}[/tex].thanks