Pulley system to balance the weight of a person

AI Thread Summary
The discussion centers on analyzing an ideal pulley system to determine the tensions in the ropes. Participants agree that the tensions T1, T2, and T3 are equal and can be represented as F, leading to the conclusion that F equals half the combined weight of the person and the smaller pulley. There is debate over the calculation of T4, with some suggesting that it should account for the weight of the upper pulley, leading to the equation T4 = 3F + Mp*g. The importance of drawing free body diagrams is emphasized for clarity in understanding the forces at play. Overall, the conversation highlights the complexities of tension calculations in pulley systems and the necessity for precise logic and clear communication.
lorenz0
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Homework Statement
If the mass of the man to be saved is ##m##, the mass of the smaller pulley ##m_p## and the mass of the bigger pulley ##M_p##, find: ##T_1, T_2, T_3, T_4, T_5, F.##
Relevant Equations
##\vec{F}=m\vec{a}##
Since we are dealing with an ideal rope, we have that ##T_1=T_2=T_3=F and T_2+T_3=2F=(m+m_p)g\Leftrightarrow F=\frac{m+m_p}{2}g.##
##T_4=3F+(m+m_p+M_p)g=\frac{3}{2}(m+m_p)g+(m+m_p+M_p)g=(\frac{5}{2}m+\frac{5}{2}m_p+M_p)g## and ##T_5=mg-2F.##

Is this correct? If not, I woould appreciate a brief explanation on how to deal with these ideal-pulley ideal-rope systems since they seem quite counterintuitive to me.
 

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I agree that ##T_1 = T_2 = T_3## and that we can call this ##F##. As you say, the pulleys are ideal so all three tensions will match.

I agree that ##T_2 + T_3 = 2F##. Since those two tensions support the man plus the smaller pulley, it follows that ##2F = (m+m_p)g##. So yes, we can conclude that ##F=\frac{m+m_p}{2}g##

You will have to persuade me that ##T_4 = 3F + (m + m_p + M_p)g##. How did you arrive at that? I think you are double dipping there.
 
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@lorenz0, in addition to what @jbriggs444 has said, can I add this...

Presumably the system is in equilibrium (acceleration = 0).

Your equations/logic could be clearer if not joined together into a single line/paragraph.

The facts that ##T_1 = T_2 = T_3## and ##T_2+T_3=2F## lead to ##T_1 = F##. Can you see a simpler way to get this relationship directly?

All parts of the system are in equilibrium. You can draw a free body diagram for any part of the system. If you draw (or just imagine) a free body diagram for the upper pulley alone, it should help you.
 
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Steve4Physics said:
@lorenz0, in addition to what @jbriggs444 has said, can I add this...

Presumably the system is in equilibrium (acceleration = 0).

Your equations/logic could be clearer if not joined together into a single line/paragraph.

The facts that ##T_1 = T_2 = T_3## and ##T_2+T_3=2F## lead to ##T_1 = F##. Can you see a simpler way to get this relationship directly?

All parts of the system are in equilibrium. You can draw a free body diagram for any part of the system. If you draw (or just imagine) a free body diagram for the upper pulley alone, it should help you.
Drawing the free body diagram for the upper pulley alone, it seems that ##T_4=3F+M_Pg.## Is this correct?
 
Last edited:
lorenz0 said:
Drawing the free body diagram for the upper pulley alone, it seems that ##T_4=3F.## Is this correct?
No. You missed something this time.
 
jbriggs444 said:
No. You missed something this time.
The mass of the upper pulley, I guess.
 
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lorenz0 said:
Drawing the free body diagram for the upper pulley alone, it seems that ##T_4=3F.## Is this correct?
Pulley.jpg
 
lorenz0 said:
The mass of the upper pulley, I guess.
You mean the weight of the upper pulley. (Hopefully it's not a 'guess', as a free body diagram must include the body's weight.) Your Post #4 edited equation is now correct.

BTW, it's best to not significantly edit a post once someone has replied to it. Better to write a new Post making clear any changes you want. That can avoid confusion and messages at cross-purposes.
 
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