How Is Tension Distributed in a Multi-Block System?

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In a multi-block system where a girl pulls three blocks with a force of 6 N, the tension in the rope between the first and second blocks is calculated to be 5 N, while the tension between the second and third blocks is 2 N. The weights of the blocks were initially misstated as Newtons but are actually in kilograms. The discussion emphasizes the importance of understanding the relationship between force and mass in calculating tension. Participants express a willingness to share the steps for solving the problem if needed. Overall, the conversation revolves around clarifying the tension distribution in the system.
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Ok - I was helping a friend out and was helping her with a problem but I am not sure if i was right.
A girl is pulling three blocks with a rope at a force of 6 N . There is two more strings between each block. The first block weighs 0.5 N the second block weighs 1.5N the third block weighs 1 N. Ignore friction. What is the amount of tension on each rope? :confused:
 
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Sorry i made a mistake the weight of the blocks are meant to be in kilograms not N .
lol idiot me...
 
The force on the rope between the first and second blocks is 5N
And the force on the rope between the second and third blocks is 2N.

I hope that helps
If you need the steps to get the answers, just ask.
 
physicsgrade11 said:
Sorry i made a mistake the weight of the blocks are meant to be in kilograms not N .
lol idiot me...
The unit of weight IS Newton, though..
 
Miracles said:
The force on the rope between the first and second blocks is 5N
And the force on the rope between the second and third blocks is 2N.

I hope that helps
If you need the steps to get the answers, just ask.

hey... how'd u do it?

thanks much
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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