Solving Homework Problem: Masses, Pulleys, and Forces

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SUMMARY

The forum discussion centers on solving a physics homework problem involving three blocks connected by strings over a frictionless, massless pulley. The blocks have masses of 1.0kg, 3.0kg, and 5.0kg, respectively. The user derived equations for acceleration and tension in the strings, ultimately calculating the acceleration as -1.25 m/s², tension T2 as 11N, and tension T1 as 44N. The normal force exerted by the table on the 5.0kg block was determined to be 6N.

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  • Understanding of Newton's Second Law of Motion
  • Familiarity with tension in strings and forces in equilibrium
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  • Ability to solve algebraic equations
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chiurox
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Homework Statement



The problem seems like this: on the left, closer to the floor, is block3 with mass 1.0kg, there is a string above it which connects it to block2 with mass 3.0kg which has a string above it that circles around a pulley and comes back down connecting it to block1 with mass 5.0kg

The pulley is frictionless and massless


Homework Equations


a. Derive an equation for the acceleration of the block m1=5.0kg and calculate the value of the acceleration.

b. What is the tensionin the string connected between the 1.0kg block (mass3) and the 3.0kg (mass2) block?

c. What is the tension in the string connected between the 3.0kg block (mass2) and the 5.0kg block (mass1)?

d. After the 5.0kg block (mass1) comes to rest on a table surface, what is the force the table exerts on the block?

The Attempt at a Solution



a. I'm not sure how I would derive an equation for it, but I just solved it using the Fnet-y of the larger block and the second block with mass 3.0k. I added both equations:
T1 - m1g = m1a ---------- T1 - 50 = 5a (plugged)
-T1 + (m2g + m3g) = m2a ---------- -T1 + 40 = 3a
After adding both equations I found a to be -1.25m/s^2 (because I established counter-clockwise movement of the pulley as being positive) Is this right?

b. 10-T2 = -1.25 so T2 = 11N I think it makes sense...

c. T1 - 50 = 5(-1.25) so T1 = 44N

d. N = m1g - T1 = 50-44 = N = 6N Well, I think it makes sense, but I'm not sure though.

Could anyone help check these please? because at first I was really lost and still am.
 
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chiurox said:
a. I'm not sure how I would derive an equation for it, but I just solved it using the Fnet-y of the larger block and the second block with mass 3.0k. I added both equations:
T1 - m1g = m1a ---------- T1 - 50 = 5a (plugged)
-T1 + (m2g + m3g) = m2a ---------- -T1 + 40 = 3a
After adding both equations I found a to be -1.25m/s^2 (because I established counter-clockwise movement of the pulley as being positive) Is this right?
It's almost right. In your second equation, what you really did was combine m2 & m3 as a single system--perfectly OK. But then you need to use the mass of that system:
-T1 + (m2g + m3g) = (m2 + m3)a

And then you'd solve for both a and T1.
 
Alright, I understand it, thank you very much. Good thing it didn't change much (because of sig figs) on the other answers.
 

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