What is the acceleration of the connected blocks on a rough table with friction?

AI Thread Summary
The discussion revolves around calculating the acceleration of three connected blocks on a rough table with a coefficient of kinetic friction of 0.420. The initial calculation yielded an acceleration of 2.74 m/s², but after reviewing the equations and considering the contributions of each block, a mistake was identified regarding the friction force calculation. The correct approach involved ensuring that the gravitational force was properly accounted for, leading to a revised acceleration of 2.212 m/s². The participant confirmed that their teacher reviewed the work and suggested the initial discrepancy might be due to an error in the online homework system. Ultimately, the correct acceleration was established through careful reevaluation of the forces involved.
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Homework Statement


Three blocks are connected on the table as shown below. The table is rough and has a coefficient of kinetic friction of 0.420. The objects have masses of 4.00 kg, 1.00 kg and 2.00 kg, as shown, and the pulleys are frictionless.

p5-50.gif

Determine the acceleration of each object and their directions.

Homework Equations


F=ma
Ffriction=mu*normal force

The Attempt at a Solution


I made positive in the direction i think everything is moving...counterclockwise.

Fsystem = msystem*a
m1g-Ffriction-m2g = (m1+m2+m3)a
m_1g-(mu*m2g)-m2g = (m1+m2+m3)a

a = (m1g-(mu*m2g)-m2g)/(m1+m2+m3)

I plugged everything in and got 2.74 m/s^2.

I think it's okay that I ignored tension and made everything one system...although maybe I'm wrong about that.

It also could be that I got the answer right, and my teacher entered the wrong answer for our online homework thing so it's telling me I got it wrong...
 
Last edited:
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Hello hm8,

Welcome to Physics Forums!
hm8 said:
I made positive in the direction i think everything is moving...counterclockwise.

Fsystem = msystem*a
m1g-Ffriction-m2g = (m1+m2+m3)a
m_1g-(mu*m2g)-m2g = (m1+m2+m3)a

a = (m1g-(mu*m2g)-m2g)/(m1+m2+m3)
Check the above subscripts. Besides the indirect frictional part, the force of gravity on the middle block isn't contributing to the acceleration of the system. But what about block 3?
 
collinsmark said:
Hello hm8,

Welcome to Physics Forums!

Check the above subscripts. Besides the indirect frictional part, the force of gravity on the middle block isn't contributing to the acceleration of the system. But what about block 3?
That was a typo, sorry...using the 2kg block (block 3) gives me the answer 2.74 m/s^2.

My teacher also looked at my work today and said it looked okay, so I think it may just be a problem with how he programmed the answer into the online homework. Unless you see anything else wrong with it?
 
hm8 said:
That was a typo, sorry...using the 2kg block (block 3) gives me the answer 2.74 m/s^2.
Okay then, your
...a = (m1g - μm2g - m3g)/(m1+m2+m3)
equation looks correct to me.

So the equation seems right to me, but something else does not (according to my calculations). See below.
My teacher also looked at my work today and said it looked okay, so I think it may just be a problem with how he programmed the answer into the online homework. Unless you see anything else wrong with it?
Try plugging the numbers back into your equation. I get something different than 2.74 m/s2.
 
Last edited:
Ahh...for some reason I was forgetting to multiply by g for the friction of block to...I was just doing μ*m

The new answer (2.212 m/s^2) works! Thanks!
 

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