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

Click For Summary

Homework Help Overview

The problem involves three blocks connected on a rough table with a coefficient of kinetic friction of 0.420. The masses of the blocks are 4.00 kg, 1.00 kg, and 2.00 kg, and the setup includes frictionless pulleys. The main question is to determine the acceleration of each object and their directions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the blocks, including gravitational forces and friction. There is an exploration of the system's acceleration using the equation F=ma and considerations of how to account for friction and the mass of each block. Some participants question the contributions of specific forces to the overall acceleration.

Discussion Status

There is ongoing dialogue about the correctness of the equations used and the calculations performed. Some participants have offered guidance on checking subscripts and the contributions of forces, while others have noted discrepancies in the calculated acceleration values. The discussion reflects a collaborative effort to verify the approach and calculations without reaching a definitive conclusion.

Contextual Notes

Participants mention potential errors in the online homework system and the importance of ensuring all forces, including friction, are correctly accounted for in the calculations. There is also a note about a typo in the initial equations presented.

hm8
Messages
16
Reaction score
0

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:
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
23
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K