What is the Correct Way to Calculate Tension Between Blocks?

  • Thread starter Thread starter TG3
  • Start date Start date
  • Tags Tags
    Blocks Tension
Click For Summary

Homework Help Overview

The problem involves calculating the tension between two blocks, one with a mass of 3 kg and the other with a mass of 2 kg, being pulled across a floor with a force of 24 N. The coefficient of friction is given as 0.19, and the acceleration has been previously determined to be 2.936 m/s².

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations for tension, including T = F - m1a and T = (m2/(m1+m2)) x F. There are attempts to calculate tension using different methods, but results are consistently incorrect. Questions arise regarding the calculation of acceleration and the net force, particularly how friction affects the overall force.

Discussion Status

Some participants have provided guidance on using free-body diagrams and applying Newton's second law, while others are exploring the calculations of net force and friction. There is an ongoing exploration of the methods used to find acceleration and how they relate to the tension calculation.

Contextual Notes

Participants note that despite using labeled units and different methods, the calculated tension remains incorrect. There is a focus on understanding the relationship between the forces acting on the blocks and the assumptions made in the calculations.

TG3
Messages
66
Reaction score
0

Homework Statement


A block of mass 3 kg pulls a block of mass 2 kg across a floor with a force of 24 N. The coefficient of friction is mew = .19. What is the tension between the blocks?
Note:
Acceleration = 2.936. (Solved for this earlier and was correct.)

Homework Equations


T =F- m1a
T = m2a
T=(m2/(m1+m2)) x F


The Attempt at a Solution


T = 2.936 X 2 = 5.872 Wrong.
T = (24-9.3195) - (3x2.936) = 5.8725 Wrong.
T = (2 / (2+3)) x (24-9.315) = 5.8722 Wrong again.

I've tried 3 different methods, come up with very similar answers that when rounded are equivalent, labeled units, and still gotten it wrong... what's up?
 
Physics news on Phys.org
T =F- m1a
Try this!
 
That was the second method I used:
(24-9.3195) is the net force (since 9.3195 is the force of friction pushing back) and (3x2.936) is m1a.
Still no luck...
 
How did you find the acceleration for the first part? I know you got the correct answer, just want to see how you got it
 
The Acceleration equals the Net force / (mass 1+ mass 2)
The net force equals 24 - resistance due to friction. [(m1+m2) x 9.81 x .19]

Still no luck with the tension though...
 
Draw a free-body diagram for m1.
Use SF=ma, as applied to the mass m1. Since we know the acceleration from part (a), we can find the net force. The net force is the sum of the forces. One of those forces is the tension T.
 
Got it. Thanks!
 

Similar threads

Replies
21
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
8K
Replies
14
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K