Find tension in string between two blocks on horizontal surface

Click For Summary
SUMMARY

The discussion centers on calculating tension in a string between two blocks on a horizontal surface, specifically addressing the normal forces acting on each block. The correct approach involves recognizing that the normal force for each block differs, impacting the friction calculations. The textbook answer of 25 N is confirmed, but participants emphasize the importance of accurately determining the normal forces rather than simply summing the weights of both blocks. This nuanced understanding is critical for solving similar physics problems effectively.

PREREQUISITES
  • Understanding of free body diagrams
  • Knowledge of normal force and its calculation
  • Familiarity with frictional forces
  • Basic principles of Newtonian mechanics
NEXT STEPS
  • Study the concept of normal force variations in multi-body systems
  • Learn about friction calculations in different scenarios
  • Explore advanced free body diagram techniques
  • Review problem-solving strategies for tension in strings
USEFUL FOR

Physics students, educators, and anyone interested in mastering mechanics, particularly in understanding forces acting on multiple objects in a system.

I_Try_Math
Messages
119
Reaction score
25
Homework Statement
Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2 m/s^2 to the right, what is the magnitude F of the applied force?
Relevant Equations
F = ma
1706249395888.png

IMG_20240126_010206.jpg

Did I draw the free body diagram incorrectly by treating the two blocks as one system? My textbook claims the answer is 25 N.
 
Physics news on Phys.org
Your mistake is not exactly that you treat as one system but that you calculate the total normal force as equal to the total weight of the two bodies. Think again.

Hint : Each box has different normal force on it, in one it is equal to its weight but on the other it is NOT!!!
 
  • Like
  • Informative
Likes   Reactions: I_Try_Math and Orodruin
Delta2 said:
Your mistake is not exactly that you treat as one system but that you calculate the total normal force as equal to the total weight of the two bodies. Think again.

Hint : Each box has different normal force on it, in one it is equal to its weight but on the other it is NOT!!!
If I understand correctly, the magnitude of the larger block's normal force will be different than its weight which will change the friction calculation. Thank you very much for your help on this question, I'll redo the problem soon.
 
  • Like
Likes   Reactions: Delta2
I_Try_Math said:
the magnitude of the larger block's normal force will be different than its weight which will change the friction calculation.
That's true, but for the purpose of answering the question in post #1 you can treat the two blocks as a unit. You just have to get the total normal force right.
Answering the question in the thread title is a different matter.
 
  • Like
Likes   Reactions: Delta2 and I_Try_Math
I_Try_Math said:
If I understand correctly, the magnitude of the larger block's normal force will be different than its weight which will change the friction calculation. Thank you very much for your help on this question, I'll redo the problem soon.
Yes. With a force of ##25N## at that angle, you might imagine that the first block is almost being pulled off the ground. It's a strange angle for the problem setter to choose.
 
Last edited:
  • Like
Likes   Reactions: I_Try_Math

Similar threads

Replies
21
Views
913
Replies
2
Views
1K
  • · Replies 47 ·
2
Replies
47
Views
3K
Replies
19
Views
3K
Replies
3
Views
558
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 37 ·
2
Replies
37
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
19
Views
4K