Given the diagram, what is the tension in each string?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the tension in strings connected to masses, specifically with a setup involving angles and forces. The masses in question are all 1 kg, and a diagram is referenced for clarity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest drawing force diagrams for each mass and applying the equations of motion in both x and y directions. There is an exploration of the relationships between the tensions T1 and T2, with some participants questioning the use of sine and cosine in the equations.

Discussion Status

The discussion is ongoing, with participants providing guidance on setting up the equations and checking assumptions about the forces involved. There is acknowledgment of uncertainty regarding the calculations for additional tensions T3 and T4.

Contextual Notes

Participants note the importance of correctly identifying the angles and the need to clarify the direction of forces in the equations. There is also mention of a diagram that is essential for understanding the problem setup.

laladude
Messages
14
Reaction score
0

Homework Statement




Given this diagram and the masses involved, what is the Tension in each string. All masses are 1kg.

Diagram provided

4.jpg


Sorry, the angle for T2 is 10°


Homework Equations





The Attempt at a Solution



I really don't know where to start.
 
Last edited:
Physics news on Phys.org
For each mass, draw a force diagram.

Then use ∑Fx=max and ∑Fy=may. (And you should know what ax and ay are :wink:)
 
Redbelly98 said:
For each mass, draw a force diagram.

Then use ∑Fx=max and ∑Fy=may. (And you should know what ax and ay are :wink:)

okay, so for T1 and T2..

x-dir would be ∑Fx = - T1 cos30 + T2 cos10 = 0

y-dir would be ∑Fy = T1 cos30 + T2 cos10 + -mg = 0

Then solve for T1 = T2 cos10/cos30 = 0

Plug into y-dir eq. (T2 cos10/cos30) cos30 + T2 cos10 = mg

T2 = mg/1.97 which is 4.97.

Then plug T2 to T1. Correct?

Now I am uncertain about T3 and T4..
 
Last edited:
laladude said:
okay, so for T1 and T2..

x-dir would be ∑Fx = - T1 cos30 + T2 cos10 = 0
Yes, good.
y-dir would be ∑Fy = T1 cos30 + T2 cos10 + -mg = 0
Almost -- for the y-dir, we use sin instead of cos.
Then solve for T1 = T2 cos10/cos30 = 0
Except for the "=0" part, that is correct.
Plug into y-dir eq. (T2 cos10/cos30) cos30 + T2 cos10 = mg
Yes, except that there should be some sin's in there because of what I said before about the y-direction forces.
T2 = mg/1.97 which is 4.97.

Then plug T2 to T1. Correct?
Yes, that is the idea.
Now I am uncertain about T3 and T4..
Have a look at the forces acting on Mass #3.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
4K
Replies
4
Views
1K
  • · Replies 22 ·
Replies
22
Views
1K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 29 ·
Replies
29
Views
2K
Replies
2
Views
1K