A Mass Hanging From Multiple Strings

Click For Summary
SUMMARY

A 6 kg mass is suspended from string 3, with string 1 attached to a wall and string 2 making a 40-degree angle with the ceiling. The problem requires calculating the tension in string 2. The user initially calculated the gravitational force (F_g) as 58.8 N and set up equilibrium equations for the forces. However, a sign error in the calculations led to an incorrect result of -91.44 N for T2. The correct approach was confirmed, but the user needs to clarify the calculations with their teacher.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of vector resolution in physics
  • Familiarity with trigonometric functions (sine and cosine)
  • Ability to set up equilibrium equations for forces
NEXT STEPS
  • Review vector resolution techniques in physics
  • Study equilibrium of forces in static systems
  • Practice solving problems involving tension in strings
  • Learn about common errors in trigonometric calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for examples of tension problems in static equilibrium.

Null42
Messages
2
Reaction score
0
A mass of 6 kg is hanging from a string, named string 3. The top of this string is connected to to other strings, string one which goes straight to the left to connect with a wall, and string two which connects to the ceiling making a 40 degree angle between the ceiling and the string. The problem is asking for the tension of string 2.
The available answers are:

EDIT: Will try and provide a picture when possible.

a) 1.2 N
b) 11 N
c) 34 N
d) 3.5 N
e) 40 N


I thought I could solve the problem by considering each string a vector, representing the force of each. Since there is no movement, the sum of the Forces should equal 0.
So , T1+T2+T3=0

I figure the force of T3 to be the F_g, which should be 6 *9.8=58.8 N

Using this I then came up with the following
T1*cos(180)+T2*cos(40)+58.8*cos(270) = 0
and
T1*sin(180)+T2*sin(40)+58.8*sin(270)=0

I figured I could use the second equation most easily, since the sin(180)=0.

So, I came up with 0+(.643)T2+58.8=0
(.643)*T2=-58.8
T2=-91.44
Of course, this is nowhere close to any of the available answers, and I can't seem to wrap my head around any other way of doing it. Any help would be greatly appreciated.
 
Last edited:
Physics news on Phys.org
Null42 said:
T1*sin(180)+T2*sin(40)+58.8*sin(270)=0

I figured I could use the second equation most easily, since the sin(180)=0.

So, I came up with 0+(.643)T2+58.8=0
(.643)*T2=-58.8
T2=-91.44
Assuming your description is accurate, the available answers are way off.

Your method is perfectly fine. You did make a sign error, but otherwise your answer is correct. (What is sin(270)?)
 
Ahhh, yes, sorry, 58.8 * -1. Thank you for that. I guess I'll just have to discuss it with the teacher. Thanks!
 

Similar threads

Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
10K
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K