Tension Problem (two ropes + weight)

  • Thread starter Thread starter DarkNightwing
  • Start date Start date
  • Tags Tags
    Tension Weight
AI Thread Summary
To determine the tension in two ropes supporting a 300kg block, the total weight is calculated as 2940 N. The system is in equilibrium, requiring that the vertical components of the tensions equal the weight, while the horizontal components cancel each other out. The equations derived from the angles of the ropes (60 degrees for T1 and 55 degrees for T2) lead to two simultaneous equations. By solving these equations through substitution, the tensions are found to be T1 = 1862 N and T2 = 1620 N. Understanding the vector components of tension is crucial for solving such equilibrium problems.
DarkNightwing
Messages
3
Reaction score
0
Oh I knew I'd have to come crawling onto the internet in search of help, help me physics friends !

1. Determine the tension in each of the ropes holding the object. Rope T1 is at an angle of 60 degrees, rope T2 is at an angle of 55 degrees, the suspended block has a mass of 300kg.
2. w = mg
3. 300kg * 9.8 m/s = 2940 N, or the weight of the block, but since this I've just divided the weight by the sin of each rope's given angle. Completely baffled and very tired, looking for a push in the right direction so my brain can worky again.

The answers are:
T1 = 1862 N , T2 = 1620 N but I have no clue how to arrive at them
 
Last edited:
Physics news on Phys.org
DarkNightwing said:
Oh I knew I'd have to come crawling onto the internet in search of help, help me physics friends !

1. Determine the tension in each of the ropes holding the object. Rope T1 is at an angle of 60 degrees, rope T2 is at an angle of 55 degrees, the suspended block has a mass of 300kg.



2. w = mg



3. 300kg * 9.8 m/s = 2940 N, or the weight of the block, but since this I've just divided the weight by the sin of each rope's given angle. Completely baffled and very tired, looking for a push in the right direction so my brain can worky again

Sum your forces to put the system in equilibrium.
 
Ok so in other words the tension in T1 and T2 adds up to the weight of the block?

T1 + T2 = 2940 N

My problem is determining the tension in each rope, I thought maybe you took:

2940 / sin(60) = but this comes out to over 3300 some and that's more force than would be required to hold the block
 
Without knowing where the ropes are attached - hence the angles with respect to the walls or the ceiling ... it doesn't really matter.

Treat the tensions as vectors. They each have vertical and horizontal components.

The vertical components of the tensions must equal the weight. (It's in equilibrium.)

And the horizontal components add to 0. (It's in equilibrium.)
 
Since the object is in equilibrium, the sum of the forces in the y-components must be balanced, and same with your x-components.

This means (if angles are above the horizontal)
T1sin60 + T2sin55 = the weight of the object

Since you have 2 unknowns you need 2 equation, so you must use the forces in the x-directions to get another equation

-T1cos60 + T2cos55 = 0 ---> One of these must be negative cause they should be in opposite directions

Rearrange to solve for either T1 or T2 and plug into the other equation to solve for the other
 
Last edited:
Doh I forgot to state that I have the answers (it's a practice test, reviewing for Friday):

T1 = 1862 N , T2 = 1620 N

Also putting these in the topical post
 
does it matter which one we make negative
 
To solve this:

We know that the y component of the 2 tension must add together to equal the downward force of the block (This makes 0 net force = block not moving).
Therefore:

h1sin55 + h2sin60 + (-9.8)(300) = 0

We also know that the x components of the 2 tension must cancel each other out because the block doesn't move left or right.

h1cos55 + (- h2cos60) = 0

Use substitution to solve the problem and your done.

Hope this helps!
 
Back
Top