How Do You Calculate Tension Forces in a Hanging Weight Problem?

[mrserv0n]

Let me try to draw this out

--------------------- <-- R2 <30degree
\ R1<50degree /
\ /
\ /
\ /
\ /
\ /
\ / T2 is acting north along the 30 degree (R2)
| T1 is acting north along the 50 degree (R1)
|
|
W=300N

Problem: a 300-N weight is hanging from the ropes R1 and R2. Obtrain the tension forces T1 and T2 acting on the weight.

I have a test in 2 hours and no clue how to solve this problem. I have figured out the 6 others study ones for the most part but need someone to help do this

It appears u can draw it doesn't keep it in tact its basically a triangle upside down with a weight hanging from it

 

[nrqed]

Let me try to draw this out

--------------------- <-- R2 <30degree
\ R1<50degree /
\ /
\ /
\ /
\ /
\ /
\ / T2 is acting north along the 30 degree (R2)
| T1 is acting north along the 50 degree (R1)
|
|
W=300N

Problem: a 300-N weight is hanging from the ropes R1 and R2. Obtrain the tension forces T1 and T2 acting on the weight.

I have a test in 2 hours and no clue how to solve this problem. I have figured out the 6 others study ones for the most part but need someone to help do this

It appears u can draw it doesn't keep it in tact its basically a triangle upside down with a weight hanging from it

Those problems are always dealt the same way:

a) Draw a y and x-axis (here the obvious choice is to put the x-axis aligned East-West)

b) Decompose all the forces into their x and y components

c) Apply $\sum F_x = m a_x$ and $\sum F_y = m a_y$. In this example, the x and y components of the forces add up to zero.

d) You have two equations and two unknowns (at most). Solve for what you are looking for.

 

[kyle8921]

To obtain the tension forces T1 and T2, we can use the principles of static equilibrium. This means that the sum of all forces acting on the weight must be equal to zero, since the weight is not accelerating.

First, we can break down the weight into its components. The vertical component, Wsin50, is balanced by the tension force T1. The horizontal component, Wcos50, is balanced by the tension force T2. We can set up the following equations:

ΣFy = Wsin50 - T1 = 0
ΣFx = Wcos50 - T2 = 0

Solving for T1 and T2, we get:

T1 = Wsin50 = 300sin50 ≈ 230.9 N
T2 = Wcos50 = 300cos50 ≈ 191.8 N

Therefore, the tension forces T1 and T2 acting on the weight are approximately 230.9 N and 191.8 N, respectively.