Net Force: Tension problems with angles

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Homework Help Overview

The problem involves a rope forming a Y shape with a block hanging at the end, where the weight of the block is given as 100N. The angles of the tensions in the ropes are specified, with one tension at 90 degrees and the other at 58 degrees. The objective is to find the tension in each rope.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up equations based on the forces in the x and y directions but questions the validity of their approach after calculating a tension value that seems erroneous. Participants inquire about the measurement of angles and the setup of the problem to clarify the scenario.

Discussion Status

Participants are actively discussing the setup of the problem, with some seeking clarification on the angles and their orientation. There is an exploration of the components of the forces involved, particularly in the x-direction, but no consensus has been reached regarding the solution.

Contextual Notes

There is a mention of a missing visual representation that could aid in understanding the problem setup. The original poster has clarified the orientation of the angles, but further details about the configuration may still be needed.

uni777
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Homework Statement


A rope forms a Y.

At the end of the Y hangs a block with a W= 100N
Whole the two upper ends are on the walls.
(Left side) [tex]T{1}[/tex] with an angle of 90 degrees
(Right side) [tex]T{2}[/tex] with an angle of 58 degrees

Find the tension of each rope on the wall.

Homework Equations


No definite formulas

The Attempt at a Solution


[tex]\Sigma[/tex]Fx=[tex]T{1}[/tex]cos90-[tex]T{2}[/tex]cos58=0
Summation of Fy=Tsub2sin58-100=0
=117.92

cos90=0
Which makes the value of [tex]T{1}[/tex] an error.
Where did I go wrong?
 
Last edited:
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Without a picture, it is hard to imagine what is going on. How are the angles measured? With respect to the vertical or with respect to the horizontal?
 
Oh sorry...
T1 is measured respect to the vertical
T2 respect to the horizontal

As for the picture ill try to provide but erm its like this:
Your left pointing finger must form a right angle with your right pointing finger such that your nails/tips are touching each other.

Then T1 is attached on your left pointing finger.
And T2 is attach on your right pointing finger.
 
p4_11alt.gif


oh its like this but instead of a ninja...its a block
 
OK. So it looks like in the x-direction you have two components of forces. Since the force on the left T1 is entirely in the x-direction (has no y-component), what is its x-component? What about the x-component of T2?
 

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