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Tension Problem (two ropes + weight)

by DarkNightwing
Tags: ropes, tension, weight
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DarkNightwing
#1
Nov10-08, 04:58 PM
P: 3
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
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asleight
#2
Nov10-08, 05:01 PM
P: 154
Quote Quote by DarkNightwing View Post
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.
DarkNightwing
#3
Nov10-08, 05:05 PM
P: 3
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

LowlyPion
#4
Nov10-08, 05:06 PM
HW Helper
P: 5,343
Tension Problem (two ropes + weight)

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.)
Physics197
#5
Nov10-08, 05:07 PM
P: 74
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
DarkNightwing
#6
Nov10-08, 05:11 PM
P: 3
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
cragar
#7
Mar21-09, 10:24 AM
P: 2,464
does it matter which one we make negative
mxhaloxm
#8
May31-09, 07:34 PM
P: 1
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 doesnt move left or right.

h1cos55 + (- h2cos60) = 0

Use substitution to solve the problem and your done.

Hope this helps!


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