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chudd88
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[STRIKE]
Two ropes are connected to a steel cable that supports a hanging weight as shown.
If the maximum tension either rope can sustain without breaking is 5000 N, determine the maximum mass m that the ropes can support.
Newton's equations #1, #3.
Let the left rope have tension T1, and the right rope have tension T2. So, the horizontal component of T1 is cos(60)*T1, which must equal the horizontal component of T2, which is cos(40)*T2. So:
cos(60)*T1 = cos(40)*T2
T1 = (cos(40) / cos(60)) *T2
T1 = 1.532*T2
So T1 has the greater tension.
The problem states that the maximum tension for either T1 or T2 is 5000N. So, since T1 has the greatest tension, we give it a tension of 5000N. This means that T2 has a tension of 3266N.
The weight of the block is equal to the sum of the vertical components of the two tensions. So:
w = sin(60)*T1 + sin(40)*T2
So we have:
w = sin(60)*5000N + sin(40)*3266N
w = 5035N
That's the answer I get. My book tells me the answer is 6400N. I'm not sure where I've gone wrong here.[/STRIKE]
Nevermind all of this. As it turns out, I was just screwing up the final calculation. sin(60)*5000N + sin(40)*3266N doesn't equal 5035N, as I indicated.
So, nothing to see here...
Homework Statement
Two ropes are connected to a steel cable that supports a hanging weight as shown.
If the maximum tension either rope can sustain without breaking is 5000 N, determine the maximum mass m that the ropes can support.
Homework Equations
Newton's equations #1, #3.
The Attempt at a Solution
Let the left rope have tension T1, and the right rope have tension T2. So, the horizontal component of T1 is cos(60)*T1, which must equal the horizontal component of T2, which is cos(40)*T2. So:
cos(60)*T1 = cos(40)*T2
T1 = (cos(40) / cos(60)) *T2
T1 = 1.532*T2
So T1 has the greater tension.
The problem states that the maximum tension for either T1 or T2 is 5000N. So, since T1 has the greatest tension, we give it a tension of 5000N. This means that T2 has a tension of 3266N.
The weight of the block is equal to the sum of the vertical components of the two tensions. So:
w = sin(60)*T1 + sin(40)*T2
So we have:
w = sin(60)*5000N + sin(40)*3266N
w = 5035N
That's the answer I get. My book tells me the answer is 6400N. I'm not sure where I've gone wrong here.[/STRIKE]
Nevermind all of this. As it turns out, I was just screwing up the final calculation. sin(60)*5000N + sin(40)*3266N doesn't equal 5035N, as I indicated.
So, nothing to see here...
Last edited:
We've all been there / done that