Statics assignment HELP! required

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speko1
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Homework Statement



QUESTION 3
Assume that the pulley at A is a small frictionless pulley. The cord AB, and the cord supporting the block are made from the same cord.
Determine the maximum weight rounded down to the nearest Newton of the block that can be suspended in the position shown if each cord is only allowed to support a maximum tension 0f 1978 Newtons, and determine the angle θ.

Homework Equations



NILL

The Attempt at a Solution


Well, so far, I have been getting different answers. There is the exact same question in the textbook, but I am not sure if the working i did, is the same as the one in the book. I tried using the equations of equilibrium, and somehow i got that the angle is 80 degrees which is right. Could anyone give me a hint how to solve this problem? I don't even understand the question. Is value of T 1978 Newtons, or of Fab?
Also, i am guessing that T is 1978 Newtons, since to find the maximum weight, we need a maximum tension in the cable.
I tried finding W(weight) by Tcos30, but it doesn't make sense.
Any help would be appreciated.

Thank you.
 

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Answers and Replies

  • #2
PhanthomJay
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I can't tell from the problem statement what the tension is, it seems to be referring to another problem '1 978' to get that number. But if you assume that is the breaking strength in Newtons for the rope, that's the max any of the ropes can handle. You should note that the tensions in the rope on both sides of a frictionless pulley are the same. Also, since these tensions are equal, I don't know wheer the 80 degrees comes from. The resultant force from the pulley cable tensions must line up with the force from the cable that's holding the pulley at 30 degrees. That leads to theta = 60 degrees. Which rope breaks first, the wrapped pulley rope or the rope holding up the pulley??
 
  • #3
speko1
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Hey Jay,

Thanks for the help.

The question is asking what the weight in the block on the other side of the rope pulling it downwards.

I am pretty sure that the theta is 80 degrees, as the lecturer confirmed it, but I just don't get which of the two ropes has the tension of 1978N. I tried using equations of equilibrium, summing all x and y components, but it doesn't seem to work. Also, I am aware that the tension in the rope holding the block should be the same on both sides, but does it mean that the T=W? because it doesn't seem like it. If we say that the tension pulling the rope to the left is unknown, how am I going to find the W?
 
  • #4
speko1
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I can't tell from the problem statement what the tension is, it seems to be referring to another problem '1 978' to get that number. But if you assume that is the breaking strength in Newtons for the rope, that's the max any of the ropes can handle. You should note that the tensions in the rope on both sides of a frictionless pulley are the same. Also, since these tensions are equal, I don't know wheer the 80 degrees comes from. The resultant force from the pulley cable tensions must line up with the force from the cable that's holding the pulley at 30 degrees. That leads to theta = 60 degrees. Which rope breaks first, the wrapped pulley rope or the rope holding up the pulley??

Also I don't get it where you got 30 degrees angle? The angle between the imaginary axis and the rope holding the pulley is 40 degrees.
 
  • #5
PhanthomJay
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Hey Jay,

Thanks for the help.

The question is asking what the weight in the block on the other side of the rope pulling it downwards.

I am pretty sure that the theta is 80 degrees, as the lecturer confirmed it,
Your sketch shows that the angle between the rope holding the pulley and the vertical is 30 degrees, which makes theta = 60 degrees. If the angle is 40 degrees between the rope holding the pulley and the vertical, then theta = 80 degrees.
but I just don't get which of the two ropes has the tension of 1978N. I tried using equations of equilibrium, summing all x and y components, but it doesn't seem to work. Also, I am aware that the tension in the rope holding the block should be the same on both sides, but does it mean that the T=W?
Yes.
because it doesn't seem like it.
Why not? Draw a free body diagram of the hanging block.
If we say that the tension pulling the rope to the left is unknown, how am I going to find the W?
That tension is W, and the line of action of the resultant force on the pulley of the (2) lower rope tensions acts along the line of action of the tension force of the rope holding the pulley. Solve for W from the equilibrium equation taken along the axis of the rope holding the pulley. Either the lower rope has a tension of 1978 N, or the upper rope has a tension of 1978 N. Which is it?
 
  • #6
speko1
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Hello Jay,

That definitely helped a lot, as I was kind of stuck on it.

Still, I don't know how to calculate the tension in the rope which is the same as the weight of the rigid body.

I am guessing that the top part is(the cable holding the pulley) has a tension of 1978 N, and the angle of 40 degrees. If I am to assume they are in equilibrium(which is obvious, since we are doing statics), and I don't know the value of T and W (T=W).

The tension holding the pulley has the magnitude of 1978 Newtons, because if it was otherwise the weight should be the same as the tension, and it would be 1978, which is not the correct answer.

I also tried using equations of equilibrium, but we also have to take in the count that the tension in the rope holding the weight, cannot exceed the magnitude of 1978.

Should I put the cable holding the pulley as 1978 N and should I draw a FBD with the pulley, or think of it as a dot, where all the forces intercept?

Thank you.
 
  • #7
PhanthomJay
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Hello Jay,

That definitely helped a lot, as I was kind of stuck on it.

Still, I don't know how to calculate the tension in the rope which is the same as the weight of the rigid body.

I am guessing that the top part is(the cable holding the pulley) has a tension of 1978 N, and the angle of 40 degrees. If I am to assume they are in equilibrium(which is obvious, since we are doing statics), and I don't know the value of T and W (T=W).

The tension holding the pulley has the magnitude of 1978 Newtons, because if it was otherwise the weight should be the same as the tension, and it would be 1978, which is not the correct answer.

I also tried using equations of equilibrium, but we also have to take in the count that the tension in the rope holding the weight, cannot exceed the magnitude of 1978.

Should I put the cable holding the pulley as 1978 N and should I draw a FBD with the pulley, or think of it as a dot, where all the forces intercept?

Thank you.
The (correct) tension of 1978 N in the cable holding the pulley is not the same as the tension in the rope wrapped around the pulley, which is W. You can draw a FBD of the pulley (treating the pulley as a dot) and identify all the forces acting on the pulley (which will include the 2 tensions from the lower rope), and using the line of action of the 1978 N force as the 'y' axis, sum the 1978N force with the y components of the other tensile forces and set the result equal to 0 to solve for W.
 
  • #8
speko1
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Don't worry.

The question is very complex but i solved it.

Now i got another question :tongue: that I am going to post, so please help if you can
 

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