Problem statement
Solution video1
Solution video2
DISCUSSION THREAD
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HOMEWORK 22.B - SUM 24
19 thoughts on “HOMEWORK 22.B - SUM 24”
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Do we need the tensions of the cable to solve this problem or can we just do it with the weight of the block? If we do need the tensions, I'm not sure how we find that.
I don't know that we need the tensions of the cables, but I think we need more than the weight of the block since some of the weight is transferred into the structure. My question is: how do we find the reaction forces on the pulleys at M and K? Clearly the pulleys themselves must be exerting some normal force on the cables, thereby the cables must be exerting a reaction force on the pulleys. But attempting to isolate the pulley and truss systems leaves too many unknowns.
Yeah, I'm not really sure if we don't account for them because if we did the problem would become harder to solve, so I just accounted for the weight of the block but I feel like I am missing something.
Okay, I think I got it. It's kind of hard to explain without pictures, but I'l try my best:
The pulley system is in static equilibrium, so on each pulley, the tension forces of the cables on both sides of the pulley will be equal and outwards from the pulley. Therefore, the pulley directly connected to the weight would have an upward force of P for each cable, and pulley M would have a force of P for both the leftward and downward cables, and so on.
However, for every force, there must be an equal and opposite reaction force. The cables do not provide reaction forces for each other; hence, the reaction forces must be exerted on the pulleys themselves and transferred into the truss.
In other words, if pulley M has a leftward cable with tension force P and a downward cable with tension force M, there must be rightward reaction force P and upward reaction force P exerted on point M, and so on for point K. This then gives us all the starting forces acting on the truss.
Would we still need to account for a downward force of 2P for the weight of the block?
Yes, you would. The bottom pulley would have two cables going up each exerting a force P, thereby cancelling out the block's weight of 2P.
would the force from the cable on m and k be the same maginitude then?
From what I understand, the force in the cable should be equivalent to the force P.
Same here, I have no idea on what to do with the pulleys at M and K. Will work on the other parts until I figure out what to do.
Am I the only one confused by
"horizontal members of length: 4d
horizontal members of length: 3d"
I assume one of them is supposed to be vertical
I think the vertical length is 3d and the horizontal length is 4d.
This is correct, I emailed the professor regarding the typo.
By just eyeballing the diagram, I think vertical should be 3d and horizontal should be 4d as the horizontal side seems to be longer than the vertical. There was likely a typo in the question.
The vertical length should be 3d and the horizontal length should be 4d.
Is there an X and a Y reaction for the cable support at the bottom?
I think at the cable support at the bottom connected to the weight W, there is only a Y reaction.
That's what I determined as well.
I agree with this as well as the weight is being held by each cable which is producing the two forces at M and K equal to P cancelling out the W of 2P.
Hello, when will the solution video be available?