Pulley question to determine the weight at a certain angle

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SUMMARY

The discussion focuses on determining the mass of box K required to maintain a 25-degree angle in a pulley system involving three cables (GH, HK, HNL) and a box L with a mass of 52 kg. The equations of static equilibrium (Fx=0, Fy=0) are utilized to analyze the forces acting on the system. The solution involves calculating the tension forces and applying Newton's first law to derive the necessary mass for box K. Participants emphasize the need for clear step-by-step guidance rather than general explanations.

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bobmarly12345
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Homework Statement


An assembly shown in the picture, GH , HK, HNL are three cables. the mass of box L is 52Kg,
a) determine the mass of box K in order to make angle theta as 25 degrees



Homework Equations


Fx=0
Fy=0


The Attempt at a Solution


Fhn x cos(25) = Fhg x cos(0)
fhn x sin(25) = 52kg
fhn(sin25 x cos25) =39kg

doesnt look right to me. steps on how to do it would be great & guidance through it rather than telling me its a static equillibrium question or just over complicating it. thank you for giving it a look
 

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bobmarly12345 said:

Homework Statement


An assembly shown in the picture, GH , HK, HNL are three cables. the mass of box L is 52Kg,
a) determine the mass of box K in order to make angle theta as 25 degrees



Homework Equations


Fx=0
Fy=0
that's Fxnet =0 and Fynet = 0. Yep!

The Attempt at a Solution


Fhn x cos(25) = Fhg x cos(0)
yes, good
fhn x sin(25) = 52kg
hmmm, if you look at the box L of mass = 52 kg, then the tension force in NL is ? And then HN = NL.
fhn(sin25 x cos25) =39kg
What's this?

Once you get HN by looking at the forces acting on L and using Newton 1 to solve forn NL. Now you calculate Fhg from your first equation.

At joint H, the vert comp of HN must equal the tension in HK, per Newton 1 in the y direction, and then look at the forces acting on K to solve for its mass .
doesnt look right to me. steps on how to do it would be great & guidance through it rather than telling me its a static equillibrium question or just over complicating it. thank you for giving it a look[/QUOTE]
 

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