Mass of box K for 25 degree angle in pulley assembly

[bobmarly12345]

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

right i started with
Fhn x cos(25) = Fhg x cos(0) as cos(o) is 1 i then wrote it out as
Fhn x cos(25) = Fhg
box L = 52Kg, i know HN = NL does the tension force NL= 52?
i know that when i get HN by looking at the forces acting on L & using Newtons 1st law to solve for NL i can calculate Fhg from the equation i started with. confused on how to get HN=NL thinks its 52.

then at joint H, the vert comp of HN must equal the tension in HK, per Newton 1st law in the y direction, and then i look at the forces acting on K to solve for its mass. i understand the theory I'm just confused about the calculations that need to be made, can anyone run through it with me please to find the mass of box K?

 

[truesearch]

The 52kg block weighs 52x9.81N...this is the tension force in the string HN.
If you look at point H can you see that the horizontal component of the tension = tension in string HG and the vertical component of the tension must be the weight of block K

 

[bobmarly12345]

oh yes i forgot about the effect of gravity on the 52Kg block.
ok i understand to a point about the mass of block k , so would the mass of block k be HN x (cos25+sin25) ? then divide by gravitational constant to get the Kg value ?

 

[truesearch]

You only need to consider the vertical component of the tension (510Sin25) to find the weight of block K

 

[bobmarly12345]

oh yeah i see now why its only the vertical component, so just 510xsin25=215.5 that's all?

 

[truesearch]

That is the WEIGHT...I think you have to find the mass (easy?)

 

[bobmarly12345]

yeah just divide by 9.81 i believe ?

 

[truesearch]

that's it !

 

[bobmarly12345]

brilliant thanks for your help! :)