Pulling masses up a tilted plane with kinetic friction

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The discussion centers on calculating the force required to pull two boxes up a tilted plane, factoring in kinetic friction. The kinetic friction was calculated to be 55.16N, and the angle of the incline was determined to be approximately 27.76 degrees. The user then calculated a force of 33.36N using sine functions. A final total force of 88.52N was presented, combining the force from sine and kinetic friction. However, there are discrepancies in the numbers, and a request for a more detailed breakdown of the calculations was made for clarity.
jamiebean
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Homework Statement
attached below
Relevant Equations
N= mg
螢幕截圖 2020-05-22 下午8.29.37.png


first, i calculated the kinetic friction:
0.77 x (weight of the 2 boxes x 9.8)= 55.16N

then i calculated the angle of the triangle:
tan^-1(2.5/4.75)=27.758

IMG-0945.jpg


then i drew this

then i used sine to find out force 3 which is 33.3556

so the final force needed is 33.3556 + kinetic friction= 88.516

is this correct?
 
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My numbers disagree with yours. Since you do not provide the equations in which you substituted the numbers, I cannot check the details of where you went wrong. The statement
jamiebean said:
then i used sine to find out force 3 which is 33.3556
is not informative enough. Please post a more complete solution.
 
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