Solving a Vertical Winch System

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The discussion focuses on calculating the tension in a cable of a vertical winch system lifting two packages: a 40-kg package and a 10-kg package. The frictional force acting on the 40-kg package is 60N, and the cable makes a 15-degree angle with the vertical. The user initially calculated the tension as 88N for the 10-kg package but received feedback indicating errors in their calculations, particularly in equations (1) and (3). A recommendation was made to draw separate force diagrams for each block to clarify the forces acting on them.

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Hey! I'm having some problems with this question:
An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the diagram. The magnitude of the frictional force acting between the wall and the 40-kg package is 60N. The angle the cable makes to the vertical is 15. If the packages are lowered with an acceleration 1.0 m/s2 [D], what is the tension in the cable between the winch and the 40.0 kg package?

We were also given a diagram:
http://i1289.photobucket.com/albums/b505/homework2012/Untitled_zpsd252a02a.jpg

How i solved it:
(1) Fnet y= Fg1+Fg2+T2+T1Y+FK
mnet(anet)= Fg1+Fg2-T2-T1Y-FK
(2) Fnet=Fg+T2
ma=m2g-T2
(10)(0.1)=(10)(9.8)-T2
T2 = 88N
(3) mnet(anet)= Fg1+Fg2-T2-T1Y-FK
(50)(0.1)= 40(9.8) + 10(9.8) -88 -T1Y- 60
T1Y= 292N
(4) Sin?=o/h
Sin75=292/h
h = 302.3 N

Is how i solved it right? If not is there another way I should be solving it?
 
Last edited:
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Draw a force diagram for each block separately. (1) and (3) are wrong. You also didn't use the right numbers when you plugged into (2).
 

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