 Problem Statement

300N Box on a frictionless inclined plane. The plane itself is in the Zdirection.
Box is held by a rope AB and a force in the direction Z. Determine the tension in the rope and the force F.
 Relevant Equations

sumFx=(3.2/6.21)AB+144N=0sumFx=(3.2/6.21)AB+144N=0
sumFy=(−4.4/6.21)AB−300N=0sumFy=(−4.4/6.21)AB−300N=0
sumFz=(−3/6.21)AB−F=0sumFz=(−3/6.21)AB−F=0
Hello
Got a following problem.
300N Box on a frictionless inclined plane. The plane itself is in the Zdirection.
Box is held by a rope AB and a force in the direction Z. Determine the tension in the rope and the force F.
The answers to this problem are 104N and 215N
My question is where do i go wrong? the components, equations of equilibrium or both?
rope AB components that i calculated are following
ABx= 3.2m, ABy=4.4m ABz=3m
Box(xdir)= 144N, Box(ydir)= 300N
Equations of equilibriums would then go as follow:
sumFx=(3.2/6.21)AB+144N=0sumFx=(3.2/6.21)AB+144N=0
sumFy=(−4.4/6.21)AB−300N=0sumFy=(−4.4/6.21)AB−300N=0
sumFz=(−3/6.21)AB−F=0sumFz=(−3/6.21)AB−F=0
144n comes from the incline itself, meaning that300N∗sin(36.87)∗cos(36.87)300N∗sin(36.87)∗cos(36.87) and 300 is just mg in the direction y
6.21 is the resultant of the force vector AB.
Personally i think the components are wrong but i would like to know for sure and i have no way of being sure of that myself so help is desperately needed. Thanks in advance
Got a following problem.
300N Box on a frictionless inclined plane. The plane itself is in the Zdirection.
Box is held by a rope AB and a force in the direction Z. Determine the tension in the rope and the force F.
The answers to this problem are 104N and 215N
My question is where do i go wrong? the components, equations of equilibrium or both?
rope AB components that i calculated are following
ABx= 3.2m, ABy=4.4m ABz=3m
Box(xdir)= 144N, Box(ydir)= 300N
Equations of equilibriums would then go as follow:
sumFx=(3.2/6.21)AB+144N=0sumFx=(3.2/6.21)AB+144N=0
sumFy=(−4.4/6.21)AB−300N=0sumFy=(−4.4/6.21)AB−300N=0
sumFz=(−3/6.21)AB−F=0sumFz=(−3/6.21)AB−F=0
144n comes from the incline itself, meaning that300N∗sin(36.87)∗cos(36.87)300N∗sin(36.87)∗cos(36.87) and 300 is just mg in the direction y
6.21 is the resultant of the force vector AB.
Personally i think the components are wrong but i would like to know for sure and i have no way of being sure of that myself so help is desperately needed. Thanks in advance
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