Hello

Got a following problem.

300N Box on a frictionless inclined plane. The plane itself is in the Z-direction.

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(x-dir)= 144N, Box(y-dir)= -300N

Equations of equilibriums would then go as follow:

$\displaystyle sum Fx=(3.2/6.21)AB+144N=0$

$\displaystyle sum Fy= (-4.4/6.21)AB-300N=0$

$\displaystyle sum Fz= (-3/6.21)AB-F=0$

144n comes from the incline itself, meaning that$\displaystyle 300N*sin(36.87)*cos(36.87)$ 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