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dbag123

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- Homework Statement
- 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.

- 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 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 215NMy 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:

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 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 215NMy 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:

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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