3D Statics problem: Box on an inclined plane constrained by a rope

AI Thread Summary
The discussion revolves around a 300N box on a frictionless inclined plane, held by a rope and a vertical force. The user is trying to determine the tension in the rope and the force F, with expected answers of 104N and 215N. Confusion arises regarding the components of forces acting on the box, particularly the horizontal component of gravity and the calculations for equilibrium equations. Participants suggest focusing on components parallel to the incline instead of the coordinate system, leading to further clarification on the resultant force vector and its dimensions. The conversation highlights the importance of correctly identifying forces and their components to solve the problem accurately.
dbag123
Messages
76
Reaction score
3
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

243620


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
 
Last edited by a moderator:
Physics news on Phys.org
dbag123 said:
Box(x-dir)= 144N,
Not sure what this represents. Gravity has no horizontal component.
 
haruspex said:
Not sure what this represents. Gravity has no horizontal component.
the 144N represents the amount of force with which the box slides in the x direction on the frictionless plane. in other words 180N attempts to slide down and the x component of that is 144N.
secondly if gravity indeed has no horizontal component then does that mean that my equilibrium equation for sum of x gets quite weird, then again they might not be correct either.
 
Last edited:
dbag123 said:
the 144N represents the amount of force with which the box slides in the x direction on the frictionless plane.
I don't think that has meaning.
There are three forces on the box. We don't know what the tension is yet, so we don't know the normal force either. All we know is gravity, which acts straight down.
Note that you also wrote
dbag123 said:
Box(y-dir)= -300N
So you are effectively counting two contributions from gravity.
What you could do is sidestep the normal force by considering components parallel to the plane.
 
haruspex said:
I don't think that has meaning.
There are three forces on the box. We don't know what the tension is yet, so we don't know the normal force either. All we know is gravity, which acts straight down.
Note that you also wrote

So you are effectively counting two contributions from gravity.
What you could do is sidestep the normal force by considering components parallel to the plane.
components parallel to the incline or the coordinate plane?. to the incline they would be x: 4m y: 2m z: 3m
to coordinate plane they would be x: 3.2m y: -4.4m z: -3m and the forces respectively: sum fx= (4m/sqrt(29))AB=0, sum fy= (2m/sqrt(29))AB-300=0, sum fz=(-3/sqrt(29))AB-F=0 and components along the coordinate system are in original post. the original equations get as close as F=205N but its quite not right and with it i can't really get AB. i am truly at a loss
 
dbag123 said:
components parallel to the incline
Yes.
dbag123 said:
they would be x: 4m y: 2m z: 3m
I thought we were discussing forces.
 
haruspex said:
Yes.

I thought we were discussing forces.
sum fx= (4m/sqrt(29))AB=0, sum fy= (2m/sqrt(29))AB-300=0, sum fz=(-3/sqrt(29))AB-F=0​
 
dbag123 said:
fx= (4m/sqrt(29))AB=0
ergo AB = 0 ?
 
BvU said:
ergo AB = 0 ?
AB= 215N and F=104N
 
Last edited:
  • #10
So there must be some other force working in the ##x##-direction. But @haruspex #4 suggests to sidestep the normal force.

Also: where does the 29 come from ? If C is point (0,3,0) then ##\angle## BCA is not 90##^\circ##
 
  • #11
dbag123 said:
The answers to this problem are 104N and 215N
Ah, but in what order :rolleyes: ?
 
  • #12
BvU said:
So there must be some other force working in the ##x##-direction. But @haruspex #4 suggests to sidestep the normal force.

Also: where does the 29 come from ? If C is point (0,3,0) then ##\angle## BCA is not 90##^\circ##
sqrt(29) would be the resultant of the AB force vector and that's on the incline plane. the same resultant on the coordinate plane is 6.21m
 
  • #13
BvU said:
Ah, but in what order :rolleyes: ?
[F,AB]
 
Last edited:
  • #14
dbag123 said:
sqrt(29) would be the resultant of the AB force vector and that's on the incline plane. the same resultant on the coordinate plane is 6.21m
Better to include dimensions from now on. sqrt(29) is a number, not a force.
On the incline plane, CB = 5 m (point C is (0,3,0), the bottom end of the pole).
Not clear to me what you mean with 'coordinate plane'. I agree the length of AB is 6.21 m

Also agree with #23 :smile: even though the problem statement had
dbag123 said:
Determine the tension in the rope and the force F.
dbag123 said:
The answers to this problem are 104N and 215N
 
  • #15
BvU said:
Better to include dimensions from now on. sqrt(29) is a number, not a force.
On the incline plane, CB = 5 m (point C is (0,3,0), the bottom end of the pole).
Not clear to me what you mean with 'coordinate plane'. I agree the length of AB is 6.21 m

Also agree with #23 :smile: even though the problem statement had
coordinate plane, meaning x-z plane instead of the incline plane
 
  • #16
You have lost me: what is the 6.21 and what are the units ?

[edit] got to go - building closes and tonight I have a training :frown:
 
  • #17
BvU said:
You have lost me: what is the 6.21 and what are the units ?
6.21 is in meters
 
  • #18
On the x-z plane the projection of AB is not 6.21 m !
 
  • #19
BvU said:
On the x-z plane the projection of AB is not 6.21 m !
correct. 6.21m i got from squaring the components separately under a sqrt so, x: 3.2m y: 4.4m z: 3m and these gave me the 6.21m which is the length of AB.
 
  • #20
BvU said:
You have lost me: what is the 6.21 and what are the units ?

[edit] got to go - building closes and tonight I have a training :frown:
ok. thanks for help.
 

Similar threads

What is the tension in a horizontal rope on an inclined plane with friction?
Replies
22
Views
3K

Hot Threads

Recent Insights

Back
Top