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

• dbag123
In summary, the problem is that the user is not sure what the wrong component is for the rope. The user is also not sure if the equilibrium equation is correct. The user needs help with solving for the tension in the rope and the force.
dbag123
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

Last edited by a moderator:
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:
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##

dbag123 said:
The answers to this problem are 104N and 215N
Ah, but in what order ?

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

BvU said:
Ah, but in what order ?
[F,AB]

Last edited:
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 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

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 even though the problem statement had
coordinate plane, meaning x-z plane instead of the incline plane

You have lost me: what is the 6.21 and what are the units ?

 got to go - building closes and tonight I have a training

BvU said:
You have lost me: what is the 6.21 and what are the units ?
6.21 is in meters

On the x-z plane the projection of AB is not 6.21 m !

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.

BvU said:
You have lost me: what is the 6.21 and what are the units ?

 got to go - building closes and tonight I have a training
ok. thanks for help.

## 1. What is a 3D statics problem?

A 3D statics problem involves analyzing the forces acting on an object in three-dimensional space while it is at rest. This includes understanding how different forces, such as gravity, tension, and friction, interact with each other to keep the object in a stable position.

## 2. How does an inclined plane affect the forces on a box?

An inclined plane introduces a component of gravity that acts parallel to the surface of the plane, in addition to the vertical component. This means that the total force of gravity on the box will be split into two vectors, one perpendicular to the plane and one parallel to it.

## 3. What role does the rope play in the 3D statics problem?

The rope acts as a constraint, limiting the movement of the box on the inclined plane. It also provides tension forces that counteract the forces acting on the box, helping to keep it in place. The angle of the rope and the tension forces it provides are important factors to consider in the problem.

## 4. How are the forces and angles calculated in a 3D statics problem?

The forces and angles can be calculated using vector mathematics, such as trigonometry and the rules of equilibrium. This involves breaking down the forces into their components and using equations such as Newton's Second Law and the sum of the forces in each direction being equal to zero.

## 5. What are some real-world applications of 3D statics problems?

3D statics problems are used in many fields, including engineering, architecture, and physics. Some examples of real-world applications include analyzing the forces on a building or bridge, designing stable structures for construction, and understanding the forces on objects in space, such as satellites or spacecrafts.

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