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

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Problem 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 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:

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:
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:

haruspex

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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
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.
 
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 cant really get AB. i am truly at a loss
 
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
 

BvU

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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##
 

BvU

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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 thats on the incline plane. the same resultant on the coordinate plane is 6.21m
 

BvU

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sqrt(29) would be the resultant of the AB force vector and thats 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
Determine the tension in the rope and the force F.
The answers to this problem are 104N and 215N
 
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
 

BvU

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

[edit] gotta go - building closes and tonight I have a training :frown:
 

BvU

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On the x-z plane the projection of AB is not 6.21 m !
 
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 lenght of AB.
 
You have lost me: what is the 6.21 and what are the units ?

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

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