Statics: Equilibrium in 3-Dimensions

  • #1
Soumalya
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2

Homework Statement



The portable reel is used to wind up and store an air hose. The tension in the hose is 100 N and a vertical 200-N force is applied to the handle in order to steady the reel frame. Determine the minimum force P which must be applied perpendicular to the handle DE and the vertical components of the force reactions at the feet A, B, and C. The diameter of the coil of reeled hose is 300 mm, and the weight of the loaded reel and its frame may be neglected. Note that force P is perpendicular to OD. State any assumptions.
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2. Homework Equations

The scalar force equilibrium equations in three mutually perpendicular directions x-,y- and z-,i.e,

∑Fx=0 ∑Fy=0 and ∑Fz=0

The scalar moment equilibrium equations about three mutually perpendicular axes through a point,i.e,

∑Mx=0 ∑My=0 and ∑Mz=0

The Attempt at a Solution



My assumption initially was that the surface on which the feet of the whole thing rests is frictionless,so that, there are only normal force reactions at A,B and C. Assuming so by balancing the horizontal components of P(which is Pcos 30°) and the 100 N tension(which is 100cos15°) in the hose the value of P obtained doesn't match the given answer(which is given as P=50N).

But If we consider the surface to be rough we will have three mutually perpendicular force reactions at each feet - a normal reaction force and two orthogonal components of the resultant frictional force acting in the plane of the surface.Therefore, we would have to deal with 10 unknowns in the problem including 9 unknown force reaction components at A,B and C and the unknown force P. For the problem to be statically determinate we may allow up to only six unknowns in case of a 3-D force system.What additional assumptions can simplify the problem?[/B]
 
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  • #2
The object is not a single rigid body so you have more than 6 constraints since each part is static. You should assume the axis is a friction-less bearing so no torque about it can be exchanged between reel and stand. Since the reel is not accelerating axially you should then be able to solve for P by simply considering the total torque about the axis.

You cannot however solve for all forces in this problem, there could be opposing forces applied at points A and B and C parallel to the base surface (imagine you compressed the square-C shape of the base a bit and glued it to the floor before it could "spring back".) Fortunately the problem only asks for P.
 
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  • #3
jambaugh said:
Fortunately the problem only asks for P
The second part asks for normal reactions at A, B and C. But you are right that P can be solved first, and quite easily, by just considering forces on the reel.
 
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  • #4
jambaugh said:
The object is not a single rigid body so you have more than 6 constraints since each part is static. You should assume the axis is a friction-less bearing so no torque about it can be exchanged between reel and stand. Since the reel is not accelerating axially you should then be able to solve for P by simply considering the total torque about the axis.

You cannot however solve for all forces in this problem, there could be opposing forces applied at points A and B and C parallel to the base surface (imagine you compressed the square-C shape of the base a bit and glued it to the floor before it could "spring back".) Fortunately the problem only asks for P.

Assuming that the reel is mounted on a frictionless bearing attached to the stand so that no torque is exchanged between the reel and the stand a moment equlibrium equation about the axis of the reel gives,

100(cos15°).0.15=P(0.3)

⇒P=48.3N (which is close to the given answer of P=50N)

But the problem also asks for the normal reactions at the feet of the stand at A,B and C.It seems that they cannot be solved for but my textbook has got answers to them!
 
  • #5
haruspex said:
The second part asks for normal reactions at A, B and C. But you are right that P can be solved first, and quite easily, by just considering forces on the reel.
Is there absolutely no way of determining the normal reactions at A,B and C with the help of some simplifying assumptions?

My textbook has got answers to them as NA = 108.6N, NB = 32.4N and NC = 58.1N
 
  • #6
You should be able to solve for them. You have 3 normal components = 3 variables. You have from the first part and given information the total force and torque applied to the base. The normal force gives you the sum of the three normal components. The two torque components about axies parallel to the ground gives the differential components. Note the fact that points A B and C have the same vertical coordinate so there can be no couplings due to forces parallel to the base surface contributing to torques other than about the vertical axis. All horizontal axis torques must come from normal components.

[EDIT: and PS, pardon me for not reading the rest of the problem.]
 
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  • #7
Soumalya said:
But the problem also asks for the normal reactions at the feet of the stand at A,B and C.It seems that they cannot be solved for but my textbook has got answers to them!

Why can't you determine the reactions at the feet? You know all of the loads on the reel. Try writing the equations of static equilibrium for the reel. If you choose wisely, you should have two equations in two unknowns to solve.
 
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  • #8
Soumalya said:
Is there absolutely no way of determining the normal reactions at A,B and C with the help of some simplifying assumptions?
I didn't say they could not be found, I just said P can found first, quite easily, by considering only the forces on the reel.
Having found P, you could then treat the reel plus holder as a single rigid body to find the normal reactions at the ground.
 
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  • #9
SteamKing said:
Why can't you determine the reactions at the feet? You know all of the loads on the reel. Try writing the equations of static equilibrium for the reel. If you choose wisely, you should have two equations in two unknowns to solve.
I solved the problem treating the reel and the stand as separate rigid bodies.:smile:
 
  • #10
Soumalya said:
I solved the problem treating the reel and the stand as separate rigid bodies.:smile:
Ok, but that does seem a long way around since it involves dealing with the forces between the two. That can be avoided if you find P by taking moments about the reel's axis, then finding the ground forces by treating the reel and stand as one.
 
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