[First Year Statics] Find equivalence location of system

[lc99]

Homework Statement

Part C ) help

Homework Equations

So far..

Moment resultant = 1800i +3200k , magnitude = 3070 N*m
Force resultant = 500i +300j + 800k , magnitude = 990 N

The Attempt at a Solution

For the location, I'm not getting x = 1.16, y =2.06.
Instead , I am getting

Mx = Fr*y --> y = Mx / Fr = 1800/990 = 1.82 in the y direction
My = Fr*x --> x = My / Fr = 3200/990 = 3.23

What am i doing wrong?

Edit: i think i found the solution somewhere, but I am not understanding it. How do i know that resultant moment is in the direction of Force resultant? wouldn't they be perpendicular?[/B]

 

[Stephen Tashi]

Edit: i think i found the solution somewhere, but I am not understanding it. How do i know that resultant moment is in the direction of Force resultant? wouldn't they be perpendicular?

How do your text materials define an "equivalence location"? That's not a hint - I genuinely want to know.

The question you quoted isn't clear because it refers to "its line of action" and we don't know what "it" is.

Just taking a guess at what's going on, perhaps you are being asked how to represent the net effect of the forces as a force F applied at a point together with a couple of forces whose moment is parallel to F. (As in Poinsot's Theorem on page 18 of the PDF http://www.cs.cmu.edu/afs/cs/academic/class/16741-s07/www/lectures/lecture14.pdf )

In that case, by some terminology used in the question, it wants the moment of the couple in the same direction as F.

 

[lc99]

How do your text materials define an "equivalence location"? That's not a hint - I genuinely want to know.

The question you quoted isn't clear because it refers to "its line of action" and we don't know what "it" is.

Just taking a guess at what's going on, perhaps you are being asked how to represent the net effect of the forces as a force F applied at a point together with a couple of forces whose moment is parallel to F. (As in Poinsot's Theorem on page 18 of the PDF http://www.cs.cmu.edu/afs/cs/academic/class/16741-s07/www/lectures/lecture14.pdf )

In that case, by some terminology used in the question, it wants the moment of the couple in the same direction as F.

Now that you mentioned, I thinked we assume that Fr and the Mr are parallel because this is a 3-D wrench problem? I think it is called "reduction to a wrench" in my textbook. Are you familiar with this , and is this probably because i have to treat them as parallel?

Equivalent System: "Sometimes it is convenient to reduce a system of forces and couple moments
acting on a body to a simpler form by replacing it with an equivalent system,
consisting of a single resultant force acting at a specific point and a resultant
couple moment. A system is equivalent if the external effects it produces on
a body are the same as those caused by the original force and couple
moment system " (Engineering Mechanics 14th edition by R.C Hibbeler).

 

[Stephen Tashi]

I think it is called "reduction to a wrench" in my textbook. Are you familiar with this , and is this probably because i have to treat them as parallel?

I haven't worked practical problems in "reduction to wrench". I just know the general concept. As to what question your problem is posing, I don't know. I haven't read the entire problem.

 

[lc99]

I haven't worked practical problems in "reduction to wrench". I just know the general concept. As to what question your problem is posing, I don't know. I haven't read the entire problem.

The question was on my mastering engineering, so it guess they worded it differently and unclearly. But, this is the full question..."Replace the three forces acting on the plate by a
wrench. Specify the magnitude of the force and couple
moment for the wrench and the point P(x, y) where the
wrench intersects the plate. "

 

[Stephen Tashi]

this is the full question..."Replace the three forces acting on the plate by a
wrench. Specify the magnitude of the force and couple
moment for the wrench and the point P(x, y) where the
wrench intersects the plate. "

By definition of a wrench, the moment of the couple must be in the same direction as the resultant force.

As I said, I haven't studied practical tricks and techniques for working this sort of problem. If you need that sort of advice, I'll look for some expert to help you.

 

[kimbyd]

I'm really not sure what the question is asking. It would probably help if you posted the entire thing, starting with point A.

 

[lc99]

I'm really not sure what the question is asking. It would probably help if you posted the entire thing, starting with point A.

A) Replace the three forces acting on the plate by a
wrench. Specify the magnitude of the force
B)find the couple moment for the wrench
C)find the point P(x, y) where the
wrench intersects the plate. "

 

[CWatters]

I might be wrong but...I believe the torque produced by the resultant force about the X axis must be the same as that produced by Fb about the X axis. However your solution for "y" uses a simple ratio. I don't think that's correct because the 990N resultant force isn't vertical but at some angle? I think you need just the vertical component of the resultant.

Also I think you made a typo. You wrote 1800i + 3200k when I think it should be 1800j + 3200k.

 

[kimbyd]

I might be wrong but...I believe the torque produced by the resultant force about the X axis must be the same as that produced by Fb about the X axis. However your solution for "y" uses a simple ratio. I don't think that's correct because the 990N resultant force isn't vertical but at some angle? I think you need just the vertical component of the resultant.

Also I think you made a typo. You wrote 1800i + 3200k when I think it should be 1800j + 3200k.

I'm pretty sure more than just the vertical component of the net force is necessary (the vertical component is trivial: 800N).

A) Replace the three forces acting on the plate by a
wrench. Specify the magnitude of the force
B)find the couple moment for the wrench
C)find the point P(x, y) where the
wrench intersects the plate. "

Thanks, that makes this clearer. Please do note the typo that CWatters pointed out. Also, are you clear on what the reduction of the system to a wrench involves?

 

[kimbyd]

Anyway, since I won't be able to respond until the morning and I just verified that my understanding is accurate, I'll lay out a partial description.

The idea of the wrench is that a wrench has a torque in the same direction as its force. Basically, you push (or pull) in one spot on the object while applying a twisting motion along the same orientation. Determining the wrench for this system involves solving a system of equations. First, you know the force applied. Then you have to find a location on the plane where if you add a torque of the correct magnitude, the total torque will equal the total torque you estimated. So you have to solve for three variables: the x and y coordinates of the intersection point, as well as the strength of the torque applied to the wrench. You also get three equations from setting the total torque equal. Solve the system of three equations, and you get the x,y location they're asking for (as well as the torque you have to apply to the wrench to make it equivalent).