Moment Question: Solving for Mb with Point B and Force F at A = (a,b)

[Minamatta]

Question on moment !

Homework Statement

Mb ( moment about b ) = 20k
B = (3,4)
F (force) = 5i-2j

Homework Equations

Find the equation .

The Attempt at a Solution

Mb =r*F
r= BA where A is a point on line of action of force
let A=(a,b)
r=A-B = (a,b) - (3,4) = (a-3,b-4)
Mb = (a-3,b-4) *(cross) (5i-2j) = (-2a-6)-(5b-20) k
( -2a-6-5b+20 )k = 20k
-2a-5b=6 ------> (1)
what to do else ?!
hope to help me

 

[Simon Bridge]

Welcome to PF
Note: there is inportant information missing from your description:
Units for force, and I don't recognize "k" as a unit for torque (moment).
I take it that point B is the location of the pivot (you didn't say)?

your problem statement does not include a problem - you have failed to state which equation you are expected to find. Without a clear description of the problem, you cannot find a solution.

It looks like you have been asked to find the length of the moment arm - in which case you need to know if the force is perpendicular or not.

 

[SteamKing]

k is the unit vector perpendicular to i and j (This is a vector problem)
Units would be nice, but are not essential to solving for a and b.
Without another piece of information, I don't think a and b can be determined.

 

[Minamatta]

okay,,
Look K represents a unit vector perpendicular to the plane containing A & B
SO how to solve equation ?!

 

[Simon Bridge]

There is not enough information to help you.
You have not stated the problem - what is it you are supposed to be doing?
Are you supposed to be finding the moment arm?

Assuming that is the idea:
You have established that A is somewhere on the line 2x+5y+6=0
Have you tried sketching this on some axis and seeing how it relates to the position of the pivot?
Trouble is, you have one equation and two unknowns - there is no unique solution to the equation. Put it another way, the line is the solution.

For a unique solution you still need to know the angle the force makes with the moment arm somehow ... and information to do that has not been provided in the problem statement. That is as far as anyone can go with the information provided. Please reread the question to make sure there is no other information provided.

 

[Minamatta]

There is not enough information to help you.
You have not stated the problem - what is it you are supposed to be doing?
Are you supposed to be finding the moment arm?

Assuming that is the idea:
You have established that A is somewhere on the line 2x+5y+6=0
Have you tried sketching this on some axis and seeing how it relates to the position of the pivot?
Trouble is, you have one equation and two unknowns - there is no unique solution to the equation. Put it another way, the line is the solution.

For a unique solution you still need to know the angle the force makes with the moment arm somehow ... and information to do that has not been provided in the problem statement. That is as far as anyone can go with the information provided. Please reread the question to make sure there is no other information provided.

thank you for replying
i don't know my teacher gave me this info only and asked about the equation that's all
anyway thanks.

 

[Simon Bridge]

Thank you - but which equation?
What exactly did your teacher ask you to do?

It's pretty unusual for a teacher to give you a problem that cannot be solved.

 

[Minamatta]

Thank you - but which equation?
What exactly did your teacher ask you to do?

It's pretty unusual for a teacher to give you a problem that cannot be solved.

hehe yes
okay ,, he want the equation of the point on the line of action

 

[Minamatta]

look i got an idea what about getting Ma=zero k
as the line pass through it
so my problem now how to get the equation
please remind me how to get the equation

 

[Simon Bridge]

Ah OK - you already realize: the moment arm is the perpendicular distance from the center of rotation to the line of action of the force.

The line you calculated is the "line of action" right?
Did you draw the picture?

 

[Minamatta]

Ah OK - you already realize: the moment arm is the perpendicular distance from the center of rotation to the line of action of the force.

The line you calculated is the "line of action" right?
Did you draw the picture?

i got the point but don't know how to get equation please tell me how

 

[Simon Bridge]

I don't understand - if you have the point, then what equation are you missing?

In physics, the line of action of a force F expresses the geometry of how F is applied. It is the line through the point at which F is applied and along the direction in which F is applied. You have been supposing that you wanted to find the point that F is applied - but from what you say it looks like your teacher expects you to find the moment arm - which is different. The moment arm is the perpendicular distance from the center of rotation to the line of action of the force.

You have the line of action.
You can find a line perpendicular to that, which passes through the center or use the definition of angular momentum. (What is the angle of the applied force to the moment arm?)

You realize that this is probably what the homework is testing?
There is a limit to what I can tell you. Though you are free to search the web:
http://www.purplemath.com/modules/strtlneq3.htm

 

[Simon Bridge]

How did you get on in the end?