# Help calculation torque, moments.

1. Dec 5, 2008

### tweety1234

1. The problem statement, all variables and given/known data

A uniform rod AB of length 4m and mass 2kg is suspended in a horizontal position by two vertical strings attached at points p and Q where AP=1m and AQ=3m. When a particle of mass 3kg is attached at point R of the rod, the rod is on the point of turning about point P. Calculate the distance AR.

3. The attempt at a solution

As the rod is on the point of tilting about P than, Rp =0

I have drawn the diagram and have attached with this post, but i am not sure it is correct, and also I am not quite sure were to put point 'R' on my diagram.

the moments equation that I have come up with is;

$$2kg + 3kg (xm-3m)=0$$

$$2 \times 9.8 + 3 \times 9.8 (xm-3m) =0$$

$$19.6 + 29.4x-88.2=0$$

$$29.4x-68.2=0$$

$$29.4x=68.2$$

$$x = 2.3$$

I think my moments equation is wrong, or my diagram, can anyone please show me where i am going wrong

thanks!!

Last edited: Dec 5, 2008
2. Dec 5, 2008

### aerospaceut10

I'm a bit confused with your notation, I guess. what is m?

3. Dec 5, 2008

### jmarcian

here is my attempt,

first off, dont forget that you are looking for WEIGHT, which is (mass)(gravity)

ok, if you draw your diagram, you will notice in order for the rod to move, the weight at point r has to be between AP, let's call A=0m and P=1m. Now draw the rods weight, and where it should be placed, in the center of the rod, or W=2m. notice now that the distance from P to the center of the rod must be 1m. since A=0m and W=2m, and since P=1m, the distance from P to W is 1m.

now just set up you equation, i used a variable "a" to represent the moment arm of the point r, but this distance measures rp not ar. but that is easy to solve, let start with the moment equation.

Mp=(3kg)(9.81kg/m2)(a)-(2kg)(9.81kg/m2)(1m)=0
solving for a we get a=2/3m

now remember what i said before a is the moment arm of the point r, measured from p. also remember for the rod to move, r has to be to the left of p. and since AP=1m, and rP=2/3m we arrive at our solution

Ar=1/3m

hopethis helps! this was my first time trying to explain something on this site hehehe

4. Dec 5, 2008

### HallsofIvy

Staff Emeritus
I presume that "xm" means x meters and "3m" means 3 m. But I don't know why you are calculating torque about Q. The problem says it is about to pivot about P, not Q. "AQ=3m" means that the distance from A to Q is 3 meters. Since the stick is 4 meters long that means that the stick has supports 1 meter from each end.

Since the mass of the rod is 2 kg, we can treat that as a downward force 2 m from A. A 3 kg mass is attached at point R, some unknown "x" meters from A.

If "the rod is on the point of turning about point P", the torque in both directions, clockwise and counter clockwise must be the same and the total torque 0 and the upward force at Q must be 0. That means we have only the weight of the rod, 2g N at 1 m from P and the 3g N force at R. Taking "to the right of P" to be positive, we have torque 2g N-m from the weight of the rod and torque 3g(x-1) (since P is itself 1 meter from A). The total torque is 3g(x-1)+ 2g= 0. Solve that for x.

It should be obvious that to offset the weight of the bar to the right of P, R must be to the left of P and x must be less than 1.

5. Dec 5, 2008

### HallsofIvy

Staff Emeritus
One of the things you should have learned from the things you were asked to read when you registered is that you should not give full solutions!

6. Dec 5, 2008

### tweety1234

Last edited: Dec 5, 2008
7. Dec 5, 2008

### jmarcian

WHOOPS! ill keep that in mind!!

in your drawing, tweety, notice r is not in the appropriate place. becuase having the weight where you placed it would not make the bar rotate about point P, but Q instead...

8. Dec 5, 2008

### tweety1234

Oh right, so should it be next to p instead?

9. Dec 5, 2008

### jmarcian

yes! (in between p and a!)