Solving the Weight of a Meterstaff with a Dynamometer

[Izekid]

A meterstaff who weights 0,12 kg is moveable around of it's ends. In the other staff it hangs a weight of 0,50 kg. The staff is hold in a line by a dynamometer, who is fastened in the middle of the staff. What does the dynamometer show?

I take 0,12*0,50*6,67*10^-11 =4,002*10^-12 / 1^2 =4,002*10^-12

Which is totally wrong this should be -11

tell me how to do this I am totally lost

 

[Integral]

I am also lost. I got that way trying to read your problem, Could you look for a better translation?

 

[HallsofIvy]

Actually, not bad English- not compared to my (put whatever language you like here!).

You have a meter stick pinned to a wall by one end (so that end cannot go up or down but the stick can swing about that point). There is a mass of 0.5 kg at the other end and there is a support at the center of the stick, which itself has mass 0.12 kg. What is the force exerted on the stick by that support?

take 0,12*0,50*6,67*10^-11 =4,002*10^-12 / 1^2 =4,002*10^-12[/quot]
Now this really loses me! I have no idea where that "6,667" came from, not to mention the "10^-11"! And why are you multiplying them all together? It would help to show some of your reasoning and not just arithmetic.

Calculate the "torque" about the pinned end of the stick. The mass 0.5 kg has weight 0.5g N (so force is downward) and, since it is 1 m from the pinned end, exerts a torque or 0.5g N-m, clockwise, around the pinned end. Let F be the force exerted by the support. That force is exerted upward and is 0.5 m from the pinned end and so exerts a 0.5F N-m torque, counter-clockwise. We can also, presuming the meter stick is uniform, treat the mass of the stick as if it were concentrated at the center- there is a torque of (0.5)(0.12g)= 0.06g, clockwise. The pinned end, since it is 0 m from itself, contributes no torque. Since the stick does not move, the total torque around any point, and, in particular, around the pinned end must be 0: the counter-clockwise and clockwise torques must be the same. That is
0.5F= 0.5g+ 0.06g= 0.56g. Now solve for F. The answer is NOT exactly 11 N but it is close.

 

[Izekid]

Gahh

You know I was using the the the F=Gm1m2/r2 and G is Constant of Gravitation = 6,667*10^-11
Yeah now that I calulated it I also got 0,56 because i thought in another way... but how do I resolv F I'm like this it stands nothing in my book!

I added a pic to those who don't understand my english !

Thx

 

[Izekid]

Is this the way

The Force for the weight is 0,56nM and then the force must be utilised on the other side to so 0,56+0,56 * 9,82 =10,984=11nM

 

[HallsofIvy]

Yes, exactly.

(It would be more correct (in English, at least) to say the "torque" rather than the force, is 0,56 Nm- in any language you shouldn't use the word "force" with two different meanings. {But who am I to talk, we use "pounds" for both mass and force!})

The formula I gave you was 0,5F= 0,56g so that F= 0.56g/0,5= (0,56)(9,81)/0,5. That gives the same thing.

I hope you realize that you are NOT calculating F by (Gm₁m₂)/r². You certainly don't need that for things happening on the surface of the earth!