Torque Problem. help please

[ChubbyChaser]

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So i am having trouble understanding this problem.


A one meter board with a uniform density, hangs in static equilibrium from a rope with tension T 0.2 m from the left edge of the board. A weight of 3 kg hangs from the left end of the board. What is the mass of the board?

It is question 60 in the examkrackers physics book if anyone is interested

For some reason the book says to take 0.2(3kg) and set it equal to 0.3X

0.2(3)=0.3X
I understand the 0.2 times 3 kg as the lever arm part, but why do we set it equal to 0.3 x. I woul dhave set it equal to 0.8x because thats the remaining length of the board, and this is the force that balances the (.2)(3kg).


Please Help

 

[wcbcruzer]

I don't get it, both the rope and weight are at the left edge of the board. The board would just flop down.

 

[ChubbyChaser]

the rope is 0.2 meters from the end of the 1 meter board, and the 3kg mass is at the veryleft edge of the board

 

[wcbcruzer]

the rope is 0.2 meters from the end of the 1 meter board, and the 3kg mass is at the veryleft edge of the board

oh pfff... For some reason I thought the torque T was 0.2. I shouldn't post so quickly.

 

[wcbcruzer]

Ok so...

3kg(10) = 30 N at left edge, (30)(0.2) = 6 N.m

6 N.m = (mass of board)(10)(0.3)

mass of board = 2kg

 

[ChubbyChaser]

correct but why is it 0.3 lol, i think im missing something big here.. is it because 0.5 would be the center of the board????? and if so is this basically how you would solve most torque problems.

haha i think im clueless when it comes to torques

 

[wcbcruzer]

correct but why is it 0.3 lol, i think im missing something big here.. is it because 0.5 would be the center of the board????? and if so is this basically how you would solve most torque problems.

haha i think im clueless when it comes to torques

oh yes. Whenever you have a long uniform object like a board or rod, you use the center to calculate torque, force, etc. It's like condensing the entire weight of the object at a point in the center. So instead of a board, you can think of it as a ball with mass 2kg connected to the rope and weight by a rigid massless rod.

 

[mshheaddoc]

I had the same question, thanks for asking it!

 

[amar21]

lol i remebmer this questino too

its because you're figuring it out from the center of the board.. at 0.5 m.. so thats how you got the 0.3

 

[ChubbyChaser]

thanks all, that clears up alot of problems cause some of the EK explanations are just crappy!

 

[Genie133]

why is it 0.3 (measured from the 0.2 to the middle) rather than 0.5 (measured from the left end)? why do you start measuring at different points for each of them?

 

[BrokenGlass]

why is it 0.3 (measured from the 0.2 to the middle) rather than 0.5 (measured from the left end)? why do you start measuring at different points for each of them?

You start measuring at the same point, i.e. the point where the rope is attached to the board. The distance from this pivot point to the left end of the board is 0.2m. The distance from this pivot point to the center of the mass of the board is 0.3m. The torques are in opposite directions and you set the 2 torques equal to each other.

 

[Genie133]

You start measuring at the same point, i.e. the point where the rope is attached to the board. The distance from this pivot point to the left end of the board is 0.2m. The distance from this pivot point to the center of the mass of the board is 0.3m. The torques are in opposite directions and you set the 2 torques equal to each other.

ahah...makes sense. thanks!

 

[naphthalene]

Can someone please explain why the pivot point is where the rope is attached? I thought the pivot point can be anything we choose it to be.

 

[eleveneleven]

Can someone please explain why the pivot point is where the rope is attached? I thought the pivot point can be anything we choose it to be.

It doesn't have to be where the rope is attached. If you set the pivot point where the weight is attached, you would have (Tension)(0.2 m.) = (mass of bar)(9.8)(0.5 m.), which would get you the same answer.

You can get an expression for tension from the force equations (T = 3g + mass of bar x g).

 

[naphthalene]

It doesn't have to be where the rope is attached. If you set the pivot point where the weight is attached, you would have (Tension)(0.2 m.) = (mass of bar)(9.8)(0.5 m.), which would get you the same answer.

You can get an expression for tension from the force equations (T = 3g + mass of bar x g).

Do you mean:

T(0.2) = mgd = 3*10*0.5 -->(it's 0.5 and not 1m because we have to choose the center because its uniform density,right?)

Then we solve for T = 75

we know that T= mass of block*g + mass of board*g = 3*10+m*10

Solve for mass of board, and I got 4.5.... 🙁 I'm dumb

 

[bambam02]

Problem has 2 unkowns---> Tension, and Mass of board... therefore we need 2 equations to figure this out.


First--> everything up = everything going down. T= 30N + W

Second, take your moments (torques) about any point on the board... the center of the board is easiest here, and set CCW = CW moments.

30(0.5) = T(0.3)

solve for T --> T=15/.3=50

then solve for W = T-30= 50-30=20N = 2kg


The other way to work this is as a center of mass problem. Clearly the center of mass is where the string is...

so just work it like a sea saw problem. 30N*.2=W*.3
solve for W --> W= 30N(2/3) = 20N = 2kg

Draw a picture if your having trouble with the dimensions

 

[drechie]

can someone explain to me the relationship between pivot point and center of mass?

 

[SN2ed]

Thread closed. These types of questions belong in the MCAT Study Q&A.