Torque of a Board with masses

[AwayFromReality]

Hey guys

I have a seemingly simple concept that I'm having trouble understanding. It has to do with masses on a board about some fulcrum, here is an example problem

Mary and Tim balance on a 10 m board. The board has a mass of 20kg. If Mary and Tim have a combined mass of 180kg, what is Mary's Mass? (Mary is 7m from the fulcrum, Tim is 3m from the fulcrum)

I know I have to set the counterclockwise torque = clockwise torque. I know Mary has counterclockwise torque, I know Tim has clockwise torque.

But I can't seem to figure out the direction for the board's torque or the distance from the fulcrum for the board. The questions is pretty simple once I can figure out the direction and distance for the board.

Any help would be much appreciated.

EDIT: Here's a picture of the question

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[Cawolf]

I am going to take it that we can say Tim is at x = 0, Mary is at x = 10, and the axis of rotation is at x = 3.

Let Mt + Mm = M = 180 kg

Torque = T(tim) + T(board) + T(mary) = 0

T = 0 = (Mt)(g)(3m) - (20kg)(g)(2m) - (Mm)(g)(7m)

(cancel all g's)

(3 m)(Mt) = 40 kg*m + (7 m)(Mm)

Mt = [40 kg*m + (7 m)(Mm)]/(3 m)

we know Mm + Mt = 180 kg

Mt = 180kg - Mm

[40 kg*m + (7 m)(Mm)]/(3 m) = 180kg -Mm

40 kg*m + (7 m)(Mm) = 540 kg*m - (3 m)(Mm)

10 m(Mm) = 500 kg*m

Mm = 50 kg so Mt = 130 kg

Mary is 50kg and Tim is 130 kg

 

[justadream]

I am going to take it that we can say Tim is at x = 0, Mary is at x = 10, and the axis of rotation is at x = 3.

Let Mt + Mm = M = 180 kg

Torque = T(tim) + T(board) + T(mary) = 0

T = 0 = (Mt)(g)(3m) - (20kg)(g)(2m) - (Mm)(g)(7m)

(cancel all g's)

(3 m)(Mt) = 40 kg*m + (7 m)(Mm)

Mt = [40 kg*m + (7 m)(Mm)]/(3 m)

we know Mm + Mt = 180 kg

Mt = 180kg - Mm

[40 kg*m + (7 m)(Mm)]/(3 m) = 180kg -Mm

40 kg*m + (7 m)(Mm) = 540 kg*m - (3 m)(Mm)

10 m(Mm) = 500 kg*m

Mm = 50 kg so Mt = 130 kg

Mary is 50kg and Tim is 130 kg

If I see this on the MCAT, I'm voiding immediately lol.

 

[AwayFromReality]

I am going to take it that we can say Tim is at x = 0, Mary is at x = 10, and the axis of rotation is at x = 3.

Let Mt + Mm = M = 180 kg

Torque = T(tim) + T(board) + T(mary) = 0

T = 0 = (Mt)(g)(3m) - (20kg)(g)(2m) - (Mm)(g)(7m)

(cancel all g's)

(3 m)(Mt) = 40 kg*m + (7 m)(Mm)

Mt = [40 kg*m + (7 m)(Mm)]/(3 m)

we know Mm + Mt = 180 kg

Mt = 180kg - Mm

[40 kg*m + (7 m)(Mm)]/(3 m) = 180kg -Mm

40 kg*m + (7 m)(Mm) = 540 kg*m - (3 m)(Mm)

10 m(Mm) = 500 kg*m

Mm = 50 kg so Mt = 130 kg

Mary is 50kg and Tim is 130 kg

I really appreciate you writing out the solution but I already know how to do it. What I don't understand is how to tell whether the board will have counterclockwise or clockwise torque (because counterclockwise torque is positive and clockwise torque is negative). I know Mary will have counterclockwise torque and Tim will have clockwise torque. Here is a picture of the original question which I probably should have included in my original post.

 

[popopopop]

If the board weighs 20kg and is 10 meters long, it's fulcrum needs to be at 5 meters for it to be balanced/no torque (if there are no weights on it). From left to right, 0 being where Mary stands, the support is at 7 meters so the board would rotate counterclockwise if you remove Mary and Tim. If Mary and Tim is balanced on the board, there wouldn't be any torque on the board because they are in equilibrium.

Is that what you're asking?

 

[Cawolf]

I really appreciate you writing out the solution but I already know how to do it. What I don't understand is how to tell whether the board will have counterclockwise or clockwise torque (because counterclockwise torque is positive and clockwise torque is negative). I know Mary will have counterclockwise torque and Tim will have clockwise torque. Here is a picture of the original question which I probably should have included in my original post.

A board of uniform density will exert it's force about it's center of mass - i.e. x = 5 m; which is 2 m from the fulcrum.

Positive or negative is arbitrary.

If you want to go by the convention then it is positive in this drawing (counterclockwise).

 

[AwayFromReality]

A board of uniform density will exert it's force about it's center of mass - i.e. x = 5 m; which is 2 m from the fulcrum.

Positive or negative is arbitrary.

If you want to go by the convention then it is positive in this drawing (counterclockwise).

Well the reason I go by positive or negative is cause it helps me determine rotational equilibrium. In this case if Mary is "+", Board is "+" and Tim is "- "then rotational equilibrium would be "TorqueMary" + "TorqueBoard" +" TorqueTim" = (+) + (+) + (-) = 0. If that makes any sense.

 

[AwayFromReality]

If the board weighs 20kg and is 10 meters long, it's fulcrum needs to be at 5 meters for it to be balanced/no torque (if there are no weights on it). From left to right, 0 being where Mary stands, the support is at 7 meters so the board would rotate counterclockwise if you remove Mary and Tim. If Mary and Tim is balanced on the board, there wouldn't be any torque on the board because they are in equilibrium.

Is that what you're asking?

Actually that was exactly what I was asking. As a followup question if Mary and Tim were both equally spaced from the fulcrum (i.e 5m each) then the rotational equilibrium equation will only consists of Mary's torque and Tim's Torque, not the Board's Torque?

 

[Cawolf]

Well the reason I go by positive or negative is cause it helps me determine rotational equilibrium. In this case if Mary is "+", Board is "+" and Tim is "- "then rotational equilibrium would be "TorqueMary" + "TorqueBoard" +" TorqueTim" = (+) + (+) + (-) = 0. If that makes any sense.

It makes sense, I was just saying it is arbitrary as long as you assign positive or negative to a certain direction.

Actually that was exactly what I was asking. As a followup question if Mary and Tim were both equally spaced from the fulcrum (i.e 5m each) then the rotational equilibrium equation will only consists of Mary's torque and Tim's Torque, not the Board's Torque?

If you mean that the fulcrum is under the board's center of mass then yes.

T = r x F and if r = 0 (standing on fulcrum or being under center of mass of board) then it exerts no torque.

 

[popopopop]

Correct, since the board

Actually that was exactly what I was asking. As a followup question if Mary and Tim were both equally spaced from the fulcrum (i.e 5m each) then the rotational equilibrium equation will only consists of Mary's torque and Tim's Torque, not the Board's Torque?

Yeah, the board has no acceleration so there is no net force acting on it so there can't be torque.

 

[AwayFromReality]

Correct, since the board


Yeah, the board has no acceleration so there is no net force acting on it so there can't be torque.

It makes sense, I was just saying it is arbitrary as long as you assign positive or negative to a certain direction.



If you mean that the fulcrum is under the board's center of mass then yes.

T = r x F and if r = 0 (standing on fulcrum or being under center of mass of board) then it exerts no torque.

Wow, this whole thing just became a lot more clear to me. Thank you to both of you for your help