EK Physics #306

[Addallat]

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Mary and Tim balance on a board 10 m as shown. The board has a mass of 20 kg. If Mary and Tim have a combined mass of 180kg, what is mary's mass?


I was approaching it as balancing torques
taking 3 meters as being from fulcrum for Tim
and 7 meters as the distance from fulcrum for Mary

I have no idea what to do with the mass of the board ... I'm guessing that the board is homogenous so it would be 2 meters away from the fulcrum?

This what I came down to since the Question states they're balancing i'm assuming torques must be in equilibrium

7meters* (mass Mary) * (10 m/s^2) + 20kg * (10m/s^2) 2 meters= 3 meters * (mass Tim) * (10 m/s^2)

I don't know where to go from here ... please help, thank you

 

[dcamkl1]

Is the answer A) 50?

Yes, you are right that whenever the mass of the board is given, the center of mass is at the equidistant point. In this case, the weight of the 10m board is at 5m, which is 2m from the fulcrum.

Let M = Mary's mass, and T = Tim's mass.

Since, M+T=180..... T=180-M

So, to solve: torques on the left = torques on the right

(7m x 10g x M) + (20kg x 10g x 2m) = (180-M)(3m)(10g)

Solve for M = 50kg.

You could have intuitively eliminated C and D, because for them to balance, Mary's mass must be less than Tim's.

 

[Addallat]

Is the answer A) 50?

Yes, you are right that whenever the mass of the board is given, the center of mass is at the equidistant point. In this case, the weight of the 10m board is at 5m, which is 2m from the fulcrum.

Let M = Mary's mass, and T = Tim's mass.

Since, M+T=180..... T=180-M

So, to solve: torques on the left = torques on the right

(7m x 10g x M) + (20kg x 10g x 2m) = (180-M)(3m)(10g)

Solve for M = 50kg.

You could have intuitively eliminated C and D, because for them to balance, Mary's mass must be less than Tim's.

ahhhhh got it thank you and yes the answer was A