EK 1001 Physics Problem 306 (Torque)

[MusicJunkie]

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I'm having the hardest time with Torque in the EK 1001 book.

So, problem 306 states: Mary and Tim balance on a 10 meter board. The board has a mass of 20 kg. If Mary and Tim have a combined mass of 180 kg, what is Mary's mass?

So, Mary is on the far left, 7 meters away from the fulcrum. Tim is on the far right, 3 meters away from the fulcrum.

So I went ahead and set up my equation like this:

(7 meters)(10)(Mass of Mary) + (20 kg)(10)(2 meters) = (3 meters)(Tim's mass)(10)

Cancelled out all of the gravities:
(7 meters)(Mass of Mary) + (20 kg)(2 meters) = (3 meters)(Tim's mass)

Further simplification:

7M + 40 = 3T
40 = -7M + 3T

... And now I don't know what to do. I don't even know if I'm on the right track, really. Help?!

 

[FutureDoctor503]

1 --- Mary = x
2 --- Tim = (180-x)

F1d1 = F2d2
m1gd1 = m2gd2
m1d1 = m2d2 (g cancels out)
(x)(7) = (180 - x)(3)
7x = 540 - 3x
10x = 540
x = 54

Mary weighs 54 kg and Tim weighs 126 kg (180 - x = 180 - 54 = 126).

 

[MusicJunkie]

1 --- Mary = x
2 --- Tim = (180-x)

F1d1 = F2d2
m1gd1 = m2gd2
m1d1 = m2d2 (g cancels out)
(x)(7) = (180 - x)(3)
7x = 540 - 3x
10x = 540
x = 54

Mary weighs 54 kg and Tim weighs 126 kg (180 - x = 180 - 54 = 126).

Thank you so much! 🙂

Is there a reason you didn't include the mass of the meter stick and it's weight? I was under the impression that it needed to be included if we aren't told to neglect the meter stick's mass.

 

[FutureDoctor503]

Thank you so much! 🙂

Is there a reason you didn't include the mass of the meter stick and it's weight? I was under the impression that it needed to be included if we aren't told to neglect the meter stick's mass.

You are right. I shouldn't have ignored the mass of the meter stick.

In that case, this is how you should set up the problem.

m1gd1 + (mass of 10 meter stick)(2 meters) = m2gd2
(x)7 + 40 = (180 - x)(3)
x = 50
Mary - 50 kg, Tim - 130 kg

The reason why I accounted for the torque generated by the 10 meter stick on the left side of the equation (with the torque generated by Mary) is because the center of mass is in the middle (at 5 meters), which is 2 meters to the left of the fulcrum.

 

[Postictal Raiden]

You are right. I shouldn't have ignored the mass of the meter stick.

In that case, this is how you should set up the problem.

m1gd1 + (mass of 10 meter stick)(2 meters) = m2gd2
(x)7 + 40 = (180 - x)(3)
x = 50
Mary - 50 kg, Tim - 130 kg

The reason why I accounted for the torque generated by the 10 meter stick on the left side of the equation (with the torque generated by Mary) is because the center of mass is in the middle (at 5 meters), which is 2 meters to the left of the fulcrum.

Why didn't you simply add the torque to both sides of the equation?

 

[FutureDoctor503]

Why didn't you simply add the torque to both sides of the equation?

You cannot add it to both sides (you can only do that when the fulcrum/pivot point is at the center of mass and adding the torque on both sides would cancel each other out). For this problem, the center of mass of the board/stick is on the left side of the fulcrum. To create an equilibrium, you have to add the torque (generated by the board) on the left side of the equation.

 

[Postictal Raiden]

You cannot add it to both sides (you can only do that when the fulcrum/pivot point is at the center of mass and adding the torque on both sides would cancel each other out). For this problem, the center of mass of the board/stick is on the left side of the fulcrum. To create an equilibrium, you have to add the torque (generated by the board) on the left side of the equation.

oh I get it now. Thank you.