Torque A meter stick weighs 200N and is supported at the centerThree weighs of 100N 150N and 300N are placed respectively at the 10cm 40cm and 60cm points of the meter stick Where should another?

Answer 1

Your set-up is so long that that you ran out of space before you could ask the

question. In fact, you got cut off leaving us with two questions . . . how

heavy should a single weight be, and where should it hang, in order to balance

the whole thing ?

I just noticed that since the stick is 1 meter long, and the three original

weights are placed at the 10cm, 40cm, and 60cm marks, the 10 and the

40 are on one side of the pivot, and the 60 is on the other side, 10 cm

from the pivot. It's all starting to come together.

On one side:

-- There is a weight at the "10cm" mark. That point is 40 cm from the pivot,

and the weight is 100N, so the moment pulling this side down is 4,000 N-cm.

-- There is another weight at the "40cm" mark, and that's 10cm from the pivot.

The weight is 150N, giving us another 1,500 N-cm pulloing this side down.

-- The total moment pulling this side down is (4,000 + 1,500) = 5,500 N-cm.

-- (Plus another 25 x 100 = 5,000 due to the weight of 1/2 the stick. But the

other half is already balancing that much, so I don't think we need to worry

about the weight of the stick at all. This is typical of these meter sticks that

are manufactured by the Red Herring Pedagogical Industries Group LLC of

Eckveldt, South Dakota, and so often used in these Physics demonstrations.)

-- At the "60cm" mark, 10cm from the pivot on the other side, is hanging a

weight of 300N. That gives us 3,000 N-cm of moment working against the two

weights on the first side.

-- In order to balance the whole thing, we have to provide only another

(5,500 - 3,000) = 2,500 N-cm on the second side.

-- All we can say is: The product of (the weight of a single weight hung from

the second side) multiplied by (its distance in centimeters from the pivot) must

be 2,500.

-- The lightest weight that can do the job is 50N, if it's hung at the endof

the stick on the second side, 50cm from the pivot. If we have more weight to

play with, then it can be hung closer to the pivot, just as long as (weight) times (distance from the pivot) = 2,500 N-cm.

Answer 2

The torque due to the weights must balance the torque due to the meter stick. To find where to place another weight (x N) for balance, set up the equation: (200N * 50cm) = (100N * 10cm) + (150N * 40cm) + (300N * 60cm) + (xN * distance). Solve for x to determine the weight and its distance to keep the meter stick balanced.