Question

A meterstick is initially balanced on a fulcrum at its midpoint. You have four identical masses. Three of them are placed atop the meterstick at the following locations: 27cm , 38cm , and 99cm .Where should the fourth mass be placed in order to balance the meterstick?

Answer 1

Sum the torques about the fulcrum, which is at \(x=50 cm\).

Let clockwise be positive. \[\sum \tau = 0 \rightarrow mg \left [ (27-50) + ( 38-50) + (99-50) +(x - 50) \right] = 0\]

Answer 2

that did not really help at all. can you explain why you use that formula? i did not even see it in my book. nor did it lead me to a valid answer.

Answer 3

Sure.

Let's start with a basic definition of torque. \[\tau = r \times F\]where r is the distance from the axis of rotation to the force and F is the force. Since both r and F are vectors, their cross product will produce a third vector which is orthogonal to both.

More simply, \[\tau= F \cdot d\]This is only valid if F is perpendicular to the axis of rotation. Here d is the perpendicular distance between this axis and the force. In this particular example, the force exerted by the masses act orthogonally to the axis of rotation.

I'm assuming that the locations of the masses is given from the meter stick's 0 mark. Should I assume instead that these distance are from the fulcrum?

Answer 4

so let me get this straight... we use the torque definition and add them all up? :O... thing is we do not know the mass. and i thought it canceled out in the equation when solving for the x. :O... but i did nto get it right :(

Answer 5

m and g don't necessarily cancel. We can divide them out though since the equation equals zero and m and g are common to all terms.

Yes, we are adding up the torques created by all four masses.

Can you clarify how I should interpret the locations given please?

If you got a wrong answer with the above equation, I'm inclined to thing the distances are defined from the fulcrum not the 0 mark on the meterstick (as that is how I set up the previous equation).

Answer 6

That doesn't answer my question.... Please refrain from unnecessary drawings. I'm trying to help you here. You need to help me now.

Answer 7

i am sorry. :/. but the distances are not from the fulcrum. it just says that they are at the marks stated above.

Answer 8

Okay.

If the fulcrum is at the middle of the meter stick, it is on the 50 cm mark.

We need to determine the distance from the fulcrum each mass is.

For mass 1: \(x_1 = 50 - 27\)
For mass 2: \(x_2 = 50 - 38\)
For mass 3: \(x_3 = 99-50\)
For mass 4: \(x_4 = x-50\) <--This assumes that the forth mass is to the right of the fulcrum (on the same side as mass 3).

Now, we can express the sum of the torques about the fulcrum as\[-m_1g x_1 - m_2g x_2 + m_3gx_3 + m_4gx_4 = 0\]
Note how the first two terms are negative. This is because these torques create counter-clockwise motion about the fulcrum, which according to our sign convention is negative.
Since \(m_1 = m_2 = m_3 = m_4 = m\)\[mg (-x_1 - x_2 + x_3 + x_4) = 0\]\[\left [ -(50 - 27) - (50 - 38) + (99-50) + (x-50) \right] = 0\]

Answer 9

sigh... v.v... thanks for the help... i think i will go somewhere else... i just used my last attempt.

Answer 10

What answer did you get?

Sorry that my solution isn't getting you the right answer.