Question

What is the mass placement for section II, part A) of this lab experiment? Assume the Center of Mass is at the 50 cm mark.

A) In this part of the laboratory you will explore the basic concept of rotational equilibrium, i.e. the balancing of torques about a fulcrum or pivot point. You will be given values for 3 masses along with given positions (Xi) on a meter stick for which two of these masses will be hung. Your task is to determine the position that the third mass must be hung in order to balance the meter stick, i.e. achieve rotational equilibrium. The location of the center of mass of the meter stick (its pivot point) will be determined by balancing the meter stick without masses added.
Xcm = ____________
M1 = 50 g @ 15 cm to left of the center of mass; M2 = 100 g @ 35 cm to the left of the center of mass; determine where to put M3 = 150 g to balance as reference from the center of mass of the stick. Show your calculations here:
X3 = ________________________________ to the LEFT / RIGHT of the Center of Mass.
Now test out your calculations with your meter stick balance. Briefly describe the results of your test here.

Answer 1

Xcm of meter stick without masses will be at 50cm

From the basic observation it is clear that M3 will be right side as other 2 masses are on left side.

Now M1×X1 + M2 ×X2 = M3×X3

SO X3 = (50×15+100×35)÷150

Or X3 = 28.33 cm to the right side