Science Olympiad

Simple & Compound Machines.

The easiest lever to analyze is the first class lever (seesaw), that is balanced by itself. The center of gravity of the lever is on the fulcrum.

cg

If a lever is not moving (rotating) then it is said to be at static equilibrium. When an object is at static equilibrium the following is true:

ΣF = 0, that is netF = 0, no unbalanced forces.Στ = 0, that is there are no unbalanced torques.

If you place a seesaw so that its center of gravity is on the fulcrum, it will balance. That is, the left side balances the right side.

The weight of the seesaw on the left creates a torque that tries to make it rotate counter-clockwise so that the left side would go down.

F

The weight of the seesaw on the right creates a torque that tries to make it rotate clockwise so that the right side would go down.

F

The two balance each other.

Another way to look at this is that we can place all the weight of the seesaw at its center of gravity.

Fg

c.g.

The center of gravity of the seesaw is at the axis of rotation (fulcrum) so the lever arm is zero and the force creates no torque.

Note: The center of gravity may not be at the geometric center.

Especially when using wooden meter sticks!

Two identical 40.0 kg twin girls are sitting on opposite ends of a seesaw that is centered on the fulcrum, that is 4.0 m long, and weighs 700 N.

c.g.

FN = ?

x1 = 2.0 m x2 = 2.0 m

First we need to draw a torque diagram of the seesaw. This is a free body diagram which includes the lever arms.

FL = 700 N

c.g..

F1 = 400 N F2 = 400 N

We place all the forces at their proper location.

We define the axis of rotation (circle with a dot in the middle) and the lever arms.

FN = ?

x1 = 2.0 m x2 = 2.0 m

Fg = 700 N

c.g..

F1 = 400 N F2 = 400 N

Στ = 0 or ΣτCCW = ΣτCW

F1x1 = F2x2

FL and FN both act through the axis of rotation, so their lever arm is zero.

(400 N)(2.0 m) = (400 N)(2.0 m)

800 Nm = 800 Nm

The torques balance so the seesaw can be in static equilibrium.

One 400 N girl sits on one end of a seesaw that is centered on the fulcrum, is 4.0 m long, and weighs 700 N. Where must her 650 N brother sit in order for the seesaw to be in static equilibrium?

c.g.

?

So, what do you do to balance the seesaw if the two people are not the same weight (mass)?

Option #1, move the heavier person closer to the fulcrum.

FN = ?

xG = 2.0 m xB = ?? m

FL = 700 N

c.g..

FG = 400 N FB = 650 N

Στ = 0 or ΣτCCW = ΣτCW

FGxG = FBxB

FL and FN both act through the axis of rotation, so their lever arms are zero.

(400 N)(2.0 m) = (650 N)xB

800 = 650xB

xB = (800 Nm)/(650 N)xB = 1.23 m

One 400 N girl sits on one end of a 4.0 m long seesaw weighing 700 N That has moved the center of gravity of the lever 0.2 meters towards her. Where must her 650 N brother sit in order for the seesaw to be in static equilibrium?

c.g.

?

Option #2, move the center of gravity of the seesaw so that more of the seesaw is on the side of the lighter person,

Now the seesaw creates a torque helping the girl.

FN = ?

xG = 2.2 m xB = ?? m

FL = 700 N

c.g..

FG = 400 N FB = 650 N

Στ = 0 or ΣτCCW = ΣτCW

FGxG + FLxL = FBxB

FN acts through the axis of rotation, so its lever arm is zero.

(400)(2.2) + (700)(0.2) = (650 N)xB

880 + 140 = 650xB

xB = (1020 Nm)/(650 N)xB = 1.57 m

xL = 0.2 m

Simple Machines. (Simple case.)

Given small mass placed on one side.

Given unknown large mass on the other.

Unless the values are not too extreme, you may not be able to move the large mass close enough to the fulcrum.

c.g.

If this setup does work, you don’t have to worry about the weight of the lever.

Simple Machines (More realistic case)

Given small mass placed on one side.

Given unknown large mass on the other.

Move the fulcrum near one end of the lever.

c.g.

Now the lever helps balance the large weight.

FN acts through the axis of rotation, so its lever arm is zero.

FN

xS = cmxB = cm

FL = N

c.g..

FS = N FB = ?? N

xL = cm

Στ = 0 or ΣτCCW = ΣτCW

FSxS + FLxL = FBxB

FN acts through the axis of rotation, so its lever arm is zero.

FN

xS = cmxB = cm

FL = N

c.g..

FS = N FB = ?? N

xL = cm

Στ = 0 or ΣτCCW = ΣτCW

FSxS + FLxL = FBxB

If you divide through by “g” you get:

mSxS + mLxL = mBxB

xS = cmxB = cm

mL = g

c.g..

mS = g mB = ?? g

xL = cm

Στ = 0 or ΣτCCW = ΣτCW

FSxS + FLxL = FBxB

If you divide through by “g” you get:

mSxS + mLxL = mBxB

High School Compound Machines.

0 50

c.g.

B

0 50

c.g.

First Class Lever Second Class Lever

0 50

c.g.

xS = cmxE = cm

mL1 = g

c.g..

mS = g mE = ?? g

xL = cm

mSxS + mLxL = mExE

First part of lever system

xE = cm

xB = cm

mL2 = g

c.g. .

mE = gmB = ?? g

xL2 = cm

mExE = mLxL + mBxBB

0 50

Second part of lever system

Things to note:• For practice I am using meter sticks and

commercial fulcrums. THEY ARE NOT ALLOWED.• You must build your own lever system.• You may want to have two set places to have you

fulcrum depending on the given masses.• You may want to have the unknown mass at a

predetermined spot.• Make sue that you know the mass of your lever

and have marked the location of its center of gravity.