Solving a Two-Body Pulley Problem

The Problem: Find the Acceleration of a Pulley System

m 1

m 2

Six Steps Summary Slide Step 1: Draw a Picture of the Situation Step 2: Draw and Label all Force Vectors Step 3: Decide on a Direction for the

Acceleration of Each Mass Step 4: Write Newton’s Law for Each Mass Step 5: Solve the System of Simultaneous

Equations that Results from Step 4 Step 6: Examine the Answer to See that it

Makes Sense Physically

Step 1: Drawing a Picture You have been given the

following problem to solve: A 2kg mass and a 4kg mass

hang on opposite sides of a mass-less, frictionless pulley. Determine the acceleration of the masses.

The first thing you should do is to draw a picture of the situation. Like this . . .

4kg2kg

Step 2: Draw and Label Force Vectors

Draw the two vectors for the force of gravity on each mass (“weights”)

W = mg. The two values are 40N and 20N.

Note that the 20 N force vector is about half as long as the 40 N one.

The two upward force vectors represent the force of string tension

on each mass - T.

4kg2kg

T T

40 N20 N

Incorrect Drawing of Force Vectors

The weights do not originatefrom the objects’ centers ofmassThe weight of the 2 kg mass is as long as (if not longerthan) the weight of the 4 kg mass. The tension force vectors areoff center and not originatingfrom the point of contact.

Such a poor drawing may not cause you to obtain an incorrect answer for this problem. But it will make it harder for someone else looking at your work to understand your thinking.

4kg 2kg

Examples of sloppy force vectors.

Step 3: Decide on the Direction of Each Mass’ Acceleration By examining the situation you

can tell that the 4 kg mass will accelerate down and the 2 kg mass will accelerate up.

We put arrows alongside the

masses to indicate this.

The direction of acceleration is taken to be the positive direction for forces acting on that mass when we write Newton’s Law in the next step.

4kg2kg

a

a

Step 4: Write Newton’s Law for The Left Hand Mass For the mass on the leftF = ma

Since the acceleration is downward the downward weight force is positive and the upward tension force is negative. Filling in Newton’s Law we get:

+VE 40 – T = 4a

4kg

T

40 N a

Step 4: Write Newton’s Law for the Right Hand Mass For the mass on the right

F = ma

Since the acceleration is upward the upward tension force is positive and the downward weight force of gravity is negative. Filling in Newton’s Law we get:

+VE T – 20 = 2a

T

20 N

2kg a

Step 5: Solve Equations for Two Variables (a and T)

Equation 1: 40 – T = 4a

Equation 2: T – 20 = 2a

Adding 20 = 6a

Solution: a = 3.33ms-2

T = 26.67N

These 2 equations are simultaneous equations which can be solved by adding.

Step 6: Examining the Answer

Check your answers to see if they are logical.

a = 3.33 ms-2. Is this number reasonable? The largest it could ever be is 9.8 m/s2

(or 10m/s2) since that is the acceleration caused by gravity in free fall. And the smallest it should ever be is 0 which would be the case if both masses were the same mass.

T= 26.67 N . This value is between the two values of weight for the two masses. Think of it this way: 26.67 N is less than 40 N and so the mass on the left accelerates down. And 26.67 N is more than 20 N so the mass on the right accelerates up.

4kg2kg

T T

40 N

20 N

Practice Problem Try this problem.

A 12 kg mass and a 3 kg mass hang on opposite sides of a pulley. Determine the acceleration of the system and the force of tension in the cable connecting the masses.

Remember the six steps.

Six Steps Summary Slide Step 1: Draw a Picture of the Situation Step 2: Draw and Label all Force Vectors Step 3: Decide on a Direction for the

Acceleration of Each Mass Step 4: Write Newton’s Law for Each Mass Step 5: Solve the System of Simultaneous

Equations that Results from Step 4 Step 6: Examine the Answer to See that it

Makes Sense Physically

Drawing With Force Vectors

12kg3kg

T T

120 N

30 Na

a

Equations120 – T = 12a

T – 30 = 3a

90 = 15a

a = 6ms-2

on 12kg

on 3kg

Adding

Solving a Two-Body Pulley Problem with

Friction

The Problem: Find the Acceleration of a Pulley System below

A 110gm mass and a 25gm mass hang on opposite sides of a mass-less, frictionless pulley. Determine the acceleration of the masses if the table is rough. = 0.2

Applet

M=25gms

M=110gms

Results

Results of Pulley Experiment

time distance

0 0

0.68 0.055

0.86 0.08

1.19 0.155

1.48 0.24

1.78 0.345

2.03 0.45

2.25 0.55

2.42 0.64

s = 0.1086t2 + 0.0004t

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3

time

displacement

Deductions from the Equation of Curve of Best Fit S = 0.1086t2 + 0.0004t V = 0.2172t + 0.0004 differentiate a = 0.2172 differentiate

So from the experimental data

a = 0.2172ms–2

Now we will analyse the experiment mathematically

Step 2: Draw and Label Force Vectors

W=0.25N

W=1.10N

R

RT

T

Step 3: Decide on the Direction of Each Mass’ Acceleration

a

W=0.25N

W=1.10N

R

R

a

T

T

Step 4: Write Newton’s Law for The Mass on the Table in a Horizontal Direction

For the mass on the tableFx = ma

Since the acceleration is right the tension force is positive and the friction force is negative. Filling in Newton’s Law we get:

+VE T – R = 0.11a

1.1N

R

RT

a

Step 4: Write Newton’s Law for The Mass on the Table in a Vertical Direction

For the mass on the tableFy = 0

+VE R – = 0 R = 1.1N

T

1.1N

R

R

a

Step 4: Write Newton’s Law for the Hanging Mass For the hanging mass

F = ma

Since the acceleration is downward the upward tension force is negative and the downward force of the weight is positive. Filling in Newton’s Law we get:

+VE 0.25 – T = 0.025a0.25N

a

T

Step 5: Solve Equations for Two Variables (a and T)

Equation 1: T – R = 0.11a R = 1.1

So T – 0.2x1.1 = 0.11aEquation 1: T – 0.22 = 0.11a

Equation 2: 0.25 – T = 0.025a

Adding 0.03 = 0.135a

Solution: a = 0.222ms-2

These 2 equations are simultaneous equations which can be solved by adding.

as=0.2 and R=1.1