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 SituationDraw a Picture of the Situation  Step 2: Draw and Label all Force VectorsDraw and Label all Force Vectors  Step 3: Decide on a Direction for the Acceleration of Each MassDecide on a Direction for the Acceleration of Each Mass  Step 4: Write Newton’s Law for Each MassWrite Newton’s Law for Each Mass  Step 5: Solve the System of Simultaneous Equations that Results from Step 4Solve the System of Simultaneous Equations that Results from Step 4  Step 6: Examine the Answer to See that it Makes Sense PhysicallyExamine 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... 4kg 2kg

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. 4kg 2kg T T 40 N 20 N

Incorrect Drawing of Force Vectors The weights do not originate from the objects’ centers of mass The weight of the 2 kg mass is as long as (if not longer than) the weight of the 4 kg mass. The tension force vectors are off center and not originating from 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. 4kg 2kg a a

Step 4: Write Newton’s Law for The Left Hand Mass  For the mass on the left F = 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/s 2 (or 10m/s 2 ) 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= N. This value is between the two values of weight for the two masses. Think of it this way: N is less than 40 N and so the mass on the left accelerates down. And N is more than 20 N so the mass on the right accelerates up. 4kg 2kg 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 SituationDraw a Picture of the Situation  Step 2: Draw and Label all Force VectorsDraw and Label all Force Vectors  Step 3: Decide on a Direction for the Acceleration of Each MassDecide on a Direction for the Acceleration of Each Mass  Step 4: Write Newton’s Law for Each MassWrite Newton’s Law for Each Mass  Step 5: Solve the System of Simultaneous Equations that Results from Step 4Solve the System of Simultaneous Equations that Results from Step 4  Step 6: Examine the Answer to See that it Makes Sense PhysicallyExamine the Answer to See that it Makes Sense Physically

Drawing With Force Vectors 12kg 3kg T T 120 N 30 N a a

Equations 120 – 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 timedistance

Deductions from the Equation of Curve of Best Fit  S = t t  V = t differentiate  a = differentiate  So from the experimental data a = ms –2 Now we will analyse the experiment mathematically

Step 2: Draw and Label Force Vectors W=0.25N W=1.10N R RR T T

Step 3: Decide on the Direction of Each Mass’ Acceleration a W=0.25N W=1.10N R 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 table F x = 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 RR T a

Step 4: Write Newton’s Law for The Mass on the Table in a Vertical Direction  For the mass on the table Fy = 0  +VE R –  = 0 R = 1.1N T 1.1N R 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 = 0.025a 0.25N a T

Step 5: Solve Equations for Two Variables (a and T) Equation 1: T – R = 0.11a R = 1.1 So T – 0.2 x 1.1 = 0.11a Equation 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