Laws of Motion Pulleys I

Pulleys Change the direction of force. Magnitude of force is unaffected as long as the pulleys are massless and frictionless, and the strings are massless and inelastic.

Example 1 Atwood’s machine: m1 = 5.0 kg and m2 = 10.0 kg a. Determine the acceleration of the system. Assess: Acceleration of compound bodies is solved more easily as a system. Diagram: Draw free body diagrams if requested, and/or add relevant vector arrows to the given diagram. Sum forces: To solve acceleration sum forces for the system, canceling tension. Solve: m1 m2 +Fg2 −Fg1 − T +T

Example 1 Atwood’s machine: m1 = 5.0 kg and m2 = 10.0 kg b. Determine the tension in the string. Assess: You must work with masses separately to solve for internal forces, such as tension. Diagram: Free body diagrams and the diagram at the right do not change. Sum forces: The same string is attached to both masses. You can sum forces for either mass to solve tension. Both are solved below to prove that either will work. Solve: m1 m2 +Fg2 −Fg1 −T +T

Example 2 m1 m2 Modified Atwood’s machine In the previous example (the standard Atwood’s machine) there is only one possible solution for acceleration and tension as long as the pulley is massless and frictionless. However, in the modified Atwood’s machine at the right there are four possible solutions depending on how friction affects mass 1. This example will start by summing forces for the system to show how the four possible solutions compare, and then it will sum forces for an individual mass to solve tension in the string. m1 m2

Example 2 N1 −f1 +T Fg1 −T +Fg2 m1 m2 Modified Atwood’s machine Assess: Compound body Diagram: Friction dictates the four possible solutions and is shown in red. Sum forces: Solve: 1. Frictionless surface +T −T +Fg2 N1 Fg1 −f1 m1 m2 This involves unbalanced force, and the object must accelerate. Note: If the problem had specified a frictionless surface, then the vector for friction would not have been included in the free body diagram or the sum of forces equation at all. It is include here to compare it to the other possible scenarios.

Example 2 N1 −f1 +T Fg1 −T +Fg2 m1 m2 Modified Atwood’s machine Assess: Compound body Diagram: Friction dictates the four possible solutions and is shown in red. Sum forces: Solve: 1. Frictionless surface 2. Friction is present & object remains stationary +T −T +Fg2 N1 Fg1 −f1 m1 m2 If static friction is strong enough, then the object will remain at rest. Calculations now involve the coefficient of static friction, μs . When an object is at rest both acceleration and net force are zero.

Example 2 N1 −f1 +T Fg1 −T +Fg2 m1 m2 Modified Atwood’s machine Assess: Compound body Diagram: Friction dictates the four possible solutions and is shown in red. Sum forces: Solve: 1. Frictionless surface 2. Friction is present & object remains stationary 3. Friction is present & object has constant velocity +T −T +Fg2 N1 Fg1 −f1 m1 m2 Calculations now involve the coefficient of kinetic friction, μk . When velocity is constant both acceleration and net force are zero.

Example 2 N1 −f1 +T Fg1 −T +Fg2 m1 m2 Modified Atwood’s machine Assess: Compound body Diagram: Friction dictates the four possible solutions and is shown in red. Sum forces: Solve: 1. Frictionless surface 2. Friction is present & object remains stationary 3. Friction is present & object has constant velocity 4. Friction is present & object accelerates +T −T +Fg2 N1 Fg1 −f1 m1 m2 If kinetic friction is weak, then the object will accelerate.

Example 2 N1 −f1 +T Fg1 −T +Fg2 m1 m2 Modified Atwood’s machine Solve for tension in the string Assess: Tension is an internal force. Summing forces for the system cancels internal forces. Therefore, you must sum your forces for one mass only. Diagram: The diagrams remain the same. Sum forces: Mass 1 depends on friction. Mass 2 is the same regardless of which scenario was solved previously. Solve: +T −T +Fg2 N1 Fg1 −f1 m1 m2 If stationary or moving at constant velocity. If accelerating

Example 3 m 2m Problems solved as variables A mass m on a frictionless surface is tied to a string which passes over a pulley. At the other end of the string is mass 2m hanging under the influence of gravity. What is the acceleration of the system? In variable problems the numerical values are not given. In this problem m1 = m and m2 = 2m The value of mass m is not known. We only know that mass 2m has twice the mass of mass m . m 2m Solve these problems exactly the same way as before. If you are missing a number leave the variables in letter form. *** Even if you know the numerical value of variables such as g and  you will leave them as letters. *** Your answer is an equation. As a result, you will not add units to these types of problems.

Example 3 Nm +T Fg m −T +Fg 2m m 2m Problems solved as variables a. Determine acceleration of the system. Assess: Solve acceleration using system Diagram: Add vectors to diagram Sum forces: Solve: Key difference with a variable problem: The answer is an equation. Leave g ( π, etc.) as letters. Normally numerical quantities are in decimal form using significant figure rules. However, the answer is an equation, which is exact. Use fractions in variable problems. DO NOT include units. Unit letters can be mistaken for variables. +T −T +Fg 2m Nm Fg m m 2m

Example 3 Nm +T Fg m −T +Fg 2m m 2m Problems solved as variables b. Determine tension in the string Assess: Solve tension using either mass Diagram: Diagram remains the same The solutions for both masses are provided below to show that either works. You may solve for either mass in your work. Sum forces: Solve: Substitute acceleration from part a. +T −T +Fg 2m Nm Fg m m 2m

Example 4 m1 m2 Modified Atwood’s machine This modified Atwood’s machine adds an inclined surface. As with the previous example, there are four solutions when summing forces for the system. The specific solution in each problem depends on whether friction is present, and its magnitude compared to other forces when it is present. IMPORTANT: In addition to issues with the incline there is one other problem. Unlike the previous modified Atwood’s machine, which moves only in one direction, this version can move in either direction. However, in order for mass 1 to slide down the incline it must be substantially larger than mass 2. One way to be sure is to solve for the force down the incline (m1g sin θ) and compare it to force of gravity on mass 2. The larger force will be the direction of motion. This example will solve for the more likely scenario where mass 2 moves downward while mass 1 will moves up the incline.

Example 4 +T − Fg1sin −T − f1 +Fg2 m1 m2 Modified Atwood’s machine a. Determine the acceleration of the system Assess: Compound body Diagram: Friction dictates the four possible solutions and is shown in red. Sum forces: Solve: 1. Frictionless surface 2. Friction present & object remains stationary 3. Friction present & object has constant velocity 4. Friction present & object accelerates

Example 4 +T − Fg1sin −T − f1 +Fg2 m1 m2 Modified Atwood’s machine +T −T − f1 +Fg2 − Fg1sin m1 m2 b. Solve for tension in the string Assess: Tension is an internal force. Summing forces for the system cancels internal forces. Therefore, you must sum your forces for one mass only. Diagram: The diagrams remain the same. Sum forces: Mass 1 depends on friction. Mass 2 is the same regardless of which scenario was solved previously. Solve: If stationary or moving at constant velocity. If accelerating