Problem Solving Strategy Draw a sketch and label all information. Draw a free-body diagram. Find components of all forces (+ and -). Apply First Condition for Equilibrium: SFx= 0 ; SFy= 0 5. Solve for unknown forces or angles.

Example 10 A 50 N sign hangs by a single wire as shown Example 10 A 50 N sign hangs by a single wire as shown. What is the tension in the wire? T2 T1 T2Y T1Y T1X T2X BEARS Step 1. Sketch Step 4 Find the sum of x and y components Step 3 Find components Step 2. FBD Step 5 Solve for T FW ΣFy = 0 T1Y + T2Y – Fw = 0 T1sinΘ +T2sinΘ – Fw = 0 T1sin40 + T2sin40 – 50N = 0 T1sin40 + T1sin40 – 50N = 0 2T1sin40 – 50N = 0 T1 = 50N 2sin40 T1 =T2 = 38.9 N ΣFx = 0 T1X - T2X = 0 T1cosΘ - T2cosΘ = 0 T1cos40 - T2cos40 = 0 T1cos40 = T2cos40 T1 = T2

Example 11 A 50 kg crate is hanging on a 4 m long rope which sags 5 Example 11 A 50 kg crate is hanging on a 4 m long rope which sags 5.0 cm. What is the tension in the rope? ΣFx = 0 T1X - T2X = 0 T1 = T2 T1 T2 T1Y 0.05m T2Y Θ T2X T1X 2m ΣFy = 0 T1Y + T2Y – Fw = 0 T1sinΘ +T2sinΘ – Fw = 0 2T1sinΘ – 490N = 0 2T1sin1.43 – 490N = 0 T1 = 490N 2sin1.43 T1 =T2 = 9803.06 N Fw How do I find the angle? tan Θ = opp/adj tan Θ = .05m/2m Θ = 1.43 Fw = 50kg(9.8m/s2) = 490N

Apparent Weight Apparent weight is a force that acts in opposition to gravitational force in order to prevent a body from going into freefall. When you stand on the floor, the floor pushes up on your feet with a force equal to your apparent weight

When you are in an elevator, your actual weight (mg) never changes. You feel lighter or heavier during the ride because your apparent weight increases when you are accelerating up, decreases when you are accelerating down, and is equal to your weight when you are not accelerating at all.

So you feel heavier FN + a - Fg Example 12: An 85-kg person is standing on a bathroom scale in an elevator. What is the person’s apparent weight a) When the elevator accelerates upward at 2.0 m/s2? So you feel heavier FN Fg a + -

b) when the elevator is moving at constant velocity between floors? c) when the elevator begins to slow at the top floor at 2.0 m/s2? So you feel lighter FN Fg - a +

- a + Fg So you feel weightless OH NO… the cable breaks… The acceleration is downward so the net force is downward - a + Fg So you feel weightless

Example 10 modified A 50 N sign hangs by a single wire as shown Example 10 modified A 50 N sign hangs by a single wire as shown. What is the tension in the wire? T2 T1 T2Y T1Y T1X T2X FW

Force Systems - Two Body Problems Objects connected in some manner Typically want to know the acceleration of the two objects and the force between the two objects. I prefer to determine acceleration of the system (the two objects are considered to be a single object moving (or accelerating) together as a whole) . I prefer to determine the force between the two objects by individual object analysis, isolate one object to determine the force acting between the two objects (Fcontact Ftension).

Tension on A String How do the readings compare from your left to your right hand. How do the readings compare if you and another person hold either scale? What is the conclusion? the tension is the same everywhere in the string. Which of Newton’s Laws is being demonstrated? 3rd Law (Forces of Interaction )

Ex A. A 7.00-kg box (m2) is attached to a 3.00-kg box (m1) by a lightweight rope. The 7.00-kg box is pulled by another rope with a force of 25.0 N. Determine the acceleration of the boxes and the tension in the connecting rope. The coefficient of friction between the ground and the boxes is 0.120.

Solve for acceleration as a system FN 10 kg (m1 + m2) 25NFP FF FP = 25N Fw = FN = ma =(10kg)(9.8m/s2) = 98N FF = µ FN = (0.12)(98N) = 11.76N FNET = FP - FF FNET = 25N - 11.76N = 13.24N a = FNET /m a = 13.24N/10kg = 1.32m/s2 Fw

Solve for FT with one object. The FT between the objects is the same Solve for FT with one object. The FT between the objects is the same. I would suggest choosing the object that has the least forces acting on it. FN 3 kg (m1) FT FF Fw = FN = ma =(3kg)(9.8m/s2) = 29.4N FF = µ FN = (0.12)(29.4N) = 3.53N FNET = ma = FT - FF FT - FF = ma FT = ma + FF FT = (3kg)(1.32m/s2) + 3.53N = 7.49N Fw

Show Atwood Machine https://www.youtube.com/watch?v=4ovhEkSIqV0 What happens when released? If there is acceleration, what do you think it is? If there is acceleration, is there a net force? What are the forces acting to make up the net force? What can you say about tension in the string? What can you say about the acceleration of M and m?

An Atwood's machine is a device where two masses, M and m, are connected by a string passing over a pulley. Assume that M > m. Our ULTIMATE question: What is the acceleration of the two masses? m M

What is the weight (in N) of the mass (M) on the right side? The left mass (m) of an Atwood machine measures 30.0 kg. The right mass (M) measures 45.0 kg. If the system is released (or set into motion), which way would it move? m M Clockwise What is the weight (in N) of the mass (M) on the right side? What is the weight (in N) of the mass (m) on the left side? 441 N 294 N

Assume the pulley is frictionless and massless. What can we say about FT on the left side portion of the string? On the right? The FT is the same everywhere 441 N m M 294 N Think about what the system will do. If the system is released from rest, the heavy mass will accelerate ___________ and the lighter one will accelerate _________. Are their rates of acceleration different or the same?__________. downward upward same

Remember our ultimate question: What is the acceleration of the two masses? How are we going to determine this? If we know one a, we know the other a m 441 N Could the answer be as easy as 9.8 m/s2? 294 N M NO!!! Why Not? Besides you didn’t get to perform equation addition on the final and I know you were sad. We are going to approach this like tension on unequal wires. 9.8 m/s2 only applies to freefall What else has to be considered besides m and ag? Force of Tension

The Atwood Plan of Attack Determine Fw of each object (done that!) Set up force equation (FNET = ma) for each object. The “a” here is the acceleration of the mass in response to the Net Force, NOT 9.8m/s2 FNET is the sum of what two forces? Fw and FT For mass M: who wins, Fw or FT ? What is the force equation for mass M? Fw rt - T = Mrt  a For mass m: who wins, Fw or FT ? What is the force equation for mass m? T - Fw lt = mlt  a Now, add these two equations together

Fw rt - T + T - Fw lt = (M rt + m lt)  a When we add the equations together we get: Fw rt - T + T - Fw lt = (M rt + m lt)  a Note the a here is NOT ag, but the acceleration of the masses. This is what we ultimately want to find out. Cancel out the T’s and then rearrange to solve for a (Fw rt - Fw lt) / (M rt + m lt) = a a = (441 N – 294 N) /(45 kg + 30 kg) a = 1.96 m/s2 Clockwise

Example B: An Atwood machine is set up so that an attached 275.0-kg block opposes motion of a 150.0- kg block. What is the acceleration of this system?

The system accelerates at 2.88 m/s2 We will assume M (275 kg) is on the right, thus rotation will be clockwise. 2695 N - T = 275.0 kg  a T - 1470 N = 150.0 kg  a (2695 N - 1470 N) / (275.0 kg + 150.0 kg) = a 2.88 m/s2 clockwise = a The system accelerates at 2.88 m/s2 An Atwood machine is set up so that an attached 275.0-kg block opposes motion of a 150.0-kg block. What is the acceleration of this system?

Fw rt - T = M  a Solve for T T = Fw rt - (M  a ) T = 2695 N – ( 275.0 kg  2.88 m/s2) T = 1903 N What if we are interested in determining the tension in the cord between the two masses? You can solve for T with either mass using their net force equation with the calculated acceleration. T - Fw lt = m  a Solve for T T = Fw lt + (m  a) T = 1470 N + (150.0 kg  2.88 m/s2) T = 1902 N

Example 13 A cord passing over a frictionless pulley has a 7 Example 13 A cord passing over a frictionless pulley has a 7.0 kg mass hanging from one end and a 9.0-kg mass hanging from the other. (This arrangement is called Atwood's machine). Find the acceleration of the masses. Find the tension in the cord

Magic Pulleys on a flat table Magic pulleys bend the line of action of the force without affecting tension. The pulley is “magic” – no mass, no friction, and no effect on the tension. T m2g N m1g -x x m1 Frictionless table m2

Fw – T + T – Ff = (mhang + msit) • a Fw – T = mhanging • a T – Ff = msitting • a Fw Fw – T + T – Ff = (mhang + msit) • a

Example 14 An Atwood off the table and friction An object m1 = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass m2 = 15 kg. If the coefficient of friction is 0.20 between the table and block A, how far will block B drop in the first 3.0 s after the system is released?

WOW is this going to be fun!