Chapter 5.3

Newton 2nd Law problems Pulling/Pushing Prob at and angle - Atwood Machines -Incline Planes -Tension Problems -Other Object Connected problems

What is Tension? Tension is defined as a force transmitted along a rope, chain, or wire. Tension will remain constant throughout the length of the rope. Tension is treated as a force in force diagrams and calculations Tension is measured in force units. (Newtons, dynes, or pounds)

Ropes- Assumptions Ignore any frictional effects of the rope Ignore the mass of the rope The magnitude of the force exerted along the rope is called the tension The tension is the same at all points in the rope Section 4.5

Tension Forces A taut rope has a force exerted on it. If the rope is lightweight and flexible the force is uniform over the entire length. This force is called tension and points along the rope. forces on the block forces on the rope FT FN Frope -FT Ffr m m Fg Frope = -FT by the law of reaction

Consider the following: The Pulling the Box Problem # 1 Consider the following: What is the force in the x and y directions What is the acceleration of the block?

The Pulling the Box Problem # 2 A box with a weight of 50.0 N is being pulled by Mr. H. Mr. H does not understand physics and vectors and as a result pulls inefficiently on the box when he pulls on the rope at an angle of 35.00 from the horizontal. If Mr. H is pulling with a force of 125 N…. A. What is the acceleration of the box? B. What could Mr. Herman do to help move the box faster. PROVE IT WITH ACTUAL NUMBERS

The Pulling the Box Problem # 3 𝑣 =+ 5.2 𝑚 𝑠 𝑎𝑛𝑑 𝑎 =+1.33𝑚/ 𝑠 2

Problems in Equilibrium Equilibrium means that the object has a net force and acceleration of zero. Since Fnet,x or Fx= 0, the x-components of tension should cancel each other. (Fright = Fleft) Since Fnet, y or Fy= 0, the y-components of tension should cancel the objects weight. (Fup = Fdown )

Hears What I am Talking About

Hanging Tension Problems (The Static problem) Physics Sign Fnet in the x and y direction =0 Or…it ain’t moving

Tension is equally distributed in a rope and can be bidirectional Tension is equally distributed in a rope and can be bidirectional. However, The Direction in a FBD always points up the rope Draw FBD block Draw FBD hand 30N Hand is Pulling down with 20N Tension does not always equal weight:

What is the weight of the sign? The Hanging Sign Problem # 1 What is the weight of the sign? The sign weighs 50 N.

Atwood Machines

Pulley A pulley uses tension to transfer a force to another direction. FT forces on the rope m1 FT m2 forces on block 1 forces on block 2 Ffr Frope m1 m1 Frope m2 m2 Fg

Connected Objects Apply Newton’s Laws separately to each object The magnitude of the acceleration of both objects will be the same The tension is the same in each diagram Solve the simultaneous equations Section 4.5

Derive a general formula for the acceleration of the system and tension in the ropes

Derive a general formula for the acceleration of the system and tension in the ropes Draw FBDs and set your sign conventions: Up is positive down is negative m1 m2 m1 m2

Derive a general formula for the acceleration of the system and tension in the ropes Draw FBDs and set your sign conventions: If up is positive down is negative on the right, the opposite is happening on the other side of the pulley. -T +m1g +T -m2g m1 m2 m1 m2 Then, do ΣF= ma to solve for a if possible.

Derive a general formula for the acceleration of the system and tension in the ropes +m1g +T -m2g m1 m2 m1 m2 Σ F2 = T – m2g = m2 a Σ F1 = - T + m1g = m1 a Add the equations Together…….

Σ F1 = - T + m1g = m1 a Σ Fnet = m1 a+ m2 a = - T + m1g + T – m2g -T and T Cancel !!!!! Σ Fnet =m1 a+ m2 a = m1g – m2g Now get all the a variables on one side and solve for a. Solve!!!!! Σ F2 = T – m2g = m2 a

g = a m1g - m1 a – m2g = m2 a m1g – m2g = m2 a + m1 a Factor out g on left and a on right g(m1– m2) = a (m2 + m1) Divide this away (m1– m2) g = a (m2 + m1)

Atwood’s Machine Problem 1 Problem: Using the diagram determine the acceleration of the system and how long it will take for the heavier weight to hit the ground. In an Atwood machine both masses are pulled by gravity, but the force is unequal. The heavy weight will move downward at (3.2 - 2.2 kg)(9.8 m/s2)/(3.2 + 2.2 kg) = 1.8 m/s2. Using y = (1/2)at2, it will take t2 = 2(1.80 m)/(1.8 m/s2) t = 1.4 s.

Inclined Planes

What Causes the Box to go down the Hill

Translating the FBD

Incline Plane Problems # 1

Incline Plane Problems # 2 A 100 kg crate is sliding down an incline plane. The plane is inclined at 300. The coefficient of friction between the crate and the incline is 0.3. a. Determine the net force and b. Acceleration of the crate.