Ipc definition: a push or a pull exerted on some object


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IPC definition: A push or a pull exerted on some object

  • IPC definition: A push or a pull exerted on some object

  • Better definition: Force represents the interaction of an object with its environment

  • The Unit for Force is a Newton



Contact Forces: Result from physical contact between two objects

  • Contact Forces: Result from physical contact between two objects

    • Examples: Pushing a cart, Pulling suitcase
  • Field Forces: Forces that do not involve physical contact

    • Examples: Gravity, Electric/Magnetic Force


The effect of a force depends on magnitude and direction

  • The effect of a force depends on magnitude and direction



Force Diagram: A diagram that shows all the forces acting in a situation

  • Force Diagram: A diagram that shows all the forces acting in a situation



Free Body Diagrams (FBDs) isolate an object and show only the forces acting on it

  • Free Body Diagrams (FBDs) isolate an object and show only the forces acting on it

  • FBDs are essential! They are not optional! You need to draw them to get most problems correct!



Situation: A tow truck is pulling a car

  • Situation: A tow truck is pulling a car

  • (p. 127)

  • We want to draw a FBD for the car only.



Step 1: Draw a shape representing the car (keep it simple)

  • Step 1: Draw a shape representing the car (keep it simple)

  • Step 2: Starting at the center of the object, Draw and label all the external forces acting on the object















Draw a free body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.

  • Draw a free body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.



The Law of Inertia

  • The Law of Inertia

    • An object at rest remains at rest, and an object in motion continues in motion with constant velocity (constant speed in straight line) unless the object experiences a net external force
    • The tendency of an object not to accelerate is called inertia


The net external force (Fnet) is the vector sum of all the forces acting on an object

  • The net external force (Fnet) is the vector sum of all the forces acting on an object

  • If an object accelerates (changes speed or direction) then a net external force must be acting upon it



If an object is at rest (v=0) or moving at constant velocity, then according to Newton’s First Law, Fnet =0

  • If an object is at rest (v=0) or moving at constant velocity, then according to Newton’s First Law, Fnet =0

  • When Fnet =0, the object is said to be in equilibrium



A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N and downward with a force of 236 N.

  • A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N and downward with a force of 236 N.

    • A. Find the net external force in the x direction
    • B. Find the net external force in the y direction
    • C. Find the magnitude and direction of the net external force on the crate.




A. 82 N + (-115 N )= -33 N

  • A. 82 N + (-115 N )= -33 N

  • B. 565 N + (-236 N) = 329 N

  • C. Find the resultant of the two vectors from part a and b.



Review Newton’s 1st Law:

  • Review Newton’s 1st Law:

  • In other words, if the net external force acting on an object is zero, then the acceleration of that object is zero



The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass

  • The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass



A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline?

  • A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline?



To calculate Fnet, we need m and a

  • To calculate Fnet, we need m and a

    • M=2.0 kg
    • What is a?
    • Vi= 0 m/s, t=0.50 s,
    • displacement=85 cm=.85 m
    • Welcome back kinematic equations! 




Forces always exist in pairs

  • Forces always exist in pairs

  • For every action there is an equal and opposite reaction







Weight= Fg = mg

  • Weight= Fg = mg

  • Normal Force= FN= Is always perpendicular to the surface.

  • Friction Force= Ff

    • Opposes applied force
    • There are two types of friction: static and kinetic


Force of Static Friction (Fs) is a resistive force that keeps objects stationary

  • Force of Static Friction (Fs) is a resistive force that keeps objects stationary

  • As long as an object is at rest:

  • Fs = -Fapp



Kinetic Friction (Fk) is the frictional force on an object in motion

  • Kinetic Friction (Fk) is the frictional force on an object in motion



The coefficient of friction (μ) is the ratio of the frictional force to the normal force

  • The coefficient of friction (μ) is the ratio of the frictional force to the normal force

  • Coefficient of kinetic Friction

  • Coefficient of Static Friction



A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the char is in motion, a 327 N horizontal force keeps it moving at a constant velocity.

  • A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the char is in motion, a 327 N horizontal force keeps it moving at a constant velocity.

    • A. Find coefficient of static friction
    • B. Find coefficient of kinetic friction


In order to get the chair moving, it was necessary to apply 365 N of force to overcome static friction. Therefore Fs = 365 N.

  • In order to get the chair moving, it was necessary to apply 365 N of force to overcome static friction. Therefore Fs = 365 N.

  • The normal force is equal to the weight of the chair (9.81 x 25= 245 N)



The problem states that the chair is moving with constant velocity, which means Fnet=0. Therefore, Fapp must equal -Fk.

  • The problem states that the chair is moving with constant velocity, which means Fnet=0. Therefore, Fapp must equal -Fk.





A woman is pulling a box to the right at an angle of 30 above the horizontal. The box is moving at a constant velocity. Draw a free body diagram for the situation.

  • A woman is pulling a box to the right at an angle of 30 above the horizontal. The box is moving at a constant velocity. Draw a free body diagram for the situation.





Since the suitcase is moving with constant velocity, Fnet=0.

  • Since the suitcase is moving with constant velocity, Fnet=0.

  • That means the forces in the x direction have to cancel out and the forces in y direction have to cancel out

    • Fk = Fapp,x
    • FN + Fapp,y = Fg
    • NOTICE THAT NORMAL FORCE DOES NOT EQUAL WEIGHT IN THIS SITUATION


A 925 N crate is being pushed across a level floor by a force F of 325 N at an angle of 25 above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.

  • A 925 N crate is being pushed across a level floor by a force F of 325 N at an angle of 25 above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.



So we need mass and Fnet.

  • So we need mass and Fnet.

  • We have weight (925 N). So what is mass?

  • How to find Fnet?

  • Find vector sum of forces acting on crate.





Is box accelerating in y direction?

  • Is box accelerating in y direction?

    • No. Therefore Fnet in y direction is 0
    • So FN + Fapp,y = Fg
    • So FN = Fg- Fapp,y= 925 N- 325sin(25)
      • FN= 787.65 N


Is box accelerating in x direction?

  • Is box accelerating in x direction?

    • Yes. Therefore Fnet,x is not 0
    • Fnet,x= Fapp,x – Ff
      • Fapp,x = Fappcos(25)=294.6 N
    • Use coefficient of friction to find Ff
      • Ff=μFN=(0.25)(787N)=197 N


Fnet,x = 294 N – 197 N= 97 N

  • Fnet,x = 294 N – 197 N= 97 N

  • So now we know that the Fnet on the box is 97 N since Fnet,y is 0



A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal.

  • A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal.

  • If μk between the floor and the box is 0.57, how long does it take to move the box 4.00 m starting from rest?





Is box accelerating in y direction?

  • Is box accelerating in y direction?

    • No. Therefore Fnet in y direction is 0
    • So FN = Fapp,y + Fg
    • So FN = 485sin(35) + 319 N= 598 N






We want to know how long it takes for the box to move 4.00 m.

  • We want to know how long it takes for the box to move 4.00 m.

  • Find vf so that you can solve for t

  • Solve for t



A block slides down a ramp that is inclined at 30° to the horizontal. Write an expression for the normal force and the net force acting on the box.

  • A block slides down a ramp that is inclined at 30° to the horizontal. Write an expression for the normal force and the net force acting on the box.





Solve for Fg,y and Fg,x

  • Solve for Fg,y and Fg,x





When a mass is sliding down an inclined plane, it is not moving in the y direction.

  • When a mass is sliding down an inclined plane, it is not moving in the y direction.

  • Therefore Fnet,y =0 and all the forces in the y direction cancel out.



So what are the forces acting in the y direction?

  • So what are the forces acting in the y direction?

  • Look at your FBD

  • We have normal force and Fg,y

  • Since they have to cancel out…

  • FN= mgcos(θ)



What is the force that makes the object slide down the inclined plane?

  • What is the force that makes the object slide down the inclined plane?

    • Gravity…but only in the x direction




So what are the forces acting in the x direction?

  • So what are the forces acting in the x direction?

  • Friction Force (Ff) and Gravitational Force (Fg,x)

  • If the box is in equlibrium

    • Fg,x = Ff
  • If the box is accelerating

    • Fnet= Fg,x - Ff


Example: a box is being pushed up an inclined plane…

  • Example: a box is being pushed up an inclined plane…



FN= mgcosθ

  • FN= mgcosθ

  • Fnet = Fapp- Fg,x – Ff

  • If the object is in equilibrium then

    • Fapp= Fg,x + Ff


A 5.4 kg bag of groceries is in equilibrium on an incline of angle. Find the magnitude of the normal force on the bag.

  • A 5.4 kg bag of groceries is in equilibrium on an incline of angle. Find the magnitude of the normal force on the bag.





The block is in equilibrium so…

  • The block is in equilibrium so…

    • Fnet=0
    • Fg,y= FN=mgcosθ=(5.4kg)(9.81)cos(15)
    • FN=51 N
  • Additionally, what is the force of friction acting on the block?



Fnet= 0

  • Fnet= 0

  • Fg,x= Ff= mgsinθ=5.4(9.81)sin(15)

  • Ff= 13.7N



A 75 kg box slides down a 25.0° ramp with an acceleration of 3.60 m/s2.

  • A 75 kg box slides down a 25.0° ramp with an acceleration of 3.60 m/s2.

    • Find the μk between the box and the ramp
    • What acceleration would a 175 kg box have on this ramp?




They give mass and acceleration

  • They give mass and acceleration

  • So Fnet= ma= 75kg x 3.60 m/s2

    • Fnet= 270 N
  • FN= mgcosθ

  • Fnet= Fg,x – Ff=mgsinθ - Ff



Fnet= Fg,x – Ff=mgsinθ – Ff

  • Fnet= Fg,x – Ff=mgsinθ – Ff

  • Ff= mgsinθ – Fnet

  • Ff = 75kg(9.8)sin(25) – 270 N

  • Ff = 40.62 N



We are trying to solve for μk

  • We are trying to solve for μk





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