## 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
## 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 ## 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)
## 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
## 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
## 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
## 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 ## 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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