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

 Sana 10.01.2019 Hajmi 1.75 Mb.   • ## The Unit for Force is a Newton • ## 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 • ## FBDs are essential! They are not optional! You need to draw them to get most problems correct! • ## We want to draw a FBD for the car only. • ## 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. • ## 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 • ## If an object accelerates (changes speed or direction) then a net external force must be acting upon it • ## 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. 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.  • ## C. Find the resultant of the two vectors from part a and b. • ## 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 • ## 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

• M=2.0 kg
• What is a?
• Vi= 0 m/s, t=0.50 s,
• displacement=85 cm=.85 m
• Welcome back kinematic equations!   • ## For every action there is an equal and opposite reaction   • ## Friction Force= Ff

• Opposes applied force
• There are two types of friction: static and kinetic • ## Fs = -Fapp • ## Kinetic Friction (Fk) is the frictional force on an object in motion • ## 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. Find coefficient of static friction
• B. Find coefficient of kinetic friction • ## 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.  • ## 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.  • ## 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. • ## Find vector sum of forces acting on crate.  • ## 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?

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

• No. Therefore Fnet in y direction is 0
• So FN = Fapp,y + Fg
• So FN = 485sin(35) + 319 N= 598 N   • ## 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.  • ## Solve for Fg,y and Fg,x  • ## Therefore Fnet,y =0 and all the forces in the y direction cancel out. • ## FN= mgcos(θ) • ## What is the force that makes the object slide down the inclined plane?

• Gravity…but only in the x direction  • Fg,x = Ff
• ## If the box is accelerating

• Fnet= Fg,x - Ff • ## Example: a box is being pushed up an inclined plane… • ## 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.  • ## 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? • ## Ff= 13.7N • ## 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?  • ## So Fnet= ma= 75kg x 3.60 m/s2

• Fnet= 270 N

• ## Fnet= Fg,x – Ff=mgsinθ - Ff • ## Ff = 40.62 N • ## We are trying to solve for μk Do'stlaringiz bilan baham:

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