Phys 201 1nd Edition Lecture 8 Outline of Last Lecture I. Solving for two dimensional forces and breaking up free body diagrams.Outline of Current Lecture II. Working Without FrictionA. Solving two dimensional forcesIII. FrictionA. Friction Coefficient x Normal forceB. Solving for minimum force IV. Static and Kinetic FrictionA. At rest or in motionV. Solving with frictionCurrent LectureWorking without frictionEx. A block slides down a frictionless plane which makes a 40◦ angle with relation to the horizontal. The block has a mass of 20kg. What is the normal force exerted on the block by the plane?This problem involves two dimensional motion. Therefore, our free body diagram needs to consider boththe forces perpendicular and parallel to the plane. This means we will divide our forces into perpendicular and parallel equations instead of x and y equations. N Weigh*Cos(40◦) Weight (mg) Weight*Sin(40◦)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. 40◦Because we are looking at the normal force exerted on the box by the plane, the weight is no longer opposite of the normal force like we are used to. To find the force opposite of the normal force, we need to make a new triangle with one force parallel to the plane and one force perpendicular to the plane.If you consider the weight as the y-component of a triangle in which the plane is the hypotenuse, then our perpendicular force is equal to the x-component of this triangle, and the parallel force is equal to theplane. Using our trigonometric equations, we know that the parallel force equals the weight x Sin(θ) and the perpendicular force equals the weight x Cos(θ). Now we can divide our forces into two equations andplug in all of the values we know.Weight=Mg=(20)(9.81)=196.2Fpurp=Weight*Cos(40)=150.3Fpar= Weight*Sin(40)=126.1∑Fpurp=mapurpn -mgCos(40)=20(0)n -150.3=0n=150.3NFrictionFriction works just like any other force and therefore needs to be added to your free body diagram if present. Friction always works as a resisting force, working opposite of the motion of an object. The maximum force of friction is equal to the friction coefficient times the normal force. The equation to represent this is; F=μN Ex. What is the minimum force required to pull of 50kg bag across the floor with a friction coefficient of μ=.3The first step for this problem is to draw a free body diagram that includes all x and y forces. N μ T Weight(mg)The next step is to find the normal force. Since our object has no motion in the y direction, the normal force equals the weight, which is just the mass times the acceleration of gravity. Normal force=490.5NNow to find the minimum force required to move the bag, we just multiply the normal force by the friction coefficient. Force=(490.5)(.3)=147.15NKinetic and Static friction:Kinetic friction: This is the kind of friction that acts on an object in motion. When a car is moving and it breaks, kinetic friction acts as the restricting force that slows its forward motion. Static friction: This is the kind of friction that acts on an object at rest. When you walk across the floor, static friction is what keeps your foot from sliding all over the floor. Though both types of friction work in the same way, an object very rarely has both types of static acting on it.You denote static or kinetic friction by using the subscript s or k. Friction works differently depending on whether or not an object is at rest.When an object is at rest: μs>μkfs≤μsNWhen an object is sliding down a ramp:Fk=μkNApplying Force of Friction;Ex. Two pots, the first pot with a mass of 6kg and the second with a mass of 3kg, are sitting at rest on a table, attached by a string. The friction coefficient between the pots and the table is μ=.35. This system of pots is attached to a 4kg mass by a string on a pulley. A) When the mass is dropped, what will be the acceleration of the pots across the table? B) What is the tension between the two pots(T1) and what is the tension between the system of pots and the mass being dropped(T2)?To start, we need to make a free body diagram of our system T16kg 3kgT24kgSo we have to make three free body diagrams to solve for this system. First, we’ll solve for the dropped object T2∑F=ma 39.24N-T2=ma mgNext, well make our free body diagram to solve for our pot closest to the pulley (pot 2) N N-mg=0 N=(3)(9.81)=29.43NT1T2 F=μN F=(.35)(29.43)=10.3NT2 - T1-10.3=maFinally, we’ll make a free body diagram for our first pot N N-mg=0 N=(6)(9.81)=58.86N T1T2 F=μN F=(.35)(58.86)=20.6N T1-21.6N=maNow that we have a set of equations to solve for our forces, we can combine these equations to help us solve for T1 and T2 and acceleration.Dropped mass: 39.24N-T2=ma  T2=39.24N-(4a)Pot 2: T2 - T1-10.3=ma  T2 - T1= 10.3N+(3a)Pot 1: T1-21.6N=ma T1=(6a)+21.6NTo solve for acceleration, we plug in our isolated values for T1 and T2 into the equation derived from the equation that combines these two tensions. mgmgT2 - T1= 10.3N+(3a)  (39.24N-4a) -(6a + 21.6N)=10.3N+3aNow we need to isolate a. So we move all of our force values to one side of the equation, and all of our avalues to the other.39.24N - 21.6N - 10.3N=3a+4a+6a  7.34N=13aa= 7.34/13= .56m/s2Now, we can get our values for tension by plugging our value for acceleration into our equations for our pot systems.Pot 1: T1=(6)(.56)+21.6N=25NPot 2: T2 - 25N=