Phys 201 1nd Edition Lecture 7 Outline of Last Lecture I. Calculating forces and free body diagrams.Outline of Current Lecture II. ForcesA. Action/reaction ForceIII. TensionA. Free Body Diagrams IV. Forces In Two DimensionsA. Solving x and y componentsCurrent LectureForces:When dealing with a problem in which velocity (V) is constant, the acceleration is equal to 0. This means that the sum of the forces acting on the object in any given direction is 0. . A force that is shared between two objects is known as an action/reaction force. This means that the pair of forces acting on two objects are equal and opposite reactions(Newton’s 3rd law). Think about playing tug-of-war. If you and your opponent are equally matched while pulling on a rope in opposite directions, neither of you will move. The tension in the rope you are pulling on is the same for you and your opponent, but the direction of that tension is opposite. This does not mean, however, that weight and normal force are an action/reaction force. Even though these are generally equal and opposite forces, they are acting on the same object, rather than two objects offsetting one another. Tension:Tension refers to a kind of action/reaction contact force in which the force of one object causes strain on another The easiest way to think of tension is as a string connecting two objects. T FThese 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. m1=10kgm2=20kgIf we have two blocks, one with a mass of 10kg and another with a mass of 20 kg, attached by a string moving at a constant acceleration of 2m/s2, what is the force of the two objects and what is the tension between them?The best way to solve any problem with forces is using a free body diagram to illustrate the forces. For this problem, we will need to make a free body diagram for each block individually. The free body diagram will divide the forces on both objects into x and y forces with x and y accelerations, meaning that they move in the x or y direction. Because our blocks are moving across the table and not up or down, we know that our ay=0 and our ax=2. n=98.1NFBD1 T=m1ax mg=-98.1N n=196.1N FBD2 T F mg=-196.2NBecause tension is the only force acting in the x direction of the first block, the easiest way to calculate tension is to use our force equation independently for the first block, meaning; Fx=T=m1axT=10kg(2m/s2)=20NTo solve for the force pulling the two blocks, we need to look at how the forces act on the second block. Because we know the force of tension, and tension is a force that is shared between the two blocks, we do not need to consider the mass of the first block to find the Force. Our equation for this will be;Fx=F-T=m2axF-20=20(2)F=60NAlternately, we can calculate the force pulling the blocks without knowing tension. For this, we just have to add the masses of the two blocks together and use our normal force equation. F=(m1+m2)axF=(10+20)2F=60N2-Dimensional ForcesJust like vectors, forces can be calculated in two dimensions (moving in the x and y direction at the same time). Example: An object with a weight of 20N is suspended from the ceiling by two ropes, one at a 30◦ angle from the horizontal and one at a 40◦ angle from the horizontal. Because the object is not moving in any direction, all of our forces in any given direction have to add up to 0. Because the only information we know is the weight of the object(mg) and the angle of the ropes, we will have to combine force equations and our trigonometric equation to solve for the x and y components of the force. Once we solve for the x and y components, we can use trigonometry to solve for the force(tension) of each rope. F1x F2x T1 T2 ∑Fy=F1y+F2y-20N=0  ∑Fy=F1y+F2y=20N∑Fx=F1x+F2x=0We want to solve for the x and y components of T1(tension in rope one) and T2(tension in rope two). Since we know the angle between the x and y components of each, we can use tangent of the corresponding angles to give us an equivalence for one of the variables we need to solve for.30◦40◦20NTan(30◦)=F1y/F1x T1y=0.577F1xTan(40◦)=F2y/F2x  F2y=0.839F2x0.577F1x+0.839F2x=20N-0.577F1x-0.577F2x=0N0.262F2x=20N  F2x=76.34NF1x+76.34N=0N  F1x=-76.34NF1y=0.577(76.34)=44.05NF2y=0.839(-76.34)=-64.05NT1=Sin(30)=44.05N/hyp hyp=88.1T2=Sin(40)=64.05/hyp hyp=99.6NT1=√(F1y2+F1x2) = 88.14NT2=√(F2y2+F2x2)=99.65NWe can plug in the values we got from our Tangent equations, in this case, we solved for the y components. We plug these values into our Fy equation. If we multiply our Fx equation by -0.577, we can put these two equations together to solve for one of our variables, in this case, for T2x. Once we’ve solved for this variable we can plug the value into one of the equations to solve for the value of another variable. We can keep plugging these values into our equations until we have solved all four variables.To solve for the tension of each rope, we just plug our values for the x and y components of the ropes into our trig equations to solve for the hypotenuse. It doesn’t matter which equation you use, as long as you know you aresolving for the hypotenuse and use the