Kinematics (Forces) PDF

Forces S.L.O 2.5  Resolution of forces It is defined as the process of splitting up the given force into a number of components , without changing its effect on the body is called resolution of a force. A force is, generally, resolved along two mutually perpendicular directions. ...

Forces S.L.O 2.5  Resolution of forces It is defined as the process of splitting up the given force into a number of components , without changing its effect on the body is called resolution of a force. A force is, generally, resolved along two mutually perpendicular directions.  Rectangular components of forces 🠶 Consider a force F represented by line OA making an angle θ with x-axis as shown in figure. Draw a perpendicular AB on x- axis from A. According to head to tail rule, OA is the resultant 🠶 of vectors represented by OB and BA. OA= OB+ BA (1) The components OB and BA are perpendicular to each other. They are called the perpendicular components of OA representing force F. Hence OB represents its x-component Fx ,and BA represents its y-component Fy-. Therefor, equation (1) can be written as F = Fx + Fy (2) The magnitudes Fx and Fy of forces Fx and Fy can found using s 🠶 Sin θ= perp / hypotenuse Y axis B 🠶 Cosθ= base/hypotenuse 🠶 Tan θ = perp/base F 🠶 Using this diagram Fy 🠶 sin θ= Fy/F or Fy= F Fy= sinθ Fsin θ 🠶 Cos θ = Fx/F 0r Fx=F cos θ Fx A 🠶 Similarly Tan θ= Fy/Fx or θ= Θ= Tan-1Fy/Fx  n is pulling a trolly on a horizontal road with a force of 200N making an angle of gree with the road. Find the horizontal and vertical components of the force. Cos 30= 6 and sin 30= 0.5 🠶 Data 🠶 Θ =30 degree 🠶 F =200N 🠶 Fx=? 🠶 Fy = ? Fx=F cos θ 🠶 Formula 🠶 and and = 200COS30 🠶 200X 0.866 =173.3N Fy=F Sin θ 200X SIN 30 200X 0.5 = 100N Determination of resultant vector Using Pythagoras theorem Hypotenuse= √(perpendicular)2 + (base)2 F = √ Fy)2 +(Fx)2 A boat travels toward south with a steady velocity of 30 m/s through a river in which there is a steady N flow of water in the direction of the east at a velocity of 15 m/s. i. Illustrate the given situation using a labelled W vector diagram. E ii. Show the direction of the motion of the boat in the diagram. S iii. Calculate the resultant velocity of the boat. First equation of motion 🠶 According to the definition of acceleration 🠶 ACCELERATION= CHANGE IN VELOCITY/TIME 🠶 a= (Vf-Vi) /t 🠶 If we cross multiply time with acceleration 🠶 at = Vf- Vi 🠶 OR at + Vi = Vf 🠶 Vf = Vi + at  numerical 🠶 The velocity of a car changes steadily from 10m/s to 30m/s in 10 s. What is the acceleration of the ca 🠶 A car travelling at 10 m/s accelerates uniformly at 2 ms 2. Calculate its velocity after 5 s? 🠶 A train slows down from 80 km/h with a uniform retardation of 2 ms 2. How long will it take to attain a km/h 🠶 1. a= (vf –vi)/t Second equation of motion 🠶 As we have studied S = Vav x t-------(eq1) 🠶 Vav = (Vf+ Vi)/2 🠶 Substitute the value of Vav in eq 1 it becomes 🠶 S = (Vf+ Vi)/2 x t -------(eq 2) 🠶 From first equation of motion Vf = vi +at 🠶 S = (Vi + Vi +at)/2 x t OR S = (2Vi + at )/2 x t 🠶 Therefore S = (2Vi/2 + at /2)x t 🠶 S = Vi t + ½ at2 1. A bicycle accelerates at 1 ms-2 from an initial velocity of 4 m/s for 10 s. Find the distance moved by it during this interval of time Data a= 1 ms-2 Vi= 4m/s t= 10s Formula S = Vi t + ½ at2 S=4x10 + ½ x1x10x10 S = 40 + 1/2x 100 = 40 +50 = 90m 3rd Equation of motion 🠶 S = Vav x t ------eq1 🠶 = Vav = (Vf+ Vi)/)/2 🠶 Therefore S= (Vf+ Vi)/)/2 x t ------------------eq2 🠶 From first equation of motion 🠶 Vf = Vi +at or t= (Vf- Vi)/)/a 🠶 S= (Vf-+ Vi)/)2 x (Vf- Vi)/a 🠶 2aS = (Vf + Vi)(Vf-Vi) 🠶 2aS = Vf2 –Vi 2 A car travels with a velocity of 5 m/s. It then accelerates uniformly and travels a distance of 50 m. If the velocity reached is 15 m/s , find the acceleration and the time to travel this distance. Data Vi= 5m/s S= 50m Vf= 15m/s a=? (ii)Vf= vi + at 15= 5 +2xt (15-5)/2=t t=? t=5s 2aS = Vf2 –Vi 2 2x ax50 = (15x15)- (5x5) 100a = 225-25 a= 200/100=2m/s2