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CHAPTER-5 Force and Motion-1 Ch 5-1 Physics Physics: Study of cause of acceleration of an object Cause: Force Force is said to act on the object to change its velocity Ch 5-3 Newton’s First Law If no force acts on a body, then the body’s velocity cannot change i.e. the body cannot accelerate. If body is at rest, it stays at rest, if it is in motion, it continues to move with the same velocity Ch 5-4 Force A force F is measured by the acceleration a it produces. Principal of superposition of forces: When two or more forces are acting on a body resultant force or net force Fnet can be obtained by adding the forces individually Fnet = F If no net force acts on a body (Fnet =0), the body velocity cannot change. Checkpoint 5-1 Which of the figure’s six arrangements correctly show the vector addition of forces F1 and F2 to yield the third vector, which is meant to represent net force Fnet? Ans: c, d and e Ch 5-5 Mass Mass is an intrinsic characteristic of a body Mass of a body relates a force acting on a body to the resulting acceleration The ratio of the masses of two bodies m0 and mx is equal to inverse of their acceleration a0 and ax when same force is applied to both bodies mx/m0 = a0/ax Ch 5-6 Newton’s Second Law The net force Fnet acting on a body is equal to the product of the body mass m and its acceleration a Fnet = ma; a= Fnet / m Acceleration component along a given axis is caused only by sum of forces component along that axis ax = Fnet,x /m ; ay = Fnet,y /m ; az = Fnet,z /m SI unit of force Newton (N) 1N= 1kg.m/s2 Checkpoint 5-2 The figure here shows two horizontal forces acting on a block on a frictionless floor. If a third horizontal force F3 also acts on the block, what are the magnitude and direction of F3 when the block is: (a) stationary (b) moving to the left with a constant speed 5 m/s? F= F3 +5-3=0 Then F3 =-5+3= -2 N in both cases (a) and (b) Checkpoint 5-3 The figure shows overhead views of four situations in which two forces accelerate the same block across a frictionless floor. Rank the situation according to the magnitudes of (a) the net force on the block and (b) The acceleration of the block, greatest first Calculate Fnet= F1 + F2 (a) (1) Fnet=5+3=8 (2) Fnet=5-3=2 (3) Fnet= (52+32)=34~6 (4) Fnet=[(5+3cos)2+(3sin)2] > 6 Ans: (1),(4), (3), (2) (b) Acceleration of block a a = Fnet/ constant block mass The order of acceleration is same as that of Fnet Ans: (1),(4), (3), (2) Ch 5-7: Some Particular Forces Gravitational Force Fg: The force with which a body is pulled towards earth. Downward force –Fg= m(-|g|) Weight: Equals the magnitude of a net force required to prevent the body from falling freely, as measured on the ground. weight |W| = |Fg| Ch 5-7: Some Particular Forces (FN and T) Normal Force FN : When a body presses against a surface, the surface pushes the body with a normal force FN, that is to the surface For a block resting on a table FN = mg Tension Force T: It is the force applied to a body through a cord, pulled taut opposite to object. For a mass less cord , the pulls at both ends of the cord has same magnitude T Checkpoint 5-3 In figure shown below, is the magnitude of the normal force FN greater than, less than, or equal to mg if the block and table are in an elevator moving upward (a) at constant speed (b) at increasing speed (a) For constant speed Fnet=0= FN – mg=0 FN = mg (b) For increasing speed Fnet=ma= FN – mg FN = ma + mg FN > mg Checkpoint 5-4 The suspended body in the figure (c) weighs 75 N. Is T equal to greater than, less than 75 N when body is moving upward at (a) constant speed (b) at increasing speed (c) at decreasing speed (a) For constant speed Fnet=0= T – mg=0 T = mg = 75 N (b) For increasing speed Fnet=ma = T – mg T = ma + mg T > mg (c) For decreasing speed Fnet=-ma = T – mg T = mg - ma T < mg Ch 5-8 Newton’s Third Law Newton’s Third Law: When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. For the book and crate FBC=-FCB FBC and FCB are called third-law force pair FBC and FCB are also called action and reaction forces pair Ch 5-9 Applying Newton’s Laws Problem-solving strategy: Draw a simple and neat diagram of the system Isolate the object whose motion is being analyzed. Draw freebody diagram for this object for the external forces acting on the object Establish convenient coordinate axes for each object. Find component of forces along these axes. Apply Newton’s Second Law F=ma in component form. Solve the component equations for the unknowns Ch 5-9 Applying Newton’s Laws For (a) Fy=FN-Fgs=May=0; FN= Fgs Fx=Max=T=Ma; T=Ma For (b) Fx=0 Fy=T-FgH=T-mg=-ma T=m(g-a) and T=Ma Solve for a a =mg/(m+M) Ch 5-9 Applying Newton’s Laws Sample Problem 5-6 Find T1, T2 and T3 ?; For (a) Fy=T3-Fg=0; T3= Fg=mg and Fx=0 For (b) Fx=T2cos 47-T1cos 28=0 T2cos 47=T1cos 28 ……(1) Fy= T2 sin 47+T1 sin 28-T3=0 T2 sin 47+T1 sin 28=T3 ….(2) Solve eq. (1) and (2) to get T1 and T2 Sample Problem 5-7 Find T and FN if = 27º Resolving force Fg=mg along and the plane and perpendicular to the plane T-mg sin = 0; T= mg sin Also FN-mg cos =0 Then FN= mg cos Checkpoint 5-7 In the figure , horizontal force F is appled to a block on a ramp. (a) Is the component of F that is perpendicular to the ramp Fcos or Fsin (b) Does the presence of F increases or decreases the magnitude of the normal force on the block from the ramp? Ans: (a) Fsin to ramp (b) F to ramp= N-mgcos- Fsin =0 Then N= mgcos+ Fsin Fsin increases the magnitude of the normal force on the block from the ramp Checkpoint 5-8 In the figure , what does the scale read if the elevator cable breaks sp that the cab falls freely ; that is what is the apparent weight of the passenger in free fall? Answer In free fall F = N-Fg=-Fg Apparent weight N N= -Fg+Fg = 0 Sample Problem 5-9 Find a? Introduce action and reaction par forces FAB and FBA Fapp=(mA+mB)a a= Fapp/(mA+mB) Then FBA =mBa