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CHAPTER-6 Force and Motion-II

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**Force and Motion-II Motion under frictional and centripetal forces:**

Force of friction ( required for car motion out of a pit and out of a curve) Centripetal force (Radial force) (required to turn car in a circle

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Ch 6-2 Friction An external force F applied to a block resting on a rough surface No motion till F is less than static frictional force fs : fs=sFN where FN =Fg Body slides if F fs When body start sliding static frictional force fs reduces to kinetic frictional force fk fk=kFN where FN =Fg If lFl=lfkl body moves at constant speed If lFl lfkl body accelerates

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**Checkpoint 6-1 Ans: (a) Zero (b) 5 N (c) No (d) Yes (e) 8 N**

A block lies on floor. (a) What is the magnitude of the frictional force on it from the floor? (b) If a horizontal force of 5 N is now applied to the block, but the block does not move, what is the magnitude of the frictional force on it? (c) If the maximum value fs,max of the static frictional force on the block is 10 N, will the block move if the magnitude of the horizontally applied force is 8 N? (d) If it is 12 N? (e) What is the magnitude of the frictional force in part (c) Ans: (a) Zero (b) 5 N (c) No (d) Yes (e) 8 N

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Checkpoint 6-2 In the figure, horizontal force F1 of magnitude 10 N is applied to a box on a floor, but the box does not slide. Then, as the magnitude of vertically applied force F2 is increased from zero but before the box begins to slide, do the following quantities increase, decrease, or stay the same. (a) magnitude of the frictional force on the box (b) the magnitude of the normal force on the box from the floor (c) the maximum value of fs,max of the static frictional force on the box? Answer (a) magnitude of the frictional force on the box remains same (b) the magnitude of the normal force on the box from the floor decreases (c) the maximum value of fs,max of the static frictional force on the box decreases

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**Example-Problem-6-2 Loaded sled being pulled at constant speed, find T**

Since v is constant, then l Tcos l= lfkl = kFN but FN+Tsin =Fg=mg FN = mg-Tsin Then Tcos = fk= k FN = k (mg-Tsin ) Solve for T

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Sample-Problem-6-3 Coefficient of static s friction for coin on verge of sliding down fs =s FN; s =fs / FN Fx= fs –Fgsin=0 fs = Fg sin =mg sin Fy= FN –Fg cos=0 FN = Fg cos=mg cos s =fs / FN = mg sin /mg cos s = tan

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**Ch 6-5: Uniform Circular Motion (Radial force and friction force)**

Circular motion under the effect of centripetal force Centripetal force FR accelerates a body by changing the direction of the body’s velocity v without changing the body’s speed Constant centripetal acceleration aR, directed toward the center of the circle with radius R aR=v2/R Centripetal force FR=maR =m (v2/R) For puck tied to a string with tension T and moving in a circle FR=T=m (v2/R)

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Checkpoint 6-3 Near the ground, is the speed of large raindrops greater than, or the same as the speed of small raindrops, assuming that all raindrops are spherical and have the same drag coefficient? Answer: It takes longer time to attain the terminal velocity after falling for the larger raindrops as compared to the small drops. Under a constant acceleration, the large raindrops achieve larger value of terminal velocity than small raindrops.

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Sample-Problem-6-8 Static frictional force fs prevent her from sliding down fs=Fg fs=sFN and normal force FN is provided by centripetal force FR Then FN = FR=m(v2/R) fs= s FN =sm(v2/R)=Fg=mg v2 =Rg/ s

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**Example (Motion of car on a un-banked curved road)**

Coefficient of static friction s when car is verge of sliding out of track -FR=-mv2/R= - fS= -s N -mv2/R= - smg s = v2/Rg

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**Example-Problem-6-10 (Motion of car on a Banked Curved Road)**

Calculate banking angle prevent sliding FX = -FR=-mv2/R= - Nr = -N sin -mv2/R= -N sin ……..(1) FY = NY -mg =0 NY = N cos = mg ….(2) Then (mv2/R)/(mg)= (N sin )/(N cos ) v2/Rg= tan = tan-1 (v2/Rg)

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