Prof. Dr. Ibraheem Nasser Examples_6 October 13, Review (Chapter 6)

Transcription

1 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Review (Chapter 6) cceleration of a loc againt Friction (1) cceleration of a bloc on horizontal urface When body i moving under application of force P, then inetic friction oppoe it motion. Let a i the net acceleration of the body. From the figure long the upward direction: may R mg 0 R mg ma P F a P F / m Note: F K R mg K K 1

2 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Note: The friction force f R ( W F in 60 ) ( mg F in 60 ) () cceleration of a bloc down a rough inclined plane When angle of inclined plane i more than angle of repoe, the body placed on the inclined plane lide down with an acceleration a. From the figure Note: 1- The friction force F R, R mg co. - For frictionle inclined plane 0 a gin (3) Retardation of a bloc up a rough inclined plane When angle of inclined plane i le than angle of repoe, then for the upward motion Note: For frictionle inclined plane 0 a gin

3 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Problem: The angle of repoe i defined a the angle of the inclined plane with horizontal uch that a body jut begin to lide. y definition i called the angle of repoe. Find the friction coefficient. nwer: F we now tan R. Thu the coefficient of limiting friction i equal to the tangent of angle of repoe. well a i.e. angle of repoe = angle of friction Problem: horizontal force of 10 N i neceary to jut hold a bloc tationary againt a wall. The coefficient of friction between the bloc and the wall i 0.. The weight of the bloc i nwer: For equilibrium: Weight (W ) = force of friction (F) W R N Problem: fireman of ma 60 g lide down a pole. He i preing the pole with a force of 600 N. The coefficient of friction between the hand and the pole i 0.5, with what acceleration will the fireman lide down (g = 10 m/') nwer: Friction: F R N Weight = W 600 N W F m/ ma W F a m Problem: ma m 1, placed on a rough horizontal plane: nother ma, hung from the tring connected by pulley, the tenion (T ) m produced in tring will try to tart the motion of ma m 1. Find the minimum value of m to tart the motion. nwer: For the ma m 1, we have: T F, F R m g T m g (1) For the ma m, we have: T = m g () Fro (1) and (), we get the limiting condition: m g m g m / m 1 1 The condition m m1 i the minimum value of m to tart the motion. 3

4 Prof. Dr. Ibraheem Naer Example_6 October 13, Problem: When a ma, placed on a rough inclined plane: m m 1 nother ma hung from the tring connected by pulley, the tenion (T ) produced in tring will try to tart the motion of ma m 1. nwer: t limiting condition thi i the minimum value of m, to tart the motion m Note: In the above condition Coefficient of friction tan. m1 co Problem: bloc with ma 100 g i reting on another bloc of ma 00 g. hown in figure a horizontal rope tied to a wall hold it. The coefficient of friction between and i 0. while coefficient of friction between and the ground i 0.3. The minimum required force F to tart moving will be: nwer: From the free body diagram one find that two frictional force will wor on bloc. F f f m g m m g min G G (300) N (Thi i the required minimum force) Q: In the figure, a boy i dragging a box (ma =8.0 g) attached to a tring. The box i moving horizontally with an acceleration a =.0 m/. If the frictional force i 1 N, calculate the applied force F at an angle θ=60. nwer: On x-axi, one find ma F co60 f 8 F x 1 16 F 56 N Q15 worer drag a crate acro a factory floor by pulling on a rope tied to the crate a hown in the figure. The worer exert a force of T 500 N on the rope, which i inclined at 30 o to the horizontal, and the floor exert a frictional force of f 150 N. Calculate the magnitude of the acceleration of the crate if it weight i W Mg 310 N. nwer: On x-axi, one find 4

5 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Ma T co30 f N x 83 ax g 9.13 g Q: 5.0 g bloc i moving with contant velocity down a rough incline plane. The coefficient of tatic and inetic friction between the bloc and the incline are 0.5 and 0.0, repectively. What i the inclination angle of the incline plane? nwer: Note that:, 1 tan ; tan tan ; tan Since the bloc i moving with contant velocity down a rough incline plane, we have to calculate Q: 3.5 g bloc i puhed along a horizontal floor by a force F of magnitude 15 N at an angle θ = 40º below the horizon a hown in the Figure. The coefficient of inetic friction between the bloc and the floor i 0.5. Calculate the magnitude of (a) the frictional force on the bloc from the floor and (b) the bloc acceleration. Solution: H. W. Draw the free body diagram We chooe +x horizontally rightward and +y upward and oberve that the 15 N force ha component F Fco and F Fin. x y (a) We apply Newton econd law to the y axi: R F in mg 0 R (15 N) in 40 (3.5 g)(9.8 m/ ) 44 N. With 0.5 and f R f = 11 N. (b) We apply Newton econd law to the x axi: 15 Nco 4011N F co f ma a 0.14 m/. 3.5 g Since the reult i poitive-valued, then the bloc i accelerating in the +x (rightward) direction. 5

6 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Extra problem Q: ody in the Figure weigh 10 N, and body weigh 3 N. The coefficient of friction between and the incline are μ = 0.56, μ = 0.5, and the angle θ = 40º. Let the poitive direction of an x-axi be up the incline. In unit- vector notation, what i the acceleration of if i initially (a) at ret (b) moving up the incline, and (c) moving down the incline? Solution Firt, we chec to ee if the bodie tart to move. We aume they remain at ret and compute the force of (tatic) friction which hold them there, and compare it magnitude with the maximum value The free-body diagram are hown below. T i the magnitude of the tenion force of the tring, f i the magnitude of the force of friction on body, FN i the F N magnitude of the normal force of the plane on body, i the force of gravity on body (with magnitude W = 10 N), and m g i the force of gravity on body (with magnitude W = 3 N). θ = 40 i the angle of incline. We are told the direction of but we aume it i downhill. If we obtain a negative reult for f, then we now the force i actually up the plane. (a) For we tae the +x to be uphill and +y to be in the direction of the normal force. The x and y component of Newton econd law become T f W in 0 F W co 0. N Taing the poitive direction to be downward for body, Newton econd law lead to W T three equation lead to 0. Solving thee f W W in 3 N (10 N)in N (indicating that the force of friction i uphill) and to F W co (10 N) co 40 78N N m g which mean that f,max = μfn = (0.56) (78 N) = 44 N. Since the magnitude f of the force of friction that hold the bodie motionle i le than f,max the bodie remain at ret. The acceleration i zero. (b) Since i moving up the incline, the force of friction i downhill with magnitude f FN. Newton econd law, uing the ame coordinate a in part (a), lead to T f W in m a FN W co 0 W T m a for the two bodie. We olve for the acceleration: f 6

7 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 a W W in W co m m 3N 10N in N co 40 3N+10N 9.8 m 3.9 m. The acceleration i down the plane, i.e., a ( 3.9 m/ )i, which i to ay (ince the initial velocity wa uphill) that the object are lowing down. We note that m = W/g ha been ued to calculate the mae in the calculation above. (c) Now body i initially moving down the plane, o the force of friction i uphill with f F N 3N+10N 9.8 m ˆ magnitude. The force equation become T f W in m a FN W co 0 W T m a which we olve to obtain W W in W co a m m 3N 10N in N co 40 The acceleration i again downhill the plane, i.e., object are peeding up. 1.0 m. a ( 1.0 m/ )i ˆ. In thi cae, the 61- bloc of ma m b m t = 4.0 g i put on top of a bloc of ma = 5.0 g. To caue the top bloc to lip on the bottom one while the bottom one i held fixed, a horizontal force of at leat 1 N mut be applied to the top bloc. The aembly of bloc i now placed on a horizontal, frictionle table (Fig. 6-47). Find the magnitude of (a) the maximum horizontal force that can be applied to the lower bloc o that the bloc will move together and (b) the reulting acceleration of the bloc. nwer: (a) Firt we have to calculate the coefficient of tatic friction uing the equation F mg, for the urface between the two bloc. F 1/ mg t 0.31, where mt g = (4.0 g)(9.8 m/ )=39. N i the weight of the top bloc. Second apply the Newton eq, Let M m m 9.0 g be the total ytem ma, then the maximum horizontal force ha a magnitude Ma = M g = 7 N. t b (b) The acceleration (in the maximal cae) i a = g =3.0 m/. 7

8 Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Example: In the figure bloc of ma M = 3.0 g lide along a floor while a force F of magnitude 1.0 N i applied to it at an upward angle θ. The coefficient of inetic friction = We can vary θ from 0 to 90 o (the bloc remain on the floor). What give the maximum value of θ the bloc acceleration magnitude a? nwer: On y-axi, one find Ma F in N Mg On x-axi, one find y a 0 N Mg F in y F F Max F co f F co N ax co g in M M To maximize a, differentiate the lat equation with repect to θ dax F F in co 0 tan. d M M For, we find the maximum a will be at the angle: tan