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Ch-6-091

Ch-6-091. Help-Session.

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Ch-6-091

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  1. Ch-6-091 Help-Session

  2. T082 Q#18: A 5 kg block is placed on top of a 10 kg block which is lying on a frictionless horizontal surface, as shown in Fig. 4. A horizontal force F of 60 N is applied to the 10 kg block. Find the static frictional force on 5 kg block from the 10 kg block such that the 5 kg block does not slip. A) 20 N to the right Acceleration a =60/15=4 m/s2 Force of friction between the two blocks is balanced by force(= ma) acting on the upper block Force of friction= ma= 5x4 N CH-6-082

  3. Q19. A crate is sliding down an incline that is 35◦ above the horizontal. If the coefficient of kinetic friction is 0.4 the acceleration of the crate is: A) 2.4 m/s2 Forces along the incline: mg sin-f=ma Forces  to the incline N-mg cos=0; N=mg cos f=kN = k mg cos Then mg sin-f=ma becomes mg sin- k mg cos =ma a = g (sin- k cos ) =9.8(sin35-0.4*cos35) = 2.4 m/s2 CH-6-082 f N a mgcos mgsin

  4. Q20. An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of static friction between tires and road is 0.5 The maximum speed with which this car can travel round this curve without sliding is: A) 12 m/s Fr=-mv2/r=-f=- sN=- smg v=(sgr)= (0.5 *9.8*30) =12 m/s CH-6-082

  5. T081 : Q19.: A 1.0 kg block, attached to the end of a string of length 2.0 m, swings in a vertical circle. When the block is at its highest point, its speed is 5.0 m/s. What is the magnitude of the tension in the string at that point? A) 2.7 N CH-6-081 V=5.0 m/s R 2 m T+mg=mv2/r; T=m(v2/r-g)=1.0*(52/2-9.8) =2.7 N

  6. Q20: A block of mass m = 10 kg is pushed up a rough 30o inclined plane by a force F parallel to the incline as shown in Figure 4. The coefficient of kinetic friction between the block and the plane is 0.4. Find the magnitude of the force F when the block is moving up at constant velocity. (Ans: 83 N) CH-6-081 • Forces along the incline: F-mg sin-f=0 • Forces  to the incline N-mg cos=0; N=mg cos f=kN = k mg cos Then F-mg sin-f=0 becomes F-mg sin- k mg cos =0 F = mg (sin+ k cos ) =10*9.8(sin30+0.4*cos30) = 83 N

  7. T071: Q17.: A 5.0 kg block is moving with constant velocity down a rough incline plane. The coefficients of static and kinetic friction between the block and the incline are 0.25 and 0.20, respectively. What is the inclination angle of the incline plane? (Ans: 11°) CH-6-071 Q18. A car rounds a flat curved road (radius = 92 m) at a speed of 26 m/s and is on the verge of sliding at this speed. What is the coefficient of static friction between the tires of the car and the road? (Ans: 0.75) -fs=-mv2/r=-s mg s =v2/rg=(26)2/92 x 9.8= = 0.75 Forces along the plane mgsin-k mgcos=0 sin- k cos =0 tan = k = tan-1(k)=tan-1(.20) = 11.3°

  8. Q19.: A box of mass 40.0 kg is pushed across a rough flat floor at the constant speed of 1.50 m/s. When the force is removed, the box slides a further distance of 1.20 m before coming to rest. Calculate the friction force acting on the box when it slides.(Ans: 37.5 N) CH-6-071 • Q20.A 10.0 kg box is pushed up an incline (= 30.0°) by a horizontal force of 298 N. The box then moves at a constant velocity as shown in Fig. 7. What is the frictional force on the box? ( Ans: 209 N) Decelerating force is force of friction fk Vi= 1.50 m/s; vf=0.0 m/s; a=-fk/m=-k mg/m=- kg a=(vf2-vi2)/2x =(0-1.52)/2*1.2=-0.94 fk  = kmg =ma = 0.94 x 40 = 37.6 N force of friction= fk= kN To calculate fk calculte forces  to plane mgsin  -Fcos-fk=0 fk= mgsin  -Fcos =10 x 9.8 xsin30-298xcos30 fk = -209.1 N

  9. CH-6-072 • T072: Q15.A constant horizontal force of 36 N is acting on a block of mass 4.0 kg, another block of mass 2.0 kg sits on the 4.0 kg block. The 4.0 kg block moves on a frictionless horizontal floor. Find the magnitude of the frictional force maintaining the 2.0 kg block in its position above the 4.0 kg block during the motion.(Ans: 12 N) Q18.In Fig. 4, a boy is dragging a box (mass =8.0 kg) attached to a string. The box is moving horizontally with an acceleration a = 2.0 m/s2. If the frictional force is 12 N, calculate the applied force F at an angle θ=60° (Ans: 56 N) Net force Fnet along x-axis Fx=Fcos60-fk=ma But fk=12 N F=(fk+ma)/cos60=(12+8x2)/cos60 F=56 N a =F/(m1+m2)= 36/(4+2)=6 m/s2 fs=m2a=2x6=12 N

  10. CH-6-072 • Q20. A 1000 kg airplane moves in straight horizontal flight at constant speed. The force of air resistance is 1800 N. The net force on the plane is: (Ans: zero) • Q19.At what angle should the circular roadway of 50 m radius, be banked to allow cars to round the curve without slipping at 12 m/s? (Ignore friction)(Ans: 16°) Constant speed I.e. Fnet is Zero Road banking angle=   =tan-1(v2/rg)= =tan-1(122/50 x 9.8) =16.4°

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