Question

3. Useful formulas: fx = MxN a. Find frictional force and acceleration in the case given below. F = 20 N 4 kg Hs = 0.6,HR = 0.4 b. Suppose an external force F = 10/2 Nis applied on a block of mass 4 kg in a direction that makes an angle of 45° with the horizontal. Find N.fr, a if coefficient of kinetic friction, 4x = 0.3. (8)
Find frictional force

Answer 1

QUESTION 3 (PART a)

7 Drawing the given Scenario Solving the problem. • given, mass= 4 kg on N F -20N lls = 0.6 lk = 0.4 F-20N E surface Wong 99-21m/s2 - Considesing the free body diagram of mass N= Normal reaction of surface W = mg = weight of the mass • Assuming that no motion i Occuring in direction perpendicular to Surface» (Fnet) =ON= N- Ls This says that the forces would be balonu'ng each other N=w-mq4x9-8 N = 39-2 N . Now, :. So from the laws of limiting friction, i) The static frictional force = fs = eis N= 0.66 39-2 = 23.52N ii) The Kinetic frictional force fx= UKN = 0.4% 34.2 = 15-68 N for the motion along x azis along the direction of applied force (F=209), F must over come the static iction. But 23.52 20 = 2352 > 20 - fs >F Then the body wouldn't be moving at au. > So, frictional foru - f = fs = 23-52N aceleration - ä m/s (no motion)

QUESTION 3 (PART b)

→ Again drawing the given scenario & solving the problem. • Resolving the force F = 10V2N in component along % 84 axi. AFY AN Aza - fx m 450 uso TL Surfau. - given, thi llx=0.3 N=? fx = ? a=7 • The component of Fare + i) Fz= Fcos 45º = 1oVzx} = lon Fu = Fsin 45º = 1012 x1 = ION x direction · Again, assuming that the motion U Occuring along only, implying that the mass stou's on the floor So, Wo motion along ý asis Hence, (Fretly = ON = (N+Fy) -W=0 (Batanud forces) N=-fytW = -107 469-8)= -10 +39-2 = 29.2N So, Kinetic fora of friction = fx= 0.38N = UxN fx = 0-3x 29.2 = 8.76N . axu = a • Now, acceleration along a 80, wing Newton's Second law of motion.

(Frnet), = man = ma= Fa- fx = 10- 8.76 a 10-8.76 = 1.24 - 0-31 m/s? m4s So, answers are N= 24-2 N fx= 8.760 first time na maji Cenging Q0.31 = 3-1x16' m/s? (along s)