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Friction Solutions of MD Dayal

Solutions on Friction chapter of MD Dayal

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Sandeep Pillai
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13K views34 pages

Friction Solutions of MD Dayal

Solutions on Friction chapter of MD Dayal

Uploaded by

Sandeep Pillai
Copyright
© © All Rights Reserved
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Solutions: Chapter 4 Friction Exercise 4.1 Blocks P1. For the following cases find force P needed to just impend the motion of the block. Take weight of block to be 100 N and coefficient of static friction at the contact surface to be 0.4. Pp Pp P (a) () (e) (d) = - 100.N Impendi Solution: Case (a) COE - Block 1 npenang Fy =0 a > N-100=0 Pp th. N=100N IF, =0 N P-0.4N=0 oe P-0.4x100=0 Or P=40N Case (b) COE - Block rR, =0 Pcos30-0.4N=0 Impending 30 _ Motion IF, =0 Psin30+N-100=0 .. Solving equations (1) and (2) we get r= N= 81.24.N and = P=37.52N . Ans. 67 Solutions Engineering Mechanics 68 Case (c) COE - Block 1008 EF, =0 IR -Pcos30+0.4N=0 - Impending lew atalen et Panag EF, =0 y -Psin30+N-100=0 . (2) h, van” j N Solving equations (1) and (2), we get N= 130.N and P=60N . Ans. Case (d) COE- Block ZF, =0 N-100cos30=0 N= 86.6N IF, =0 P-0.4N-100sin30=0 P-0.4x86.6-50=0 P2. A block weighing 800 N has to rest on an incline of 30°, If the angle of limiting friction is 18°, Find the least and greatest force that need to be applied on the block, parallel to the plane so as to keep the block in equilibrium. Solution: Given angle of limiting friction = 18° Knowing tang=} tanl8= Or = 0.3249 Case (1) Feast When least force P is appli move down. COE - Block IF, =0 N~800c0s30 =0 + N#692.8N ied to keep the block in equilibrium, the block impends to EF, =0 P+0,3249N-800sin30=0 P +0.3249x 692.8400 =0 Pyeast =174.9 N Case (2) Pyreatest When greatest force P is applied for keeping the block in equilibrium, the block impends to move up the incline. COE - Block IF, =0 P-800sin30-0.3249N=0 P-800sin 30 -0.3249x 692.8 =0 Pyreatest = 625.1 N P3. A block of weight 800 N is acted upon by a horizontal force P as shown in figure. If the coefficient of friction between the — block = and__—iincline~ are hy =0.35 and fl, =0.25, determine the value of P for impending motion up the plane. (MU Dec 15) a Solution: To impend the motion up the plane, the frictional force i.e 1gN will act down the inclined plane. COE - Block LFy =0 Ncos 25 - 0.35Nsin 25-800 =0 N=1054.9N =F, =0 P-Nsin25-0.35Ncos 25 =0 P -1054.9sin 25 - 0.35 x 1054.9xcos 25 =0 P=780.4N Ans. P4. A support block is acted upon by two forces as shown. Knowing #. = 0.35 and px = 0.25, determine the force P required @. to start the block moving up the incline. b. to keep it moving up. ¢. to prevent it from sliding down. Solution: a) To start the block moving up the incline. In this situation, the max. static frictional force i.e. yl’ will act down the incline. COE - Block IF, =O Ncos 25 — 0.35Nsin25 - 800 =0 N=1054.9N 70 IF, =0 Nsin25 +.0.35Ncos25-P=0 1054.9sin35 +.0.35x1054.9xcos25-P=0 P= 780.4 N .. Ans. b) To keep it moving up the incline. Solutions Engineering Mechang In this situation, the kinetic frictional force i.e. 1,.N_ will act down the incline. COE - Block ZF =0 Neos 25 - 0.25Nsin 25 - 800 =0 N=999.18N IF, =0 -P +0.25Ncos 25 +Nsin25=0 -P +0.25x999.18xcos 25 +999.18sin25=0 P=648.6N w. Ans. ¢) To prevent it from sliding down In this situation, the block impends to move down the incline and hence max. stati frictional force jgN will act up the incline. COE - Block ZF, =0 Noos 25 + 0.35Nsin25 - 800 = 0 N=758.85 N Nsin 25 -0.35Ncos25-P=0 758.85 sin 25 - 0.35 x 758.85cos 25 -P =0 P=80N .. Ans. PS. A block of weight 200 N rests on a horizontal surface. The co-efficient of friction between the block and the horizontal surface is 0.4. Find the frictional force acting on the block if a horizontal force of 40 N is applied to the block. (M4. U. Dec 09) ee Solution: Let F be the frictional force acting at the horizontal rough surface, required t¢ keep the block in equilibrium. Solutions : Friction 7 COE - Block 200N ZR, =0 40N N-200=0 or N=200N [ —- Nl IF, =0 [FB] 40- or F=40N (s5) F = 40 N is required frictional force to keep the block in equilibrium. Now, the max. frictional force the surface can produce = Finax =HsN : Fax =9-4*200 or Fax =80 N Since Frequired +ve -0.333N, +P¢os30 - 0.25N =0 ~0.333N, +Pcos 30 - 0.25x1000=0 =0.333N, + Pos 30 = 250 (2) F BO of block & Solving equations (1) and (2) we get N,=2395N and P=1210.4N Ans. P16. Find force required to pull block B as shown. Coefficient of friction between A and B is 0.3 and between B and floor is 0.25. Mass of A = 40 kg and B = 60 kg. (MU Dec 15) Solution: This is a system of two connected blocks, resting over each other. Isolating the blocks as shown. 80 Solutions Engineering Mechania COE - Block A IF, =0 ++ve -Te0s30+0.3N2 =O sess (1) LF, =0 T+ve Tsin30 +N, -392.4=0 Solving equations (1) and (2) T=115.86 N and Np =334.47 N COE - Block B LF, =0 T+ve N, -N2 -588.6=0 * N, 334.47 - 588.6 =0 or Ny, = 923.07 N IF, =0 ++ve P-0.25N, -0.3N2 =0 * P-0,25x 923.07 -0.3x334.47=0 or = P=331.1N Ans. P17. Block A of mass 27 kg rests on block B of mass 36kg. Block A is restrained from moving by a horizontal rope to the wall at C. What force P, parallel to the plane inclined at 30° with the horizontal is necessary to start B down the plane? Assume 4 for all surfaces = 0.33. (VJTT May 08) Solution: This is system of two blocks resting on each other. Isolating the blocks as shown. The contact surface between the two blocks is rough. COE- Block A ZF, =0 Tcos 30 -0.33N, ~ 264.87 sin30=0 ... EF, =0 -Tsin30 + Np —-264.87c0s 30 =0 Solving equations (1) and (2) we get T=296.9N and N,=377.8N COE -Block B IF, =0 Ny —N2 -—353.16cos30=0 81 Solutions N, -377.8-353.16cos 30 =0 Or N, = 683.6 N EF, =0 ~P +0.33No +0.33N; -353.168in30=0 2. -P+0.33x 377.8 +0,33 x 683.6 - 353.16sin 30 =0 Or P=173.68N P18. What should be the value of ‘8' so that the motion of oD block A impends down the plane? Take ma = 40 kg and mp = 13.5 kg. = 1/3 for all surfaces. (Mf. U. Dec 11) “ee Solution: This is a system of two blocks resting on each other. Isolating the blocks as shown. COE - Block B LFy =0 Nz -132.4c08@=0 N2 =132.4cos® ZF, =0 T-132,4sin®-0.33N, =0 T-132.4sin6 - 0.33x132.4¢c0s0=0 T=132.4sin@ + 43.6900 0=0........ COE -Block A ZF, =0 N, —Nz -392.4c080=0 N, -132.4cos @-392.4c08@=0 524.8.cos® or N, IF, ~392.4sin@+0.33N, +0.33N2 =0 392.4 sin0+.0.33x524.8.c0s 0 + 0.33x132.4c0s6=0 392.4sin@ = 219.06cos6 tan@ = 0.5582 or = =29.17° 82 Solutions Engineering Mechanin P19. Block A has a mass of 25 kg and block B has a mass of 15 kg. Knowing #. = 0.2 for all surfaces, determine value of @ for which motion impends. Assume frictionless pulley. Solution: This is a system of two blocks connected to each other by a rope. Let us assume at a particular value of 8, block A impends to slide down, this initiates block B to slide up. Isolating the blocks COE = Block A EF, =0 T+0.2Ny ~-245.25sin@=0... + (1) EFy =0 Nz -245.25cos0=0 Or Np =245.25c0s0 ......... (2) Eliminating Nz we get ~ T +0.2(245.25 cos 6)~245.25sin8 =0 oie was et 7.5N Or T= 245.25sin0-49.05cos@ Xo ih sesseeee (A) COE - Block B ° T-0.2N, - 0.2Np -147.15sin@=0 .., (3) Substituting for No Nj -245.25c08 6-147.15cos8=0 Or NN, =392.4c0s0 sevnseee Eliminating N, and No from equation (3), we get T-0.2(392.4cos6) —0.2(245.25cos0)-147.45sin6=0 Or = T=127.53cos@+147.15sin@ .. (B) IF, =0 ‘Ny, -Nz -147.15cos8=0 + (4) Equating equations (A) and (B), we get 245.25 sin@ — 49.05cos@ = 127.53cos@+147.15sin0 0 =60.94° Solutions : Friction P20. A block A of mass 10 kg rests on a rough inclined plane as shown, This block is connected to another block B of mass 30 kg resting on a rough horizontal plane by a rigid bar inclined at an angle of 30°. Find the horizontal P required to be applied to block B to just move block A in the upward direction, Take 1 = 0.25 for all contact surfaces, (SPCE Nov 12) Solution: This is a system of two blocks connected by a rod. As the block B is pushed to the right, the block A slides up the slope. Isolating the blocks by cutting the rod. On doing so, the force F in the rod is exposed as shown. Let us assume the force F in the rod is tensile. ~Feos 30 — Np cos 45 — 0.25N2 cos45 = 0 =. -Fcos30~0.884Nz =0 DF, =0 T+ve —Fsin 30 + No sin 45 - 0.25N2 sin45-98,1=0 «. -Fsin30+0.53Nq = 98-1 (2) Solving equations (1) and (2), we get N)=94.26N and F=-96.22N or F = 96.22 N (Compressive) COE - Block B IF, =0 Ttve N, +Fsin30 - 294.3 =0 * N, +(-96.22)sin30 - 294.3 = 0 or N, =342.4N or P21. Find the maximum value of Ws for the rod AB to remain horizontal. Also find the corresponding axial force in the rod. Take y = 0.2 for all contact surfaces. Solution: This is a system of two blocks connected by a rod. 1) IF =0 4 +ve P-0.25N, + Fcos 30 =0 P-0.25x342.4 +(-94.26)cos 30=0 P=167.2N Wy 20 kN Solutions Engineering Mechanis 84 weight of block dition that Ble Weis, keeping the rod Bie, Wn is maximum keepi ‘AB horizontal, the block B impends to slide down, causing impending motion of block A up the slope. isolating the blocks by cutting the rod. On doing so, the force F ; hs assumed tensile). in the rod is 1 eed ae chown; Fanaraea] COE - Block A EF, =0 wv Nj, cos30 -0.2N, sin30-20=0 os N, =26.11 IF, =0 F +Nj sin30 +0.2N; cos30 =O F+26.11sin30+0.2x 26.1 1cos 30 =0 F =-17.576 kN Or F#=17.576 kN (Comp.) COE -Block B IF, = -F-Ny sin45+0.2N,cos45=0 -(-17.576)-Ng sin45 + 0.2xN2 cos 45 =0 Nz =31.07 KN ZF =0 Wg + 0.2Ny sin45 +N cos45=0 “Wp +0.2x31.07 sin 45 +31.07c0s 45 =0 Wp = 26.36 kN... P22. Two blocks connected by a horizontal link AB are supported on two rough planes, p: between block A and horizontal surface is 0.4. The limiting angle of friction between block B and inclined plane is 20°. What is the smallest weight W of the block A for which equilibrium of the system can exist, if the weight of block B is 5 kN? (WJTI Mar 11) Solution: This is a system of two blocks connected by a rod. For the condition that weight of block A Wa is least, the block B impends to slide down, causing impending motion of block A to the left. itnensing [. == Wa Isolating the blocks by cutting the rod. On doing so, the force Fin the rod is exposed as shown. Let us assume the force F in the rod is tensile in nature. . — nl oan Solutions : Friction 85 Knowing lang=-. tan20=q Or 41=0.3639 ... for Block B COE -Block B ZF, =0 N2sin30 +0.3639N,cos30-5=0 .. Ny =6.194 kN LF =0 -F - No cos 30 + 0.3639N, sin30=0 -F - 6.134c0s 30 +0.3639x 6.134sin30=0 =-4.196 kN Or — F=4.196 kN (compressive) .... COE - Block A LF, =0 LR, =0 F+0.4N, =0 N, -Wa =0 4.196 +0.4N, =0 10.49-W, =0 Or —-N, =10.49 kN = Wy =10.49 KN P23. For the figure shown mass of block A is 200 kg Linkages are smooth. Rod BC is horizontal. The coefficient of friction between block A and plane is 0.2. Calculate momentum to just start the motion of block A up the plane. (EJS May 15) Solution: This is a system of connected bodies consisting of block A and Red AB and BC. Isolating Block A by cutting rod AB. Let F be the force in rod AB. COE - Block LF, =0 Feos 30- 0.2N -19628in30=0 ... (1) Ly =0 Fein30+N-1962cos30=0__ ... (2) Solving equations (1) and (2), we get F = 1367.3N and N=1015.5N COE - Rod BC EMc=0 VPtve +(Fsin60x2)-M=0 +(1367.3sin 60x 2)-M=0 M = 2368.2 Nm 86 Solutions Engineering Me, Exercise 4.2 Wedges P1. The horizontal position of the 1000 kg block is adjusted by 6° wedge. If coefficient of friction for all surfaces is 0.6, LS determine the least value of Kee] 1000 force P required to move the |_| block. (NMIMS June 07) Solution: In this wedge problem we need to find force P required to just shift the blog to the right. Isolating the wedge and block as shown. - Bl COE ~ Block Pp f,. 1000x981 EF, =0 pete .E 06 No -0.6N, =0 ‘1 ‘ EF, =0 wh -0.6Ng +N; -9810 ¥ Solving equations (1) and (2) we get inpenng. Ng =9196.9N and N,=15328N [FaDotviedge] COE - Wedge LF, =0 ZFy =0 -Nz +N3 cos6-0.6N3 sin6=0 -9196.9+0.9318 Ny =0 —P +0.6Np +Ng sin 6 +0,6N3 cos6=0 —-P+0.6x9196.9+9870sin6 +0.6x9870cos6=0 P=12439N Ans. Ng =9870N P2. A block weighing 1000 N is raised against a surface inclined at 60° to the horizontal by means of a 15° wedge as shown in figure. Find the horizontal force P which will just start the block to move if the coefficient of friction between all the surfaces of contact be 0.2. Assume the wedge to be of negligible weight. Solution: In this wedge problem we need to find force P requiréd to just lift the blo™ up. Isolating the wedge and block as shown. Solutions : Friction 87 COE - Block 4 impending EF, =90 O2N, -Np sin15 - 0.2Nz cos15 + Ng sin60 Block +0.2N3 cos 60=0 ~0.452Nq +0.966Nq =O ER, =0 Np.cos 15 -0.2Np sin15 +Ng cos 60 —0.2Ng sin60 - 1000 =0 0.9142Np + 0.3268N3 = 1000 Solving equations (1) and (2) we get Np =937.1N and Ng =438.5N COE - Wedge EFy =0 EF, =0 N, -Nqcos15 +0.2Nosin15=0 =P +0.2N; +No sin15 +0.2N cos15=0 N, -937.1c0s 15 + 0.2*937.1sin15 =O -P +0.2x856.7 +937. 1sin15 N, =856.7 N +0.2x937-1c0s15 P=594.9N P3. The vertical position of the 200 kg mass I section is being adjusted by two 15° wedges as shown. Find force P to just raise the mass. Take p = 0.2 for all surfaces. Solution: In this wedge problem we need to find force P required to just lift the 200 kg I section up. Isolating the wedges A and B as shown. COE - Wedge A IF, <0 Ng sin15 + 0.2N3 coa15—Nq =O 0.452Ng - Na =0 () ER, =0 Ng cos 15 ~0.2Ng sin 15 -0.2Nz -1962=0 0.9142Ng -0.2Ng =1962 (2) Solving equations (1) and (2) we get N3 =2381.6N and Nz =1076.5N 88 Solutions Engineering Mecho COE - Wedge B IF, =0 N, -N3 cos15 +0.2N, sin15=0 N, - 2381.6c0s15 +0.2x2381.6sin15=0 N, =2177.2N IF, =0 P-0.2N, -0.2Nq cos15—Nq sin15=0 P-0.2x2177.2- 0.2x2381.6c0s 15 ~2381.6sin15 =0 P=1512N / P4, Wedge A supports a load of W = 5000 N which is to be raised by forcing the wedge B under it. The angle of friction for all surfaces is 15°. Determine the necessary force P to initiate upward motion of the load. Neglect the weight of the wedges. Hint: Wedge A comes in contact with the right wall, as the load gets lifted. Solution: In this wedge problem we need to find force P required to just lift the 50008 load up. Isolating the wedges A and B as shown. Knowing tang=p -. tan1S=, or = 0.268 | COE - Wedge A LF, =0 . Ng sin 20 + 0.268Ng cos 20-Nz =0 0.5938N3 -N2 =0 2) IF =0 N3 cos 20 - 0.268Ng sin 20 ~ 0.268N2 - 5000 =0 0.848Ng - 0.268N2 = 5000 (2) Solving equations (1) and (2) we get Ng =7258.2N and Np =4310N COE - Wedge B LFy =0 N, Nj cos 20 + 0.268Ny sin20=0 Nj ~7258.2cos 20 + 0.268% 7258.2sin 20 =0 Ny =6155.2N 8g Solutions : Friction EF, =0 P-0.268N - 0.268N3 cos 20-N; sin20=0 P-0.268x 6155.2 -0.268x 7258.2cos 20 - 7258.2s8in20=0 P=5959.9N Ans. P5. Two blocks A and B are resting against the wall and floor as shown in figure. Find minimum value of P that will hold the system in equilibrium. p= 0.25 at the floor, u = 0.3 at the wall and p = 0.2 between the blocks. 5 (WTI Nov 09, M.U. Dec 12} ‘1000 2 Solution: In this wedge problem wedge A under is own yoo weight impends to move down pushing wedge B to the right. Force P prevents this and hold the system in equilibrium. COE - Wedge A LF, =0 No +0.2N3 cos 60 ~Ng sin 60 =0 2. Ng -0.766Ng =0 (1) IF, =0 0.3N2 +0.2Ng sin60 +Ng cos 60 -500 =0 0.3Nq +0.6732N3 =500 : (2) Solving equations (1) and (2) we get N2=424N and N3=553.7N COE - Wedge B EF, =0 Ng cos 60 - 0.213 sin60 +N, -1000=0 -553,7c0s 60 - 0.2% 553.7 sin60 +N, -1000=0 N, = 1372.7 N IF, = -P ~0.25N, +Ng sin60- 0.2N3 cos60=0 ~P -0.25x 1372.7 +553.7 sin 60 -0.2x 553.7cos 60 =0 . Ans. P=80.97N 90 Solutions Engineering Me, Exercise 4.3 Ladders Pl. A person of weight P = 600 N ascends the 5 m ladder of weight 400 N as shown, e How far up the ladder may the person climb before sliding motion of ladder takes place. ‘Smooth wall 4 Solution: For maximum distance ‘’ the person climes, the ladder impends to slip {\ down and move away from the wall. i= 03 COE ~ ladder EF, =0 T+ve N-400-600=0 N=1000 N IF, =0 ++tve Rg -0.3N=0 Ry — 0.31000 =0 Rg =300N IM, =0 +ve (Rp x5sin 60) + (400x2.5cos 60) + (600xxcos 60) =0 -(300x 5sin 60) + 500+ 300x =0 x=2.663 m on . Ans. P2. A 100 N uniform ladder AB is held in equilibrium as shown. If p = 0.15 at A and B, calculate the range of values of P for which equilibrium is maintained. Solution: Range of values of P implies we need to calculate minimum and maximum value of P required for equilibrium of ladder. Case (1) Pmisinum : When P is minimum, the ladder impends to slip down and away from the wall. COE - ladder EFy =0 T+ve Nj +0.15Nz-100=0 2% ~ (1) Solutions : Friction 91 IMg =0 U tve 0.1SN x20 - Ny X8+N2 x 20+0,15No x8 =0 -SNj +21.2Ny =0 ++ (2) Solving equations (1) and (2) we get, N,=96.58N and Nz =22.78N IF, =0 +t+ve 0.15N, ~No+P=0 0.15x96.58 - 22,78 +P =0 Pminimum = 8.293 N Case (2) Pmaximum : When P is maximum, the ladder impends to slip up and towards the wall. COE - ladder EF, =0 T+ve N, -0.15Ny -100=0 (3) EMg=0 UY #ve =N, X8—0.15N, x20 +Nz x20-0.15N2x8=0 -11N, +18.8N2 =0 (4) Solving equations (3) and (4) we get, N,=109.6N and Nz =64.14N IF, =0 ++ve P-0.15N, -Nz =0 P-0.15x109.6 -64.14=0 Praximum = 80.58 N 92 Solutions Engineering Mecha. P3. A ladder AB of length 3 m and mass 25 kg is resting against a vertical wall ay.) horizontal floor. The Indder makes an angle 50° with the floor. A man of mass 69 4 tries to climb the Indder. (a) How much distance along the ladder he will be abi climb if jt between ladder and floor is 0.2 and between ladder and wall is 0.3. “ {b) also find the angle the ladder should make with the horizontal such that the | climb till the top of the ladder. (1.0 Dee tgs Solution: (a) Let x be the maximum distance the 93n,, person of mass 60 kg = 588.6 N weight climbs, before causing the ladder to slip down and away Nz from the wall. COE - ladder LF, =0 + +ve -0.2N; +No =0 2a EFy =0 T+ve Ny + 0.3N2 ~588.6-245.25=0 ... FED - Person at x from bottom Solving equations (1) and (2), we get | N,=786.6N and Ng=157.3.N K | EIM,=0 U +ve —(Nz x 3sin 50) - (0.3N2 x3 cos 50) + (588.6 x xcos 50) + (245.25 x 1.Scos50)=0 | Substituting Nz = 157.3 N, we get x=0.571m . Ans. b) Let the person reach the top of ladder kept at minimum angle @. The impending motion of ladder is 9 Net away from the wall. i COE - ladder IM, =0 U +ve -(No x3sin®) -(0.3N2 x 3cos 6) 4+ (588.6x3cos 0) + (245.25x1.5cos@) =0 Substituting No =157.3 N, we get -(157.3x3sin 6) — (0.3x157.3x3cos@) FBD- +(588.6x 3cos 0) + (245.25x1.5cos)=0 -471.98in0+1992cos0=0 tano=4.22 or «0 = 76.67° Ans. | Solutions : Friction 93 P4. A 150 N uniform ladder 4 m long supports a 500 N weight person at its top. Assuming the wall to be smooth, find the minimum coefficient of friction which is required at the bottom rough surface to prevent the ladder from slipping. Solution: For minimum value of y, the ladder impends to slip down and away from the wall. The frictional force at A is uN. COE - ladder LF, =0 T+ve N-150-S00=0 N=650N EM, =0 U +e —(150x2cos 60) - (500x 4c0s 60) + (Rg x4sin60)=0 Rg =332 N IF, =0 + tve BN-Rp =0 px 650-332=0 H=0.5108 —.., PS. The ladder shown is 6m long and is supported by a horizontal floor and a vertical wall. 1 between floor and ladder is 0.4 and between wall and ladder is 0.25. The weight of ladder is 200 N, The ladder also supports a vertical load of 900 N at C. Determine the greatest value of @ for which the ladder may be placed without slipping. (VJTI Nov 10) Solution: For greatest value of 0, the ladder impends to slip down and away from the wall. COE - ladder LF, =0 ++Vve -0.4N, +N2 =0 EF, =0 T+ve N, + 0.25Nq -900-200=0 Solving equation (1) and (2), we get Nj=1000N and N2=400N 94 Solutions Engineering Mecha IMp=0 U +ve —(900x sin 0) -(200x3sin@) -(0.4N, x 6cos 6) +(N, x6sing) =0 -900sin 0 - 600sin 0- 2.4x 1000xcos@ + 1000x6sin@=0 4500 sin® = 2400cos8 5333, .. Ans. P6. A 6.5 m ladder AB of mass 10kg leans against a wall as shown. Assuming that the coefficient of static friction j1 is the same at both surface of contact, determine the smallest value of yi for which equilibrium can be maintained. (SPCE Dec 10) Solution: For smallest value of p, the ladder impends to slide down and away from the wall. COE - ladder A IF, =0 + +ve WN, -N2 =0 or No =HNy LF, =0 T+ve Ny +HN2 -98.1=0 N, +1(uN,)-98.1=0 or Ny +p?N, -98.1=0 Mg =0 UF +ve +(HNy x6) -(N; x2.5) +(98.1x1.25) Ny (61 -2.5) =-122.625 122.625 +p? x Gu -2.5 -122.625 —122.625y7 = 588.61 - 245.25 122.625? + 588.6" - 122.625 =0 Solving the quadratic equation, we get =0.2 ceeeeees ANB, Solutions : Friction 95 7. A ladder shown is 4 m long and is supported by a horizontal floor and vertical wall. The coefficient of friction at the wall is 0.25 and that at the floor is 0.5. The weight of ladder is 30 N. It also supports a vertical load of 150 N at 'C’. Determine the least value of ‘a’ at which the ladder may be placed without slipping to the left. (VJTI Nov 09) Solution: For the least value of a, the ladder impends to slip down and away from the wall. COE - ladder IF, =0 ++ve 0.5N, -N2 =0 a) ER, =0 T+ve N, +0.25Nz -30-150=0 N, +0.25No =180 Solving equation (1) and (2), we get Nj =160N and Ng=80N EM, =0 U +ve ~(30x 2c0s «) - (150 3cos a) + (0.25N2 x4cosa)+(N2 x4sina)=0 -60 cosa - 450cos a + 0.25x80x4cos¢+80x4sina =0 -430cosa+320sina=0 tana = 1.3437 or @=53.34° Ans. 96 Solutions Engineering Mech, P8. A uniform ladder of length 2.6m and weight 240 N is placed against a sm, vertical wall at A, with its lower end 1 m from the wall on the floor at B. The coef” of friction between the ladder and the floor is 0.3. Find the frictional force actin : ladder at the point of contact between the ladder and the floor. Will the ladder r, equilibrium in this position? Give reason. (VITI May 06) Solution: Let F be the frictional force acting at the floor, required to keep the ladder in equilibrium. COE - ladder EF, =0 T+ve N-240=0 N=240N IM, =0 U +ve +(Fx2.4)—(Nx1)+(240x0.5)=0 2.4F -(240x1)+120=0 F=50N . Frictional force required for equilibrium Now the maximum frictional force the rough floor can produce is Fraax =Ha*N Finax =0-3%240 or Figg = 72.N Since F

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