54. From the figure, cos $ = AC/AB

implies AC = AB cos $ implies (0.5) × (0.8) = 0.4.

So, CD = (0.5) – (0.4) = (0.1) m

Energy at D = energy at B

1/2 mv² = mg (CD)

v² = 2 × 10 × (0.1) = 2

So, the tension is given by,

T = mv2/r + mg = 0.1*(2/0.5 + 10) = 1.4 N

53. a) Let the velocity at B be v₂

1/2 mv₁² = 1/2 mv₂² + mgL

implies 1/2 m (10 gL) = 1/2 mv₂² + mgL

V₂²=8gL

So, the tension in the string at horizontal position T=MV²/R = 8mgL/L = 8mg   (B) Let the velocity at C be V₃   1/2 mv₁² = 1/2 mv₃² + mg(2L)  implies  1/2m(10*g*L) = 1/2 mv₃² + mg(2L)  implies V₃² = 6mgL .So, the tension in the string is given by T_c = mv²/L - mg = 6gLm/L * mg = 5mg  (C) Let the velocity at point D be v₄ 1/2 mv₁² = 1/2 mv₄² + mgh  1/2 * m * (10 gL) = 1.2 mv₄² + mgL (1 + cos 60°) impliesV₄² = 7gL  So, the tension in the string is T = (mv²/L) – mg cos 60°  implies  m*7gL/L– l – 0.5 mg   implies 7 mg – 0.5 mg = 6.5 mg.

18) Given, v = 36km/hr = 10m/s, r = 20m, (nu) = 0.4 , road is banked with an angle$=tan^-1(v²/rg) = tan^-1(100/20*10) = tan$= 0.5  When the car travels at max. speed so that it slips upward, (nu)R₁ acts downward So, R₁ – mg cos$ –mv₁²sin$/r = 0 ....(1)   And (nu)R₁ + mg sin$ –mv₁²cos$/r = 0  ...(2)  Solving  the  equations ,V₁ =[root rg*tan$ – (nu)/1+(nu)tan$]  = root[20*10*(0.1/0.2)] = 4.08 m/s = 14.7 km/hr  So, the possible speeds are between 14.7 km/hr and 54km/hr.

17) a) Net force on the spring balance.

R = mg – mw²r                  (w=omega)

So, fraction less than the true weight (3mg) is  mg - (mg - mw²r)/mg = w²/g implies [(2pi/24*3600)]² * [6400*10³]/10 =3.5 * 10^–3  (b)  When the balance reading is half the true weight, mg - (mg - mw²r)/mg=1/2 therfore w =(rootg/2r) = root(10/2*6400*10³) rad/sec  therefore duration of day is T=2pi/w  implies 2pi*(root2*6400*10³/9.8) sec implies 2pi*8000/7*3600 hr which is equal to 2hrs.

6 (a)  Stress in lower rod= T₁/A₁ implies m₁g+wg/A₁ implies W = 14kg Stress in upper rod =T₂/A₂ implies m₂g+m₁g+wg/A₂ implies W = 18kg . For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break first.

b) T₁/A₁ implies m₁g+wg/A₁ = 8 * 10⁸ implies W =14 kg and T₂/A₂ implies m₂g+m₁g+wg/A₂ = 8 * 10⁸ implies W =2kg. The maximum load that can be put is 2 kg. Upper wire will break first if load is increased.?

5) [(delta L)/L)]_st = F/AY_st and [(delta L)/L)]_{cu =}  F/AY_{cu ,} strain in steel wire/strain in copper wire =  (F/AY_st)* AY_cu/F as A_cu =A_st  therefore  Y_cu/Y_st implies 1.3*10¹¹/2*10¹¹  =0.65 

12) From the free body diagram

(nu)₁ R + 1 – 16 = 0

implies (nu)₁ (2g) –15 = 0

 (nu)₁ = 15/20 = 0.75

 (nu)₂ R₁ + 4 × 0.5 + 16 – 4g sin 30° = 0

 (nu)₂ [20 (root3)] + 2 + 16 – 20 = 0 (nu)₂=0.06 therfore Co-efficient of friction (nu)₁ = 0.75 & (nu)₂ = 0.06 From the free body diagram T + 15a – 15g = 0 implies T = 15g – 15 a ...(i)    T-(T₁ + 5a+ (nu)R)= 0 implies T = 5g + 10a + (nu)R …(ii)? implies T₁=5g + 5a …(iii) , From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g) implies a= 3.6m/s²  Equation (ii)implies T = 5 *10 + 10 * 3.6 + 0.2 * 5 * 10

implies 96N in the left string

Equation (iii) T₁ = 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string

11) From the free body diagram

T + 0.5a – 0.5 g = 0 ...(1)

(nu)R + a + T₁ – T = 0 ...(2)

(nu)R + a – T₁ = 0

(nu)R + a = T₁ ...(3)

From (2) & (3) implies (nu)R + a = T – T₁

therefore T – T₁ = T₁

implies T = 2T₁

Equation (2) becomes (nu)R + a + T₁ – 2T₁ = 0

implies (nu)R + a – T₁ = 0

implies T₁ = (nu)R + a = 0.2g + a ...(4)

Equation (1) becomes 2T₁ + 0/5a – 0.5g = 0 implies T₁ = = 0.25g – 0.25a ...(5) From (4) & (5) 0.2g + a = 0.25g – 0.25a implies a =0.4 m/s² a) Accln of 1kg blocks each is 0.4m/s²

b) Tension T₁ = 0.2g + a + 0.4 = 2.4N

c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N

28) Let the block m₁ moves upward with acceleration a, and the two blocks m₂ an m₃ have relative

acceleration a₂ due to the difference of weight between them. So, the actual acceleration at the blocks

m₁, m₂ and m₃ will be a₁.

(a₁ – a₂) and (a₁ + a₂) as from f.b.d

T = g – a₂ = 0 ...(i)

T/2 – 2g – 2(a₁ – a₂) = 0 ...(ii)

T/2 – 3g – 3(a₁ + a₂) = 0 ...(iii)

From eqn (i) and eqn (ii), eliminating T we get, g + a₂ = 4g + 4(a₁ + a₂) implies 5a₂ – 4a₁ = 3g (iv)

From eqn (ii) and eqn (iii), we get 2g + 2(a₁ – a₂) = 3g – 3(a₁ – a₂) implies 5a₁ + a₂ = (v) Solving (iv) and (v) a₁ =2g/29 and a₂=g – 5a₁implies a₂=19g/29 therefore a₁ - a₂=  -17g/29 , a1+a₂=21g/29 ,therfore acceleration of m₁, m₂, m₃ are 19g/29(up) , 17g/29(down) , 21g/29(down) , Again, for m_1, u = 0, s= 20cm=0.2m and a₂ =19g/29 as s=ut+0.5at²  0.2=0.5*(19/29)gt²  implies t= 0.25 sec

16) Max frictional force = (nu)R

From the free body diagram

R – mg cos$ =0

 R = mg cos$ ...(i)

and (nu)R + ma – mg sin$ ) = 0 …(ii)

implies (nu)mg cos$ + ma – mg sin$ = 0

implies (nu)g cos$ + a – 10 * (1/2) = 0

implies ?a = 5 – {1 – (2 3 )} * 10 ( (root3) / 2) = 2.5 m/s²

S = 12.8m, u = 6m/s

SO, velocity at the end of incline

V = root(u²+? 2as) = root[6² + 2(2*5)(12*8)] = 10m/s = 36km/h

25) (a)R₁ + ma – mg =0

implies R₁ = m(g–a) = mg – ma ...(i)

Again, F – T – (nu) R₁ =0 implies F – {(nu)(mg –ma)} – u(mg – ma) = 0

 F – (nu) mg + (nu) ma – (nu) mg + (nu) ma = 0

 F = 2 (nu) mg – 2(nu) ma implies F = 2(nu) m(g–a)

b) Acceleration of the block be a₁ therfore R₁ = mg – ma ...(i) 2F – T – (nu)R₁ – ma₁ =0 implies 2F – t – (nu)mg + (nu)a – ma₁ = 0 ...(ii) T – (nu)R₁ – M a₁ = 0? implies T = (nu)(mg – ma) + Ma₁ Subtracting values of F & T we have 2(2(nu)m(g – a)) – 2((nu)mg – (nu)ma + Ma₁) – (nu)mg + (nu) ma – (nu) a₁ = 0 implies a₁=2(nu)m(g - a)/M+m

27)  Because the block slips on the table, maximum frictional force acts on it.

       From the free body diagram

       R = mg

      therefore F – (nu) R = 0 imp;ies F = (nu)R = (nu) mg

      But the table is at rest. So, frictional force at the legs of the table is not ? R1. Let be

      f, so form the free body diagram.

      f_o – (nu) R = 0 implies f_o = (nu)R = (nu) mg.

     Total frictional force on table by floor is (nu) mg

34 a) 5a + T – 5g = 0 implies T = 5g – 5a ...(i) (From FBD)

Again (1/2) – 4g – 8a = 0  implies T = 8g – 16a ...(ii) (from FBD)

From equn (i) and (ii), we get

5g – 5a = 8g + 16a  implies 21a = –3g  implies a = – 1/7g

So, acceleration of 5 kg mass is g/7 upward and that of 4 kg mass

is 2a = 2g/7 (downward).

b)4a – t/2 = 0 implies 8a – T = 0 implies T = 8a … (ii) [From FBD]

Again, T + 5a – 5g = 0 implies 8a + 5a – 5g = 0

implies 13a – 5g = 0 implies a = 5g/13 downward. (from FBD)

Acceleration of mass (A) kg is 2a = 10/13 (g) & 5kg (B) is 5g/13.

c)T + 1a – 1g = 0 implies T = 1g – 1a …(i) [From FBD]

Again,T/2– 2g – 4a = 0 implies T – 4g – 8a = 0 …(ii) [From FBD]

implies1g – 1a – 4g – 8a = 0 [From (i)]implies a = –(g/3) downward.

Acceleration of mass 1kg(b) is g/3 (up)

Acceleration of mass 2kg(A) is 2g/3 (downward)

35) m₁ = 100g = 0.1kg

m₂ = 500g = 0.5kg

m₃ = 50g = 0.05kg.

T + 0.5a – 0.5g = 0 ...(i)

T₁ – 0.5a – 0.05g = 0 ...(ii)

T₁ + 0.1a – T + 0.05g = 0 ...(iii)

From equn (ii) T₁ = 0.05g + 0.05a ...(iv)

From equn (i) T₁ = 0.5g – 0.5a ...(v)

Equn (iii) becomes T₁ + 0.1a – T + 0.05g = 0

implies 0.05g + 0.05a + 0.1a – 0.5g + 0.5a + 0.05g = 0 [From (iv)

and (v)]implies 0.65a = 0.4g implies a =8/13g downward

37)Suppose the monkey accelerates upward with acceleration ’a’ & the block, accelerate downward with

acceleration a₁. Let Force exerted by monkey is equal to ‘T’

From the free body diagram of monkey

therefore T – mg – ma = 0 ...(i)

implies T = mg + ma.

Again, from the FBD of the block,

T = ma1 – mg = 0.

implies mg + ma + ma₁ – mg = 0 [From (i)] implies ma = –ma₁ implies a = a₁.

Acceleration ‘–a’ downward i.e. ‘a’ upward.

TherforeThe block & the monkey move in the same direction with equal acceleration.

If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they

have same weight (same mass). Their separation will not change as time passes

38)Suppose A move upward with acceleration a, such that in the tail of A maximum tension 30N produced T – 5g – 30 -5a = 0 ...(i) 30 – 2g – 2a = 0 ...(ii)

implies T = 50 + 30 +(5 × 5) = 105 N (max) implies 30 – 20 – 2a = 0 implies a = 5 m/s²

So, A can apply a maximum force of 105 N in the rope to carry the monkey B with it.For minimum force there is no acceleration of monkey ‘A’ and B. implies a = 0

Now equation (ii) is T'₁ – 2g = 0 implies T'₁ = 20 N (wt. of monkey B)

Equation (i) is T – 5g – 20 = 0 [As T'₁ = 20 N]

implies T = 5g + 20 = 50 + 20 = 70 N.

therfore the monkey A should apply force between 70 N and 105 N to carry the monkey B with it.