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ROTATIONAL MECHANICS:--
EXERCISES-(Q31 to Q45)
31. A light rod of length 1 m is pivoted at its center and two masses of 5 kg and 2 kg are hung from the ends as shown in figure (10-E3). Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.
ANSWER: The moment of inertia of the weights about the pivot,
I = 2*0.5²+5*0.5² = 7*0.5² = 7*0.25 =1.75 kg-m²
Net torque about the pivot, T = 5g*0.5-2g*0.5
= 3g*0.5 =1.5g N-m
Hence the angular acceleration α = T/I = 1.5g/1.75 =6g/7
=6*9.8/7 =8.4 rad/s²
32. Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length.
(a) Find the initial angular acceleration of the rod.
(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.
ANSWER: (a) Since the mass of the rod is uniformly distributed over its length, it will make no difference in the torque produced about the pivot. Hence the torque T =1.5g N-m
But the moment of inertia of the system will change. The moment of inertia of the rod will have to be added. The moment of inertia of the rod about the pivot =ml²/12 = 1*1²/12 = 1/12 kg-m²
Hence total moment of inertia = 1.75 + 1/12 kg-m²
= 1.75+0.083 =1.833 kg-m²
Hence the initial angular acceleration α = T/I
=1.5*9.8/1.833 =8.0 rad/s²
(b) Linear acceleration at the end of the rod, a = αr =8.0*0.5 =4.0 m/s²
Consider the 2.0 kg mass. Weight 2g N is downwards, tension T is upwards and the acceleration a is upwards. Hence,
T-2g=2a
→T=2g+2a =2*9.8+2*4.0 =19.6+8 = 27.6 N
For 5 kg mass, acceleration a is downwrds, hence
5g-T =5a
→T=5g-5a =5*9.8-5*4 =49-20 =29 N
33. Figure (10-E4) shows two blocks of masses m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley.
ANSWER: Let the accelerations of blocks M and m be a and the tension in the string be T and T' respectively. For the block M, we have,
Mg-T =Ma
T = M(g-a)
And for the block m,
T' = ma
Net torque on the pulley = Tr-T'r =(T-T')r
If the angular acceleration of the pulley be α,
Then, a =αr → α = a/r
For the pulley, Torque =Iα
→(T-T')r = Ia/r
→M(g-a)-ma =Ia/r²
→Mg = Ma+ma+Ia/r²
→ a = Mg/(M+m+I/r²)
34. A string is wrapped on a wheel of moment of inertia 0.20 kg-m² and radius 10 cm and goes through a light pulley to support a block of mass 2.0 kg as shown in figure (10-E5). Find the acceleration of the block.
ANSWER: Let the acceleration of the block be a m/s² and the tension in the string be T Newton.
2g-T = 2a
→T = 2g-2a
Torque on the pully = T*0.10 N-m
=0.10T N-m
=0.10(2g-2a)
=0.20(g-a)
Moment of inertia of the pulley, I= 0.20 kg-m²
The angular acceleration α = Torque/M.I.
→a/r = 0.20(g-a)/0.20 =g-a
→a = gr-ar
→a+ar = gr
→a = gr/(1+r)
→a=9.8*0.10/(1+0.10)
→a = 0.98/1.10
→a = 98/110 = 0.89 m/s²
35. Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m². Find the tension in the part of the string joining the pulleys.
ANSWER: Let the required tension be T'.
Now the torque on the bigger pulley = T'*0.10 N-m
Angular acceleration a/r =0.10*T'/0.20 =T'/2
→a = rT'/2 =0.10*T'/2 = 0.05T' -------- (i)
Net torque on the smaller pulley = (T-T')r' =(T-T')*0.05 N-m
Angular acceleration a/r' = 0.05(T-T')/I'
0.05T'/0.05 =0.05(T-T')/0.10 {Value of a from (i)}
→T' =0.5T - 0.5T'
→1.5T' =0.5T ---------------------------(ii)
For the weight 2g -T = 2a
→T = 2g-2a = 2g-2*0.05T' = 2g-0.10T'
Putting this value in (ii), we get
1.5T' = 0.5 (2g-0.10T')
→1.5T' = 9.8-0.05T'
→1.55T' =9.8
→T' = 9.8/1.55 = 6.32 N
36. The pulleys in figure (10-E6) are identical, each having radius R and moment of inertia I. Find the acceleration of the block M.
ANSWER: Let the acceleration be a.
Angular acceleration of the pulleys = a/R
Let the tension in the string connected to M be T and in the string connected to m be T'. Also the tension in the string between the pulleys be T".
Now Mg-T =Ma
→T = M(g-a)
Also, T'-mg = ma
→T' = m(g+a)
For pulleys,
α = a/R = (T-T")R/I
→ a =(T-T")R²/I ------------ (i)
Also, a/R =(T"-T)R/I
→a = (T"-T')R²/I ----------- (ii)
Adding (i) and (ii)
2a = (T-T')R²/I = (Mg-Ma -mg-ma)R²/I
→2aI = (Mg-Ma -mg-ma)R² =(M-m)gR²-(M+m)aR²
→2aI+(M+m)aR² = (M-m)gR²
→a{2I+MR²+mR²} = (M-m)gR²
→a = (M-m)gR²/2I+MR²+mR²
→a = (M-m)g/M+m+2I/R²
{Dividing the numeretor and the denominetor by R²]
37. The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0.20 kg-m². The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.
ANSWER: Let the tension in the string on the plane be T and on the fixed side be T'.
T =ma = 1.0*a = a {where a is the acceleration of the block}
Acceleration of the pulley = a/2
Angular acceleration = a/2r = a/0.4 m/s²
The torque on the pulley = (T'-T)*0.20
I = 0.20 kg-m²
α = T/I
→ a/0.4 = (T'-T)*0.20/0.20 = T'-T
→ T-T' = -a/0.4 = -2.5a --------------------- (i)
Let the mass of the pulley be M
Mr²/2 = I
→M = 2*0.20/0.2*0.2 =10 kg
Now Mg -T-T' = Ma/2
→T+T' =Mg-Ma/2 =10*9.8 - 10a/2 =98 - 5a ----------(ii)
Adding (i) and (ii)
2T = 98 - 5a - 2.5a
→2a = 98 -7.5a
→9.5a = 98
→a = 98/9.5 = 10.31 m/s² ≈ 10 m/s²
38. The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.50 kg-m² about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.
ANSWER: Let the acceleration of the 4 kg block be a and the tension in the string T. If the tension in the 2 kg block be T' then we have,
4g.sin45° - T = 4a
→T = 4g/√2 -4a
And T'-2g.sin45° = 2a {Same acceleration for 2 kg block}
T' = 2a + 2g/√2
Net torque on the pulley = (T-T')*0.10
=0.10(4g/√2-4a-2a-2g/√2) =0.10(2g/√2-6a)
Angular acceleration α =a/r = a/0.10
From Torque = Iα
0.10(2g/√2-6a) = 0.50*a/0.10 =5a
→ 2g/√2-6a = 50a
→56a = 2g/√2 = 2*9.8*√2/2 =9.8√2
→a = 9.8*1.41/56 = 0.247 m/s² ≈ 0.25 m/s²
39. Solve the previous problem if the friction coefficient between 2.0 kg block and the plane below it is 0.50 and the plane below the 4.0 kg block is frictionless.
ANSWER: Given µ = 0.50
Hence the equation for 2 kg block is now
T' -2g.sin45° -µ(2g.cos45°) = 2a
→T' = 2a + 2g/√2+µ*2g/√2
→T' = 2a+√2g+√2gµ
Now the torque is = Tr-T'r (T-T')r
= (4g/√2-4a -2a-√2g-√2gµ)*0.10
= 0.10 (4*9.8/1.41-6a -1.41*9.8*1.5)
= 0.10(27.80-20.73-6a) = 0.10(7.07-6a)
Angular acceleration α = a/r
Since Torque = Iα
0.10(7.07-6a) = 0.50*a/0.10 =5a
→0.707-0.60a = 5a
→5.6a = 0.707
→a = 0.707/5.6 = 0.126 ≈ 0.125 m/s²
40. A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.
ANSWER: Let us draw a diagram as below:-
The weight of the stick may be taken as acting at the CoM of the rod, i.e. 50 cm from the left. W =mg = 200g/1000 =0.20g N.
The weight of the small object W' = 20g/1000 = 0.02g N, acting at 70 cm i.e. 0.70 m from the left end.
Let tension in the left string = T and in the right string = T'
Since the stick is in balance, adding the forces in the vertical direction, we get
T+T' = W+W' = 0.20g+ 0.02g = 0.22g ---------- (i)
Now the sum of the torques about end B,
T*1.0 -W*0.50-W'*0.30 = 0 {Since the stick has no rotation}
→T = 0.20g*0.50+0.02g*0.30
→T = 0.1g + 0.006g =0.106g = 0.106*9.8 = 1.04 N
From (i),
T' = 0.22g-1.04 =0.22*9.8-1.04 = 2.16 - 1.04 = 1.12 N
41. A uniform ladder of length 10.0 m and mass 16.0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the electrician to work Safely?
ANSWER: Let the normal force on the ladder at the lower end = N and at the upper end = N' as shown in the figure below.
The force of friction on the lower end = µN
The weight of ladder acts at the middle D = W = 16g N
The weight of the electrician P = 60g N
Since the rotation of the ladder is zero, hence the sum of torque about B = 0
µN*10cos37°-N*10sin37°+16g*5sin37°+60g*2sin37° = 0
→µN*40/5-N*30/5+80*9.8*3/5+120*9.8*3/5 =0
→8µN-6N+470.4+705.6 = 0
→8µN-6N+1176 = 0
→4µN-3N+588 = 0 ------------------(i)
The sum of forces in the vertical direction,
N-16g-60g = 0
→ N = 76g = 76*9.8 = 744.8 Newton ≈ 745 N
Hence the normal force on the lower end = 745 N
The force of friction on the lower end =µN = (3N-588)/4
{From (i)}
= (3*745-588)/4 = 411.75 ≈ 412 N
Since µN = 412
→µ = 412/N = 412/745 = 0.533
42. Suppose the friction coefficient between the ground and the ladder of the previous problem is 0.540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder.
ANSWER: Let the maximum weight of the mechanic be Mg.
Equating the forces in the vertical direction
mg+Mg = N
Equating the sum of the torques about B
µN*10cos37°-N*10sin37°+16g*5sin37°+Mg*2sin37°=0
→N(0.54*40/5-30/5)+16*9.8*5*3/5+M*9.8*2*3/5=0
→N(4.32-6)+470.4+11.76M = 0
→-1.68N +470.4+11.76M = 0
43. A 6.5 m long ladder rest against a vertical wall reaching a height of 6.0 m. A 60 kg man stands halfway up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.
ANSWER: Let us draw a diagram as below,
The distance of the lower end of the ladder from the wall, AO
=√(6.5²-6.0²) = 2.5 m
The torque of the of the force exerted by the man on the ladder about the upper end = 60g*2.5/2 =75g =75*9.8 = 735 N-m
Let the Normal force by the ground be R and the frictional force = F. Normal reaction by the wall = R'.
Equating forces in the vertical direction,
R = 60g = 60*9.8 = 588 N ≈ 590 N
Equating the torques about end B,
F*6+60g*2.5/2-R*2.5 =0
→6F =2.5*588-30*9.8*2.5
→6F = 735
→F = 735/6 = 122.5 N ≈ 120 N
44. The door of an almirah is 6 ft high, 1.5 ft wide and weighs 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.
ANSWER: Let the forces exerted by each of the hinges on the door be F in horizontal direction and R in the vertical direction.
Equating the torques about the lower hinge,
F*4/3.28 =8g*½*1.5/3.28 {Since 1 m =3.28 ft}
→F = 2g*½*1.5 =9.8*1.5 = 14.7 N
Equating the forces in the vertical direction,
2R-8g = 0
→R = 4g
→R = 4*9.8 = 39.2 N
Since F and R are mutually perpendicular, the magnitude of the resultant force by each of the hinges = √(F²+R²) =√(14.7²+39.2²)
=√1752.73 ≈ 42.0 N
45. A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is θ.
ANSWER: Let us draw a diagram showing forces on the rod. If m is mass of the rod then its weight mg acts at the middle point i.e. L/2 distance from the end. Let R be the normal reaction by the roller and P by the ground. If µ be the coefficient of the friction between the rod and the ground then a force of friction, F=µP, acts at the lower end of the rod to check the movement.
Since the rod has no linear motion, hence the sum of forces in horizontal and vertical directions will be zero.
In the horizontal direction,
F = R.sinθ
→µP = R.sinθ
→R =µP/sinθ ---------------------- (i)
In the vertical direction,
R.cosθ+P = mg
→µP.cosθ/sinθ + P = mg
→mg = P(µcosθ+sinθ)/sinθ ------- (ii)
Since the rod does not rotate, the sum of torques of all the forces will be zero. Hence taking moment about the lower end A, we get
R*h/sinθ-mg(L/2)cosθ = 0
→Rh-½mgL.cosθ.sinθ = 0
→2Rh = mgL.cosθ.sinθ
→2h*µP/sinθ = {P(µcosθ+sinθ)/sinθ}*L.cosθ.sinθ
[Putting the values of R and mg from (i) and (ii)]
→2hµ = (µcosθ+sinθ)*L.cosθ.sinθ
→2hµ = µcosθ*L.cosθ.sinθ + sinθ*L.cosθ.sinθ
→2hµ - µcosθ*L.cosθ.sinθ = L.cosθ.sin²θ
→ µ(2h - L.cos²θ.sinθ) = L.cosθ.sin²θ
→ µ = L.cosθ.sin²θ/(2h - L.cos²θ.sinθ)
 |
| Figure for Q-31 |
ANSWER: The moment of inertia of the weights about the pivot,
I = 2*0.5²+5*0.5² = 7*0.5² = 7*0.25 =1.75 kg-m²
Net torque about the pivot, T = 5g*0.5-2g*0.5
= 3g*0.5 =1.5g N-m
Hence the angular acceleration α = T/I = 1.5g/1.75 =6g/7
=6*9.8/7 =8.4 rad/s²
32. Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length.
(a) Find the initial angular acceleration of the rod.
(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.
ANSWER: (a) Since the mass of the rod is uniformly distributed over its length, it will make no difference in the torque produced about the pivot. Hence the torque T =1.5g N-m
But the moment of inertia of the system will change. The moment of inertia of the rod will have to be added. The moment of inertia of the rod about the pivot =ml²/12 = 1*1²/12 = 1/12 kg-m²
Hence total moment of inertia = 1.75 + 1/12 kg-m²
= 1.75+0.083 =1.833 kg-m²
Hence the initial angular acceleration α = T/I
=1.5*9.8/1.833 =8.0 rad/s²
(b) Linear acceleration at the end of the rod, a = αr =8.0*0.5 =4.0 m/s²
Consider the 2.0 kg mass. Weight 2g N is downwards, tension T is upwards and the acceleration a is upwards. Hence,
T-2g=2a
→T=2g+2a =2*9.8+2*4.0 =19.6+8 = 27.6 N
For 5 kg mass, acceleration a is downwrds, hence
5g-T =5a
→T=5g-5a =5*9.8-5*4 =49-20 =29 N
33. Figure (10-E4) shows two blocks of masses m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley.
 |
| Figure for Q-33 |
ANSWER: Let the accelerations of blocks M and m be a and the tension in the string be T and T' respectively. For the block M, we have,
Mg-T =Ma
T = M(g-a)
And for the block m,
T' = ma
Net torque on the pulley = Tr-T'r =(T-T')r
If the angular acceleration of the pulley be α,
Then, a =αr → α = a/r
For the pulley, Torque =Iα
→(T-T')r = Ia/r
→M(g-a)-ma =Ia/r²
→Mg = Ma+ma+Ia/r²
→ a = Mg/(M+m+I/r²)
34. A string is wrapped on a wheel of moment of inertia 0.20 kg-m² and radius 10 cm and goes through a light pulley to support a block of mass 2.0 kg as shown in figure (10-E5). Find the acceleration of the block.
 |
| Figure for Q-34 |
ANSWER: Let the acceleration of the block be a m/s² and the tension in the string be T Newton.
2g-T = 2a
→T = 2g-2a
Torque on the pully = T*0.10 N-m
=0.10T N-m
=0.10(2g-2a)
=0.20(g-a)
Moment of inertia of the pulley, I= 0.20 kg-m²
The angular acceleration α = Torque/M.I.
→a/r = 0.20(g-a)/0.20 =g-a
→a = gr-ar
→a+ar = gr
→a = gr/(1+r)
→a=9.8*0.10/(1+0.10)
→a = 0.98/1.10
→a = 98/110 = 0.89 m/s²
35. Suppose the smaller pulley of the previous problem has its radius 5.0 cm and moment of inertia 0.10 kg-m². Find the tension in the part of the string joining the pulleys.
ANSWER: Let the required tension be T'.
Now the torque on the bigger pulley = T'*0.10 N-m
Angular acceleration a/r =0.10*T'/0.20 =T'/2
→a = rT'/2 =0.10*T'/2 = 0.05T' -------- (i)
Net torque on the smaller pulley = (T-T')r' =(T-T')*0.05 N-m
Angular acceleration a/r' = 0.05(T-T')/I'
0.05T'/0.05 =0.05(T-T')/0.10 {Value of a from (i)}
→T' =0.5T - 0.5T'
→1.5T' =0.5T ---------------------------(ii)
For the weight 2g -T = 2a
→T = 2g-2a = 2g-2*0.05T' = 2g-0.10T'
Putting this value in (ii), we get
1.5T' = 0.5 (2g-0.10T')
→1.5T' = 9.8-0.05T'
→1.55T' =9.8
→T' = 9.8/1.55 = 6.32 N
36. The pulleys in figure (10-E6) are identical, each having radius R and moment of inertia I. Find the acceleration of the block M.
 |
| Figure for Q-36 |
ANSWER: Let the acceleration be a.
Angular acceleration of the pulleys = a/R
Let the tension in the string connected to M be T and in the string connected to m be T'. Also the tension in the string between the pulleys be T".
Now Mg-T =Ma
→T = M(g-a)
Also, T'-mg = ma
→T' = m(g+a)
For pulleys,
α = a/R = (T-T")R/I
→ a =(T-T")R²/I ------------ (i)
Also, a/R =(T"-T)R/I
→a = (T"-T')R²/I ----------- (ii)
Adding (i) and (ii)
2a = (T-T')R²/I = (Mg-Ma -mg-ma)R²/I
→2aI = (Mg-Ma -mg-ma)R² =(M-m)gR²-(M+m)aR²
→2aI+(M+m)aR² = (M-m)gR²
→a{2I+MR²+mR²} = (M-m)gR²
→a = (M-m)gR²/2I+MR²+mR²
→a = (M-m)g/M+m+2I/R²
{Dividing the numeretor and the denominetor by R²]
37. The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0.20 kg-m². The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.
 |
| Figure for Q-37 |
ANSWER: Let the tension in the string on the plane be T and on the fixed side be T'.
T =ma = 1.0*a = a {where a is the acceleration of the block}
Acceleration of the pulley = a/2
Angular acceleration = a/2r = a/0.4 m/s²
The torque on the pulley = (T'-T)*0.20
I = 0.20 kg-m²
α = T/I
→ a/0.4 = (T'-T)*0.20/0.20 = T'-T
→ T-T' = -a/0.4 = -2.5a --------------------- (i)
Let the mass of the pulley be M
Mr²/2 = I
→M = 2*0.20/0.2*0.2 =10 kg
Now Mg -T-T' = Ma/2
→T+T' =Mg-Ma/2 =10*9.8 - 10a/2 =98 - 5a ----------(ii)
Adding (i) and (ii)
2T = 98 - 5a - 2.5a
→2a = 98 -7.5a
→9.5a = 98
→a = 98/9.5 = 10.31 m/s² ≈ 10 m/s²
38. The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0.50 kg-m² about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4.0 kg block.
 |
| Figure for Q-38 |
ANSWER: Let the acceleration of the 4 kg block be a and the tension in the string T. If the tension in the 2 kg block be T' then we have,
4g.sin45° - T = 4a
→T = 4g/√2 -4a
And T'-2g.sin45° = 2a {Same acceleration for 2 kg block}
T' = 2a + 2g/√2
Net torque on the pulley = (T-T')*0.10
=0.10(4g/√2-4a-2a-2g/√2) =0.10(2g/√2-6a)
Angular acceleration α =a/r = a/0.10
From Torque = Iα
0.10(2g/√2-6a) = 0.50*a/0.10 =5a
→ 2g/√2-6a = 50a
→56a = 2g/√2 = 2*9.8*√2/2 =9.8√2
→a = 9.8*1.41/56 = 0.247 m/s² ≈ 0.25 m/s²
39. Solve the previous problem if the friction coefficient between 2.0 kg block and the plane below it is 0.50 and the plane below the 4.0 kg block is frictionless.
ANSWER: Given µ = 0.50
Hence the equation for 2 kg block is now
T' -2g.sin45° -µ(2g.cos45°) = 2a
→T' = 2a + 2g/√2+µ*2g/√2
→T' = 2a+√2g+√2gµ
Now the torque is = Tr-T'r (T-T')r
= (4g/√2-4a -2a-√2g-√2gµ)*0.10
= 0.10 (4*9.8/1.41-6a -1.41*9.8*1.5)
= 0.10(27.80-20.73-6a) = 0.10(7.07-6a)
Angular acceleration α = a/r
Since Torque = Iα
0.10(7.07-6a) = 0.50*a/0.10 =5a
→0.707-0.60a = 5a
→5.6a = 0.707
→a = 0.707/5.6 = 0.126 ≈ 0.125 m/s²
40. A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.
ANSWER: Let us draw a diagram as below:-
 |
| Figure for Q-40 |
The weight of the stick may be taken as acting at the CoM of the rod, i.e. 50 cm from the left. W =mg = 200g/1000 =0.20g N.
The weight of the small object W' = 20g/1000 = 0.02g N, acting at 70 cm i.e. 0.70 m from the left end.
Let tension in the left string = T and in the right string = T'
Since the stick is in balance, adding the forces in the vertical direction, we get
T+T' = W+W' = 0.20g+ 0.02g = 0.22g ---------- (i)
Now the sum of the torques about end B,
T*1.0 -W*0.50-W'*0.30 = 0 {Since the stick has no rotation}
→T = 0.20g*0.50+0.02g*0.30
→T = 0.1g + 0.006g =0.106g = 0.106*9.8 = 1.04 N
From (i),
T' = 0.22g-1.04 =0.22*9.8-1.04 = 2.16 - 1.04 = 1.12 N
41. A uniform ladder of length 10.0 m and mass 16.0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the electrician to work Safely?
ANSWER: Let the normal force on the ladder at the lower end = N and at the upper end = N' as shown in the figure below.
 |
| Figure for Q -41 |
The force of friction on the lower end = µN
The weight of ladder acts at the middle D = W = 16g N
The weight of the electrician P = 60g N
Since the rotation of the ladder is zero, hence the sum of torque about B = 0
µN*10cos37°-N*10sin37°+16g*5sin37°+60g*2sin37° = 0
→µN*40/5-N*30/5+80*9.8*3/5+120*9.8*3/5 =0
→8µN-6N+470.4+705.6 = 0
→8µN-6N+1176 = 0
→4µN-3N+588 = 0 ------------------(i)
The sum of forces in the vertical direction,
N-16g-60g = 0
→ N = 76g = 76*9.8 = 744.8 Newton ≈ 745 N
Hence the normal force on the lower end = 745 N
The force of friction on the lower end =µN = (3N-588)/4
{From (i)}
= (3*745-588)/4 = 411.75 ≈ 412 N
Since µN = 412
→µ = 412/N = 412/745 = 0.533
42. Suppose the friction coefficient between the ground and the ladder of the previous problem is 0.540. Find the maximum weight of a mechanic who could go up and do the work from the same position of the ladder.
ANSWER: Let the maximum weight of the mechanic be Mg.
Equating the forces in the vertical direction
mg+Mg = N
Equating the sum of the torques about B
µN*10cos37°-N*10sin37°+16g*5sin37°+Mg*2sin37°=0
→N(0.54*40/5-30/5)+16*9.8*5*3/5+M*9.8*2*3/5=0
→N(4.32-6)+470.4+11.76M = 0
→-1.68N +470.4+11.76M = 0
→-1.68mg-1.68Mg +470.4+11.76M = 0
→-16.46M+11.76M+470.4-1.68*16*9.8 =0
→-4.70M+470.4-263.42 = 0
→M = 206.98/4.70 = 44.03 ≈44.0 kg
43. A 6.5 m long ladder rest against a vertical wall reaching a height of 6.0 m. A 60 kg man stands halfway up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.
ANSWER: Let us draw a diagram as below,
 |
| Diagram for Q-43 |
The distance of the lower end of the ladder from the wall, AO
=√(6.5²-6.0²) = 2.5 m
The torque of the of the force exerted by the man on the ladder about the upper end = 60g*2.5/2 =75g =75*9.8 = 735 N-m
Let the Normal force by the ground be R and the frictional force = F. Normal reaction by the wall = R'.
Equating forces in the vertical direction,
R = 60g = 60*9.8 = 588 N ≈ 590 N
Equating the torques about end B,
F*6+60g*2.5/2-R*2.5 =0
→6F =2.5*588-30*9.8*2.5
→6F = 735
→F = 735/6 = 122.5 N ≈ 120 N
44. The door of an almirah is 6 ft high, 1.5 ft wide and weighs 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.
ANSWER: Let the forces exerted by each of the hinges on the door be F in horizontal direction and R in the vertical direction.
 |
| Diagram for Q -44 |
F*4/3.28 =8g*½*1.5/3.28 {Since 1 m =3.28 ft}
→F = 2g*½*1.5 =9.8*1.5 = 14.7 N
Equating the forces in the vertical direction,
2R-8g = 0
→R = 4g
→R = 4*9.8 = 39.2 N
Since F and R are mutually perpendicular, the magnitude of the resultant force by each of the hinges = √(F²+R²) =√(14.7²+39.2²)
=√1752.73 ≈ 42.0 N
45. A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is θ.
 |
| Figure for Q-45 |
ANSWER: Let us draw a diagram showing forces on the rod. If m is mass of the rod then its weight mg acts at the middle point i.e. L/2 distance from the end. Let R be the normal reaction by the roller and P by the ground. If µ be the coefficient of the friction between the rod and the ground then a force of friction, F=µP, acts at the lower end of the rod to check the movement.
 |
| Figure for Q-45 |
Since the rod has no linear motion, hence the sum of forces in horizontal and vertical directions will be zero.
In the horizontal direction,
F = R.sinθ
→µP = R.sinθ
→R =µP/sinθ ---------------------- (i)
In the vertical direction,
R.cosθ+P = mg
→µP.cosθ/sinθ + P = mg
→mg = P(µcosθ+sinθ)/sinθ ------- (ii)
Since the rod does not rotate, the sum of torques of all the forces will be zero. Hence taking moment about the lower end A, we get
R*h/sinθ-mg(L/2)cosθ = 0
→Rh-½mgL.cosθ.sinθ = 0
→2Rh = mgL.cosθ.sinθ
→2h*µP/sinθ = {P(µcosθ+sinθ)/sinθ}*L.cosθ.sinθ
[Putting the values of R and mg from (i) and (ii)]
→2hµ = (µcosθ+sinθ)*L.cosθ.sinθ
→2hµ = µcosθ*L.cosθ.sinθ + sinθ*L.cosθ.sinθ
→2hµ - µcosθ*L.cosθ.sinθ = L.cosθ.sin²θ
→ µ(2h - L.cos²θ.sinθ) = L.cosθ.sin²θ
→ µ = L.cosθ.sin²θ/(2h - L.cos²θ.sinθ)
===<<<O>>>===
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