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Answers
30.Let the accelerations of blocks be 'a' m/s² and tension in the string be 'T' N. See the Free Body Diagram for both the blocks below (Forces are taken in the direction of acceleration only):-
FBD of blocks
For the block on the surface, we get, T=ma,
For the hanging block mg-T = ma
→ mg - T = ma (Replace ma =T)
→mg-T=T →2T=mg →T =mg/2 (given m=1 kg and g=10 m/s²)
→T =10/2 = 5 N
31.From the arrangement it is clear that pulley B moves half the distance the block M moves. So acceleration of B is half than M. Let acceleration of block M be 'a' and that of pulley B be a/2. If tension in the string be T, foe block M we have,
Mg-T = Ma → T = Mg - Ma --------- (A)
and for block 2M,
2T = 2M.a/2 =Ma ----------- (B)
→ 2Mg - 2Ma = Ma (Replacing T)
→ 2g-2a = a → 3a = 2g → a = 2g/3 m/s²
(a) So acceleration of block M is 2g/3 m/s²
(b) From (B), Tension in the string 2T = Ma = M . 2g/3 → T=Mg/3
(c) First calculate the force on the pulley by the strings. As shown in the figure below forces on pulley are T in horizontal and T downwards by strings.
Forces on the pulley and the clamp
Resultant = √(T²+T²) = √(2T²) = √2T = √2Mg/3. The strings exert this force on the pulley and the pulley on the clamp. As per third law of motion equal and opposite force = √2Mg/3 is applied by the clamp on the pulley. The direction of this force will be tanθ =T/T =1 → θ = 45° (From the horizontal downward by the pulley on the clamp) and 45° from the horizontal upward by the clamp on the pulley.
32.Let us first find out the tension in the string. Let the tension be T. Consider the mass 2M, it is hanging from a moving pulley and the tension in string on both side is T. So the upward force on this block is 2T and the downward force is its weight 2Mg. It is clear from the arrangement that the displacement of block 2M will always be half the displacement of block M. So acceleration of block 2M will also be half the acceleration of block M. Let the acceleration of block M be 'a' (up the plane) and that of 2M be 'a/2' (Downward). See the 'Free body diagram' of these two blocks in the figure below (Normal force on the block not shown because its component along the slope is zero):-
F.B.D showing Forces and accelerations
Net force in the direction of acceleration for the block 2M is 2Mg-2T and the acceleration is 'a/2'. It gives,
2Mg-2T = 2M. a/2 =Ma
→ 2T = 2Mg - Ma
→ T = Mg - Ma/2 ........................... (A)
Similarly net force in the direction of acceleration for block M is T-Mg.cos60° = T-Mg/2 and the acceleration is 'a'. It gives,
T-Mg/2 = Ma
→ T = Mg/2 + Ma ..............................(B)
Equating (A) and (B) we get,
Mg/2 + Ma = Mg - Ma/2
→ g/2+a = g - a/2
→ a+a/2 = g-g/2
→ 3a/2 = g/2
→ 3a = g
→ a = g/3. (Up along the plane as assumed initially)
28. Let acceleration of m1 and moving pulley be 'a' and tension in the string attached to m1 and moving pulley be T. Taking Upward direction positive, for m1 we get,
T -m1g = m1a
→T= m1g + m1a = m1 (g + a) =1 x (g+a) N = g+a N
Let the tension in the string joining m2 and m3 be F. Considering the forces on the moving pulley, we get,
2F = T → F =T/2.
Let the acceleration of blocks m2 and m3 be a' with respect to the moving pulley. Acceleration of the block m2 with respect to the upper fixed pulley = a-a'. Similarly acceleration of the block m3 with respect to upper fixed pulley = a+a'.
Consider the forces on block m2,we get,
m2 g - T/2 = m2(a-a') , → a-a' = g - T/2m2 ------------------(A)
Consider the forces on block m3,we get,
m3 g - T/2 = m3(a+a') ,→ a+a' = g - T/2m3 ------------------(B)
Add (A) and (B),
2a = 2g-½T(1/m2+ 1/m3) = 2g -½T(½+1/3) = 2g -5T/12, ...(C)
(Multiply both side by 12)
→24a =24g -5T =24g-5(g+a) =24g-5g-5a = 19g - 5a,
→24a+5a = 19g,
→29a = 19g, → a=19g/29 m/s² . (Acceleration of m1)
From (B), a' = g-a -T/2m3 =g-a -(g+a)/6 =(5g-7a)/6
=(5g-133g/29)/6 =(145g-133g)/174 = 12g/174 m/s².
Acceleration of m2 = a-a' = 19g/29 - 12g/174 = (114g- 12g)/174
=102g/174 =17g/29 m/s².
Acceleration of m3 = a+a' = 19g/29 + 12g/174
= (114g + 12g)/174 = 126g/174 = 21g/29 m/s².
Second Part :--
Length of string of m1 (distance to be travelled) = s = 20 cm 0.20 m
Initial velocity u = 0, acceleration =a= 19g/29 m/s²,
To find time t =?,
From, s=ut+½at², → 0.20 = 0 +½ x(19g/29).t²,
19gt² = 0.2 x2 x29 = 0.4 x29,
t² = 11.6/19x9.8 = 0.623 → t ≈ 0.25 s
So m1 will hit the pulley in 0.25 s.
29. In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m so that it remains at rest ?
Answer: From (C),
2a = 2g -5T/12 and T =m1 (g+a), For m1 to be in rest a=0. It gives,
T=m1g. Putting these in (C) → 0=2g-5m1g/12 →5m1/12=2
m1 =24/5 = 4.8 kg
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