H C Verma Solutions

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  • Topic: Friction, Force, Free body diagram
  • Pages : 27 (3577 words )
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  • Published : February 6, 2013
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6.1
SOLUTIONS TO CONCEPTS
CHAPTER 6
1. Let m = mass of the block
From the freebody diagram,
R – mg = 0  R = mg ...(1)
Again ma –  R = 0  ma =  R =  mg (from (1))
 a = g  4 = g   = 4/g = 4/10 = 0.4
The co-efficient of kinetic friction between the block and the plane is 0.4 2. Due to friction the body will decelerate
Let the deceleration be ‘a’
R – mg = 0  R = mg ...(1)
ma –  R = 0  ma =  R =  mg (from (1))
 a = g = 0.1 × 10 = 1m/s2.
Initial velocity u = 10 m/s
Final velocity v = 0 m/s
a = –1m/s2 (deceleration)
S =
2a
v2  u2
=
2( 1)
0 102


=
2
100
= 50m
It will travel 50m before coming to rest.
3. Body is kept on the horizontal table.
If no force is applied, no frictional force will be there
f  frictional force
F  Applied force
From grap it can be seen that when applied force is zero,
frictional force is zero.
4. From the free body diagram,
R – mg cos  = 0  R = mg cos  ..(1)
For the block
U = 0, s = 8m, t = 2sec.
s = ut + ½ at2  8 = 0 + ½ a 22  a = 4m/s2
Again, R + ma – mg sin  = 0
  mg cos  + ma – mg sin  = 0 [from (1)]
 m(g cos  + a – g sin ) = 0
  × 10 × cos 30° = g sin 30° – a
 × 10 × (3 / 3) = 10 × (1/2) – 4
 (5 / 3)  =1   = 1/ (5 / 3) = 0.11
 Co-efficient of kinetic friction between the two is 0.11. 5. From the free body diagram
4 – 4a – R + 4g sin 30° = 0 …(1)
R – 4g cos 30° = 0 ...(2)
 R = 4g cos 30°
Putting the values of R is & in equn. (1)
4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0
 4 – 4a – 0.11 × 4 × 10 × ( 3 / 2) + 4 × 10 × (1/2) = 0  4 – 4a – 3.81 + 20 = 0  a  5 m/s2
For the block u =0, t = 2sec, a = 5m/s2
Distance s = ut + ½ at2  s = 0 + (1/2) 5 × 22 = 10m
The block will move 10m.
R
mg
ma R
a
R
mg
ma R
velocity a
F
p
o
mg
R
30°
R

R
mg
ma
4kg
30°
4N
R

R
mg
ma
velocity a
Chapter 6
6.2
6. To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline =  R + 2 g sin 30° = 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)]
= 3.39 + 9.8 = 13N
With this minimum force the body move up the incline with a constant velocity as net force on it is zero.
b) Net force acting down the incline is given by,
F = 2 g sin 30° – R
= 2 × 9.8 × (1/2) – 3.39 = 6.41N
Due to F = 6.41N the body will move down the incline with acceleration. No external force is required.
 Force required is zero.
7. From the free body diagram
g = 10m/s2, m = 2kg,  = 30°,  = 0.2
R – mg cos  - F sin  = 0
 R = mg cos  + F sin  ...(1)
And mg sin  + R – F cos  = 0
 mg sin  + (mg cos  + F sin ) – F cos  = 0  mg sin  +  mg cos  +  F sin  – F cos  = 0  F =
( sin cos )
(mgsin mgcos )
   
   
 F =
0.2 (1/ 2) ( 3 / 2)
2 10 (1/ 2) 0.2 2 10 ( 3 / 2)
 
     
=
0.76
13.464
= 17.7N  17.5N
8. m  mass of child
R – mg cos 45° = 0
 R = mg cos 45° = mg /v2 ...(1)
Net force acting on the boy due to which it slides down is mg sin 45° - R = mg sin 45° -  mg cos 45°
= m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 )
= m [(5/ 2 ) – 0.6 × (5 / 2 )]
= m(2 2 )
acceleration =
mass
Force
=
m
m(2 2)
= 2 2 m/s2
9. Suppose, the body is accelerating down with acceleration ‘a’. From the free body diagram
R – mg cos  = 0
 R = mg cos  ...(1)
ma + mg sin  –  R = 0
 a =
m
mg(sin  cos )
= g (sin  –  cos )
For the first half mt. u = 0, s = 0.5m, t = 0.5 sec.
So, v = u + at = 0 + (0.5)4 = 2 m/s
S = ut + ½ at2  0.5 = 0 + ½ a (0/5)2  a = 4m/s2 ...(2) For the next half metre
u` = 2m/s, a = 4m/s2, s= 0.5.
 0.5 = 2t + (1/2) 4 t2  2 t2 + 2 t – 0.5 =0
(body moving us)
R
R

F
mg
(body moving down)
R
R

mg
R
45°
R
mg
F
30°
R
R
30°
mg
mg
ma
R
R
Chapter 6
6.3
 4 t2 + 4 t – 1 = 0
 =
2 4
4 16 16

  
=
8
1.656
= 0.207sec
Time taken to cover next half meter is 0.21sec.
10. f  applied force
Fi  contact force
F  frictional force
R  normal...
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