DC Pandey Solutions: Laws of Motion

Sol.

N = Normal force on cylinder by plank

R = Normal force on cylinder by ground,
f = Force of friction by ground by cylinder,
w = Weight of cylinder.
N cosθ = f
w + N sinθ = R,
N(R) = Reaction to N,

(Force acting on plank)
i.e., normal force on plank by cylinder
R' = Normal force on plank by ground,
w = Weight of plank,
f' = frictional force on plank by ground.
Resultant of f' and R', N(R) and w pass through point O.

Sol.

R = Normal force on sphere A by left wall,
N = Normal force on sphere A by ground,
N' = Normal force on sphere A by sphere B,
w_A = Weight of sphere A.
N' cos θ = R
N' sin θ + w_A = N

R' = Normal force on sphere B by right wall,
N ( R) = Reaction to N i.e. normal force on sphere B by sphere A,
w_B = Weight of sphere B,
R', N(R) and w_B pass through point O, the centre sphere B.

Sol.
N = Normal force on sphere by wall,

w = Weight of sphere,
T = Tension in string.

Sol.

1. F_1x = 2√3 N
2. F₂x = -2 N
3. F₃x = 0
4. F₄x = 4 N
5. F_1y = 2N
6. F_2y = 2√3 N
7. F_3y = -6 N
8. F_4y = 0

Explanation:
Component of
along x-axis : 4 cos 30° = 2√3 N
along y-axis : 4 sin 30° = 2 N
Component of
along x-axis : 4 cos 120° = - 2 N
along y-axis : 4 sin 120° = 2√3 N
Component of
along x-axis : 6 cos 270° = 0 N
along y-axis : 6 sin 270° = - 6 N
Component of
along x-axis : 4 cos 0° = 4 N
along y-axis : 4 sin 0° = 0 N

Sol. 
Taking moment about point A

AB = 1
(T sin 30) 1 = w 1/2
⇒ T = w

Sol.

Explanation:
See figure (answer to question no. 3)


T cos 30° = w
or
or

Sol. F = 10.16 newton, R = 2.4 newton

Explanation:
R cos 30° + 3 = f cos 60°

i.e.,
or  
R√3 + 6 = f ...(i)
and R sin 30° + f sin 60° = 10
i.e.,
or R + f√3 = 20 ...(ii)
Substituting the value of f from Eq. (i) in Eq. (ii)
R + ( R√3 + 6)√3 = 20
4 R + 6√3 = 20
⇒
∴   f = (2.4) √3 + 6
f = 10.16 N

Sol.

Explanation:
At point B (instantaneous vertical acceleration only)

∴  mg - T sin 45° = ma ...(i)
At point A (instantaneous horizontal acceleration only)
∴  T cos 45° = ma ...(ii)
Combining Eqs. (i) and (ii)
mg - ma = ma
⇒  

T cos 45° = m(g/2)

T/√2 = m(g/2)

T = mg/√2

Sol.  (a) 10 ms^-2 (b) 110 N (c) 20 N

Explanation:
Acceleration of system

= 10 m/s²
Let normal force between 1 kg block and 4 kg block = F₁
∴ Net force on 1 kg block = 120 - N
∴
or 10 = 120 - F₁
i.e., F₁ = 110 N
Net force on 2 kg block = 2*a
= 2*10
=20 N

Sol.  zero
Explanation:
 As, 4 g sin 30° > 2 g sin 30°
The normal force between the two blocks will be zero.

Sol.  3g/4

Explanation:

∴ N = mg/4
As lift is moving downward with acceleration a , the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift.
N + ma = mg
or
⇒

Sol.  30°, ION

Explanation:
 Angle made by the string with the normal to the ceiling = θ = 30°
As the train is moving with constant velocity no pseudo force will act on the plumb-bob.
Tension in spring = mg
= 1*10
= 10 N

Sol. 

Explanation:
 Pseudo force (= ma) on plumb-bob will be as shown in figure
T cos φ = mg + ma cos (90° - θ)
i.e., T cos φ = mg + ma sin θ ...(i)
and       T sin φ = ma cos θ
Squaring and adding Eqs. (i) and (ii),
T² = m² g² + m² a² sin² θ + 2m² ag sin θ + m² a² cos² θ ...(iii)
T² = m² g² + m² a² + m² ag (∴ θ = 30°)


or

Dividing Eq. (i) by Eq. (ii),



i.e.,

Sol.  4 N, 6 N

Explanation: 
Net force on 1 kg mass = 8 - T₂
∴ 8 - T₂ = 1*2
⇒ T₂ = 6 N
Net force on 1 kg block = T₁
∴ T₁ = 2a = 2*2 = 4 N

Sol. 3 kg

Explanation:
 F = 2 g sin 30° = g
For the system to remain at rest
T₂ = 2 g ...(i)
T₂ + F = T₁ ...(ii)
or T₂ + g = T₁ ...[ii (a)]
T₁ = mg ...(iii)
Substituting the values of T₁ and T₂ from Eqs. (iii) and (i) in Eq. [ii(a)]
2 g + g = mg
i.e., m = 3 kg

Sol.  4

Explanation:
 As net downward force on the system is zero, the system will be in equilibrium
∴ T₁ = 4 g
and T₂ = 1 g
∴

Sol.  1/3 s

Explanation:
2 g - T = 2a
T - 1 g = 1a
Adding above two equations
1 g = 3 a
∴ a = g/3
Velocity of 1kg block 1 section after the system is set in motion
v = 0 + at

On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( t' ) taken by the 1 k g block to attain zero velocity will be given by the equation.

⇒
If the 2 kg block is stopped just for a moment (time being much-much less than 1/3 s), it will also start falling down when the stopping time ends.
In  time upward displacement of 1 kg block

Downward displacement of 2 kg block

As the two are just equal, the string will again become taut after time 1/3 s.

Sol.

Explanation:
 F + 1g - T = 1a      ... (i)
and T - 2g = 2a ...(ii)
Adding Eqs. (i) and (ii),
F - 1g = 3a
∴

Sol.  (a) 2g/3, (b) 10/3 N

Explanation:
 2T = 2*a            ... (i)
and 1g - T = 2a ...(ii)
Solving Eqs. (i) and (ii),
α = g/3
∴  Acceleration of 1 kg block

Tension in the string

Sol. 

Explanation:
 Mg - T = Ma             ...(i)
T = Ma ...(ii)
Solving Eqs. (i) and (ii)
α = g/2
and T = Mg

Sol.  4 . 8 kg

Explanation:
Block of mass M will be at rest if
T = Mg ...(i)
For the motion of block of mass 3 kg
....(ii)
For the motion of block of mass 2 kg
....(iii)
Adding Eqs. (ii) and (iii),
g = 5a
i.e.,  α = g/5
Substituting above value of a in Eq. (iii),
T/2 = 2(g+a)
or
= 24g/5
Substituting value of above value of T in Eq. (i),
M = 24/5
= 4.8 kg

Sol. 

Explanation:
 T/2 = m1.2a
i.e., T = 4 m₁a ...(i)
F - T = m₂a ...(ii)
or F - 4m₁a = m₂a
or

∴ T = 4 m₁a
= 4*0.3* 2/7
= 2.4/7
= 12/35 N

Sol.  6 .83 kg

Explanation:
 Block on triangular block will not slip if

m₁a cos θ= m₁g sin θ
i.e., a = g tanθ ...(i)
N = m₁g cosθ + m₁asinθ ...(ii)
For the movement of triangular block
T - N sinθ = m₂α ...(iii)
For the movement of the block of mass M
Mg - T = Ma ...(iv)
Adding Eqs. (iii) and (iv),
Mg - N sinθ = ( m₂ + M )α
Substituting the value of N from Eq. (ii) in the above equation
Mg - (m₁g cosθ + m₁α sinθ) sinθ
=( m₂α + Mα)
i.e., M (g - α) = m₁g cosθ sinθ + (m₂ + m₁ sin²θ)α
Substituting value of a from Eq. (i) in the above equation,
M(1 - tanθ) = m₁ cosθ sinθ + ( m₂ + m₁ sin² θ) tanθ

Substituting θ = 30° , m₁ = 1 kg and m₂ = 4 kg


= 6.82 kg

Sol. (a) x = x₀ + 10t - 2.5t² , v = 10 - 5t (b) t = 4s

Explanation:
(a) Using
Displacement of block at time t relative to car would be


Velocity of block at time t (relative to car) will be

(b) Time (t) for the block to arrive at the original position (i.e., x = x₀) relative to car
x₀ = x₀ + 10 t - 2.5 t²
⇒ t = 4s

Sol. (a) x = x₀ - 2.5t², z = z₀ + 10t, v_x = - 5t, v_z = 10 ms^-1
(b) x = x₀, z = z₀ + 10t, v_x = 0, v_z = 10 ms^-1

Explanation:
(a) In car's frame position of object at time t would be given by      
In car's frame

x = x₀ + 0*t + 1/2(-5)t²
i.e., x = x₀ - 2.5 t² ...(i)
and z = z + 10t ...(ii)
Velocity of the object at time t would be

and

(b) In ground frame the position of the object at time t would be given by
In ground frame

x = x₀
and z = z₀ + 10t
Velocity of the object at time t would be

ans

Sol.  x = x₀ + 10f - 4t² , K = 10 - 8 t for 0<f<1.25 s object stops at t = 1.25 s and remains at rest relative to car.

Explanation:
 m = 2 kg

Normal force on object = mg
Maximum sliding friction = μ_smg
= 0.3*2*10 = 6 N
Deceleration due to friction = 6/2 =3 m/s²
Deceleration due to pseudo force = 5 m/s²
∴ Net deceleration = (3 + 5) m/s²
= 8 m/s²
∴ Displacement of object at any time t (relative to car)

Thus, velocity of object at any time t (relative to car)

The object will stop moving relative to car when
10 - 8t = 0 i.e., t = 1.25s
∴ v_x = 10 - 8t for 0<t<1.25 s

Sol.  9/25 mg

Explanation:
For block not to slide the frictional force ( f ) would be given by

f + ma cosθ = mg sinθ
or
f = mg sinθ - ma cosθ