Introductory Exercise 5.5
Ques 1: In figure m_{1} = 1 kg and m_{2} = 4 kg. Find the mass M o f the hanging block which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction less and the strings and the pulleys are light
Note In exercises 2 to 4 the situations described take place in a box car which has initial velocity v = 0 but acceleration
Ans: 6 .83 kg
Sol: Block on triangular block will not slip if
m_{1}a cos θ= m_{1}g sin θ
i.e., a = g tanθ …(i)
N = m_{1}g cosθ + m_{1}asinθ …(ii)
For the movement of triangular block
T  N sinθ = m_{2}α …(iii)
For the movement of the block of mass M
Mg  T = Ma …(iv)
Adding Eqs. (iii) and (iv),
Mg  N sinθ = ( m_{2} + M )α
Substituting the value of N from Eq. (ii) in the above equation
Mg  (m_{1}g cosθ + m_{1}α sinθ) sinθ
=( m_{2}α + Mα)
i.e., M (g  α) = m_{1}g cosθ sinθ + (m_{2} + m_{1} sin^{2}θ)α
Substituting value of a from Eq. (i) in the above equation,
M(1  tanθ) = m_{1} cosθ sinθ + ( m_{2} + m_{1} sin^{2} θ) tanθ
Substituting θ = 30° , m_{1} = 1 kg and m_{2} = 4 kg
= 6.82 kg
Ques 2: A 2 kg object is slid along the friction less floor with initial velocity (10 m/s)î (a) Describe the motion of the object relative to car (b) when does the object reach its original position relative to the box car.
Ans: (a) x = x_{0} + 10t  2.5t^{2} , v = 10  5t (b) t = 4s
Sol: (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 = x_{0}) relative to car
x_{0} = x_{0 }+ 10 t  2.5 t^{2}
⇒ t = 4s
Ques 3: A 2 kg object is slid along the friction less floor with initial transverse velocity (10 m/s). Describe the motion (a) in car’s frame (b) in ground frame.
Ans: (a) x = x_{0}  2.5t^{2}, z = z_{0} + 10t, v_{x} =  5t, v_{z} = 10 ms^{1}
(b) x = x_{0}, z = z_{0} + 10t, v_{x} = 0, v_{z} = 10 ms^{1}
Sol: (a) In car’s frame position of object at time t would be given by
In car’s frame
x = x_{0} + 0*t + 1/2(5)t^{2}
i.e., x = x_{0}  2.5 t^{2} …(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 = x_{0}
and z = z_{0} + 10t
Velocity of the object at time t would be
ans
Ques 4: A 2 kg object is slid along a rough floor (coefficient of sliding friction = 0.3) with initial velocity (10 m/s)î. Describe the motion of the object relative to car assuming that the coefficient of static friction is greater than 0.5.
Ans: x = x_{0} + 10f  4t^{2} , K = 10  8 t for 0<f<1.25 s object stops at t = 1.25 s and remains at rest relative to car.
Sol: m = 2 kg
Normal force on object = mg
Maximum sliding friction = μ_{s}mg
= 0.3*2*10 = 6 N
Deceleration due to friction = 6/2 =3 m/s^{2}
Deceleration due to pseudo force = 5 m/s^{2}
∴ Net deceleration = (3 + 5) m/s^{2}
= 8 m/s^{2}
∴ 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
Ques 5: A block is placed on an inclined plane as shown in figure. What must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at
Ans: 9/25 mg
Sol: For block not to slide the frictional force ( f ) would be given by
f + ma cosθ = mg sinθ
or
f = mg sinθ  ma cosθ
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