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PHXI05:LAWS OF MOTION

363255 If block $B$ is moving towards right side with acceleration $b$ . Find the acceleration of $A$ w.r.to ground.
supporting img

1 3

b

2 4

b

3 2

b

4

b

Explanation:

Let $x$ be the distance of block $B$ from rigid support and $y$ is the position of block $A$ w.r.to lower pulley connected to block $B$ .The length of the string is $L = 4 x + y + c$
By differentiating the equation w.r.to $t$ is
$0 = 4 \frac{d x}{d t} + \frac{d y}{d t} & 0 = 4 \frac{d^{2} x}{d t^{2}} + \frac{d^{2} y}{d t^{2}}$
$| \frac{d^{2} y}{d t^{2}} | = 4 b = | a_{A B} |$
$a_{A} = 4 b$ (As the wedge $B$ is moving away from $A$ and block $A$ will lose contact with $B$ hence the horizontal component of acceleration of $A$ is zero. Since there is no horizontal force on the block $A$ at $t = 0$ )
supporting img

PHXI05:LAWS OF MOTION

363256 The accelerations of the two blocks are given then the acceleration of block $C$ is
supporting img

1

1 m / s^{2}

upward

2

1 m / s^{2}

downward

3

2 m / s^{2}

upward

4

3 m / s^{2}

upward

Explanation:

Express the position of moveable pulleys and blocks with respect to fixed frame as shown in the figure
String – 1:
$l_{1} = y_{A} + y_{P_{1}} + y_{p_{1}} + c = y_{A} + 2 y_{P_{1}} + c$
$0 = V_{A} + 2 V_{p_{1}}$
$\Rightarrow 0 = a_{A} + 2 a_{p_{1}} \to (1)$
String – 2:
$l_{2} = (y_{B} - y_{p_{2}}) + (y_{C} - y_{p_{2}})$
$l_{2} = y_{B} - 2 y_{p_{2}} + y_{C}$
$0 = V_{B} - 2 V_{p_{2}} + V_{C}$
$0 = a_{B} - 2 a_{p_{2}} + a_{C} \to (2)$
From the diagram $a_{p_{1}} = a_{p_{2}}$
Since two pulleys are connected by a vertical string
From (1) and (2) we get
$a_{B} + a_{C} + a_{A} = 0$
We assume an arbitrary sign convention.Take downward direction as $(+) v e$
Given $a_{A} = 5 m / s^{2}$ , $a_{B} = - 4 m / s^{2}$
$\Rightarrow a_{C} = - 1 m / s^{2}$ i.e., block $C$ goes upward with $1 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363257 Find the acceleration of block $A$ .
supporting img

1

b

2

4 b

3

\sqrt{3} b

4

\sqrt{2} b

Explanation:

The positions of block $A$ and the pulley are shown in the figure.
The total length of the string is $l = x + y$
On differentiating the equation
$| \frac{d^{2} y}{d t^{2}} | = | \frac{d^{2} x}{d t^{2}} |$
The acceleration of the wedge is $\frac{d^{2} x}{d t^{2}} = b$
The $y$ -component of acceleration of the block is $\Rightarrow \frac{d^{2} y}{d t^{2}} = - b$ , the block has two components of accelerations. The net acceleration of block $A$ is
$\sqrt{b^{2} + b^{2}} = \sqrt{2} b$
supporting img

PHXI05:LAWS OF MOTION

363258 The acceleration of $m_{1}$ and $m_{2}$ are $a_{1}$ and $a_{2}$ then
supporting img

1

a_{1} = a_{2}

2

a_{1} = 2 a_{2}

3

2 a_{1} = a_{2}

4

a_{1} \leq a_{2}

Explanation:

$a_{2} = \frac{0 + a}{2}, a_{1} = 2 a_{2}$

PHXI05:LAWS OF MOTION

363255 If block $B$ is moving towards right side with acceleration $b$ . Find the acceleration of $A$ w.r.to ground.
supporting img

1 3

b

2 4

b

3 2

b

4

b

Explanation:

Let $x$ be the distance of block $B$ from rigid support and $y$ is the position of block $A$ w.r.to lower pulley connected to block $B$ .The length of the string is $L = 4 x + y + c$
By differentiating the equation w.r.to $t$ is
$0 = 4 \frac{d x}{d t} + \frac{d y}{d t} & 0 = 4 \frac{d^{2} x}{d t^{2}} + \frac{d^{2} y}{d t^{2}}$
$| \frac{d^{2} y}{d t^{2}} | = 4 b = | a_{A B} |$
$a_{A} = 4 b$ (As the wedge $B$ is moving away from $A$ and block $A$ will lose contact with $B$ hence the horizontal component of acceleration of $A$ is zero. Since there is no horizontal force on the block $A$ at $t = 0$ )
supporting img

PHXI05:LAWS OF MOTION

363256 The accelerations of the two blocks are given then the acceleration of block $C$ is
supporting img

1

1 m / s^{2}

upward

2

1 m / s^{2}

downward

3

2 m / s^{2}

upward

4

3 m / s^{2}

upward

Explanation:

Express the position of moveable pulleys and blocks with respect to fixed frame as shown in the figure
String – 1:
$l_{1} = y_{A} + y_{P_{1}} + y_{p_{1}} + c = y_{A} + 2 y_{P_{1}} + c$
$0 = V_{A} + 2 V_{p_{1}}$
$\Rightarrow 0 = a_{A} + 2 a_{p_{1}} \to (1)$
String – 2:
$l_{2} = (y_{B} - y_{p_{2}}) + (y_{C} - y_{p_{2}})$
$l_{2} = y_{B} - 2 y_{p_{2}} + y_{C}$
$0 = V_{B} - 2 V_{p_{2}} + V_{C}$
$0 = a_{B} - 2 a_{p_{2}} + a_{C} \to (2)$
From the diagram $a_{p_{1}} = a_{p_{2}}$
Since two pulleys are connected by a vertical string
From (1) and (2) we get
$a_{B} + a_{C} + a_{A} = 0$
We assume an arbitrary sign convention.Take downward direction as $(+) v e$
Given $a_{A} = 5 m / s^{2}$ , $a_{B} = - 4 m / s^{2}$
$\Rightarrow a_{C} = - 1 m / s^{2}$ i.e., block $C$ goes upward with $1 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363257 Find the acceleration of block $A$ .
supporting img

1

b

2

4 b

3

\sqrt{3} b

4

\sqrt{2} b

Explanation:

The positions of block $A$ and the pulley are shown in the figure.
The total length of the string is $l = x + y$
On differentiating the equation
$| \frac{d^{2} y}{d t^{2}} | = | \frac{d^{2} x}{d t^{2}} |$
The acceleration of the wedge is $\frac{d^{2} x}{d t^{2}} = b$
The $y$ -component of acceleration of the block is $\Rightarrow \frac{d^{2} y}{d t^{2}} = - b$ , the block has two components of accelerations. The net acceleration of block $A$ is
$\sqrt{b^{2} + b^{2}} = \sqrt{2} b$
supporting img

PHXI05:LAWS OF MOTION

363258 The acceleration of $m_{1}$ and $m_{2}$ are $a_{1}$ and $a_{2}$ then
supporting img

1

a_{1} = a_{2}

2

a_{1} = 2 a_{2}

3

2 a_{1} = a_{2}

4

a_{1} \leq a_{2}

Explanation:

$a_{2} = \frac{0 + a}{2}, a_{1} = 2 a_{2}$

PHXI05:LAWS OF MOTION

363255 If block $B$ is moving towards right side with acceleration $b$ . Find the acceleration of $A$ w.r.to ground.
supporting img

1 3

b

2 4

b

3 2

b

4

b

Explanation:

Let $x$ be the distance of block $B$ from rigid support and $y$ is the position of block $A$ w.r.to lower pulley connected to block $B$ .The length of the string is $L = 4 x + y + c$
By differentiating the equation w.r.to $t$ is
$0 = 4 \frac{d x}{d t} + \frac{d y}{d t} & 0 = 4 \frac{d^{2} x}{d t^{2}} + \frac{d^{2} y}{d t^{2}}$
$| \frac{d^{2} y}{d t^{2}} | = 4 b = | a_{A B} |$
$a_{A} = 4 b$ (As the wedge $B$ is moving away from $A$ and block $A$ will lose contact with $B$ hence the horizontal component of acceleration of $A$ is zero. Since there is no horizontal force on the block $A$ at $t = 0$ )
supporting img

PHXI05:LAWS OF MOTION

363256 The accelerations of the two blocks are given then the acceleration of block $C$ is
supporting img

1

1 m / s^{2}

upward

2

1 m / s^{2}

downward

3

2 m / s^{2}

upward

4

3 m / s^{2}

upward

Explanation:

Express the position of moveable pulleys and blocks with respect to fixed frame as shown in the figure
String – 1:
$l_{1} = y_{A} + y_{P_{1}} + y_{p_{1}} + c = y_{A} + 2 y_{P_{1}} + c$
$0 = V_{A} + 2 V_{p_{1}}$
$\Rightarrow 0 = a_{A} + 2 a_{p_{1}} \to (1)$
String – 2:
$l_{2} = (y_{B} - y_{p_{2}}) + (y_{C} - y_{p_{2}})$
$l_{2} = y_{B} - 2 y_{p_{2}} + y_{C}$
$0 = V_{B} - 2 V_{p_{2}} + V_{C}$
$0 = a_{B} - 2 a_{p_{2}} + a_{C} \to (2)$
From the diagram $a_{p_{1}} = a_{p_{2}}$
Since two pulleys are connected by a vertical string
From (1) and (2) we get
$a_{B} + a_{C} + a_{A} = 0$
We assume an arbitrary sign convention.Take downward direction as $(+) v e$
Given $a_{A} = 5 m / s^{2}$ , $a_{B} = - 4 m / s^{2}$
$\Rightarrow a_{C} = - 1 m / s^{2}$ i.e., block $C$ goes upward with $1 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363257 Find the acceleration of block $A$ .
supporting img

1

b

2

4 b

3

\sqrt{3} b

4

\sqrt{2} b

Explanation:

The positions of block $A$ and the pulley are shown in the figure.
The total length of the string is $l = x + y$
On differentiating the equation
$| \frac{d^{2} y}{d t^{2}} | = | \frac{d^{2} x}{d t^{2}} |$
The acceleration of the wedge is $\frac{d^{2} x}{d t^{2}} = b$
The $y$ -component of acceleration of the block is $\Rightarrow \frac{d^{2} y}{d t^{2}} = - b$ , the block has two components of accelerations. The net acceleration of block $A$ is
$\sqrt{b^{2} + b^{2}} = \sqrt{2} b$
supporting img

PHXI05:LAWS OF MOTION

363258 The acceleration of $m_{1}$ and $m_{2}$ are $a_{1}$ and $a_{2}$ then
supporting img

1

a_{1} = a_{2}

2

a_{1} = 2 a_{2}

3

2 a_{1} = a_{2}

4

a_{1} \leq a_{2}

Explanation:

$a_{2} = \frac{0 + a}{2}, a_{1} = 2 a_{2}$

PHXI05:LAWS OF MOTION

363255 If block $B$ is moving towards right side with acceleration $b$ . Find the acceleration of $A$ w.r.to ground.
supporting img

1 3

b

2 4

b

3 2

b

4

b

Explanation:

Let $x$ be the distance of block $B$ from rigid support and $y$ is the position of block $A$ w.r.to lower pulley connected to block $B$ .The length of the string is $L = 4 x + y + c$
By differentiating the equation w.r.to $t$ is
$0 = 4 \frac{d x}{d t} + \frac{d y}{d t} & 0 = 4 \frac{d^{2} x}{d t^{2}} + \frac{d^{2} y}{d t^{2}}$
$| \frac{d^{2} y}{d t^{2}} | = 4 b = | a_{A B} |$
$a_{A} = 4 b$ (As the wedge $B$ is moving away from $A$ and block $A$ will lose contact with $B$ hence the horizontal component of acceleration of $A$ is zero. Since there is no horizontal force on the block $A$ at $t = 0$ )
supporting img

PHXI05:LAWS OF MOTION

363256 The accelerations of the two blocks are given then the acceleration of block $C$ is
supporting img

1

1 m / s^{2}

upward

2

1 m / s^{2}

downward

3

2 m / s^{2}

upward

4

3 m / s^{2}

upward

Explanation:

Express the position of moveable pulleys and blocks with respect to fixed frame as shown in the figure
String – 1:
$l_{1} = y_{A} + y_{P_{1}} + y_{p_{1}} + c = y_{A} + 2 y_{P_{1}} + c$
$0 = V_{A} + 2 V_{p_{1}}$
$\Rightarrow 0 = a_{A} + 2 a_{p_{1}} \to (1)$
String – 2:
$l_{2} = (y_{B} - y_{p_{2}}) + (y_{C} - y_{p_{2}})$
$l_{2} = y_{B} - 2 y_{p_{2}} + y_{C}$
$0 = V_{B} - 2 V_{p_{2}} + V_{C}$
$0 = a_{B} - 2 a_{p_{2}} + a_{C} \to (2)$
From the diagram $a_{p_{1}} = a_{p_{2}}$
Since two pulleys are connected by a vertical string
From (1) and (2) we get
$a_{B} + a_{C} + a_{A} = 0$
We assume an arbitrary sign convention.Take downward direction as $(+) v e$
Given $a_{A} = 5 m / s^{2}$ , $a_{B} = - 4 m / s^{2}$
$\Rightarrow a_{C} = - 1 m / s^{2}$ i.e., block $C$ goes upward with $1 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363257 Find the acceleration of block $A$ .
supporting img

1

b

2

4 b

3

\sqrt{3} b

4

\sqrt{2} b

Explanation:

The positions of block $A$ and the pulley are shown in the figure.
The total length of the string is $l = x + y$
On differentiating the equation
$| \frac{d^{2} y}{d t^{2}} | = | \frac{d^{2} x}{d t^{2}} |$
The acceleration of the wedge is $\frac{d^{2} x}{d t^{2}} = b$
The $y$ -component of acceleration of the block is $\Rightarrow \frac{d^{2} y}{d t^{2}} = - b$ , the block has two components of accelerations. The net acceleration of block $A$ is
$\sqrt{b^{2} + b^{2}} = \sqrt{2} b$
supporting img

PHXI05:LAWS OF MOTION

363258 The acceleration of $m_{1}$ and $m_{2}$ are $a_{1}$ and $a_{2}$ then
supporting img

1

a_{1} = a_{2}

2

a_{1} = 2 a_{2}

3

2 a_{1} = a_{2}

4

a_{1} \leq a_{2}

Explanation:

$a_{2} = \frac{0 + a}{2}, a_{1} = 2 a_{2}$