Informative line

### A Trick To Find Relation Between Accelerations

Learn acceleration of block by virtual work method & visual idea of constraint motion. Practice to find constraint acceleration between two blocks and an accelerating pulley.

# Visual Idea of Constraint Motion

• Constraint motion means that the motion of one body depends on the motion of other body.
• Displacement, velocity and acceleration of one body is related to other body because of some physical constraint.
• These ideas help us to solve problems in Newton's law of motion.
• Consider an example of two blocks, attached to each other, as shown in figure.  • As force (F) is applied at one end of A, such that block A is accelerated with acceleration a. Due to physical contact of block B with block A, the acceleration of B will also be 'a'.  #### In the given arrangement, if the acceleration of block A is 'a', what will be the acceleration of block B?

A $$a$$

B $$2a$$

C $$\dfrac{a}{2}$$

D zero

×

As block A and block B are in physical contact with each other so, they are in constrained motion.

Hence, acceleration of A = acceleration of B = a

### In the given arrangement, if the acceleration of block A is 'a', what will be the acceleration of block B? A

$$a$$

.

B

$$2a$$

C

$$\dfrac{a}{2}$$

D

zero

Option A is Correct

# Finding Constraint Relation between Acceleration of Blocks

• To find the constraint relation between the motion of different bodies, virtual work method is used.
• According to virtual work method,

$$\sum \;\vec{T.}\vec{v} = 0$$

where $$\vec{T}$$ is the tension in the string

$$\vec{v}$$ is the velocity of block

• Generally virtual work method is used in string block problems.
• Consider a situation in which two blocks are attached with a common string, as shown in figure.  • If force F is applied on block B, then block A moves with velocity v1 and block B moves with velocity v2.  • For block A, the direction of tension in string T and direction of velocity $$\vec{v}_1$$ is same. So, product of T and v1 is positive.

T.v1 = Tv1 cos 0° = Tv1  • For block B, the direction of tension in string is opposite to direction of velocity $$\vec v_2$$. So, product of tension and velocity will be negative.

$$\vec{T.} \vec{v_2} = T \,v_2\; cos 180° = -Tv_2$$  • Applying virtual work method,

$$\sum T v = 0$$

$$\Rightarrow Tv_1 – Tv_2 = 0$$

$$v_1 = v_2$$

• Similarly, $$\dfrac{dv_1}{dt}$$ = $$\dfrac{dv_2}{dt}$$

$$\Rightarrow a_1 =a_2$$

#### In the given arrangement, if accelerations of blocks A, B and C are a1, a2 and a3 respectively, then find the constraint relation between accelerations.

A 2a1 + a2 + a3 = 0

B 2a1 + a2 – a3 = 0

C 2a1 – a2 – a3 = 0

D 2a1 – a2 + a3 = 0

×

Let T1 and T2  are the respective tension forces in upper string and lower string, as shown in figure. Relation between T1 and T2,

At pulley 2

T1 – 2T2 = 0

T1 = 2T2 Applying virtual work method,

For block A

⇒ 2Ta1  For Block B

⇒ – Ta2  For block C

⇒ Ta3  According to virtual work method,

$$\sum\;Ta = 0$$

2Ta1 – Ta2 + Ta3 = 0

2a1 – a2 + a3 = 0 ### In the given arrangement, if accelerations of blocks A, B and C are a1, a2 and a3 respectively, then find the constraint relation between accelerations. A

2a1 + a2 + a3 = 0

.

B

2a+ a2 – a3 = 0

C

2a– a2 – a3 = 0

D

2a– a2 + a3 = 0

Option D is Correct

#### In the given arrangement, if accelerations of blocks A, B and C are a1, a2 and a3 respectively, then find the constraint relation between them.

A 2a1 – a2 – a3 = 0

B a1 – a2 – a3 = 0

C a1 + a2 – a3 = 0

D 2a1 – 2a2 = a3

×

Let T1 and T2 are the respective tension forces in upper and lower string, as shown in figure. Relation between T1 and T2,

At pulley 2

T1 – 2 T2 = 0

T1 = 2 T  Applying virtual work method,

For block A,

Tension × acceleration = + (2T) a1  For block B,

T × a

= – Ta2  For block C

T × a

= – Ta3  According virtual work method,

$$\sum \;Ta = 0$$

2Ta1 – Ta2 – Ta3 = 0

2a1 – a2 – a3 = 0 ### In the given arrangement, if accelerations of blocks A, B and C are a1, a2 and a3 respectively, then find the constraint relation between them. A

2a1 – a2 – a= 0

.

B

a1 – a2 – a= 0

C

a1 + a2 – a= 0

D

2a1 – 2a2 = a

Option A is Correct

# Method of Virtual Work

• To find the constraint relation between the motion of different bodies, virtual work method is used.
• According to virtual work method,

$$\sum \;\vec{T.}\vec{v} = 0$$

where $$\vec{T}$$ is the tension in the string

$$\vec{v}$$ is the velocity of block

• Generally virtual work method is used in string block problems.
• Consider a situation in which two blocks are attached with a common string, as shown in figure.  • If force F is applied on block B, then block A moves with velocity v1 and block B moves with velocity v2.  • For block A, the direction of tension in string T and direction of velocity $$\vec{v}_1$$ is same. So, the product of T and v1 is positive.

T.v1 = Tv1 cos 0° = Tv1  • For block B, the direction of tension in string is opposite to direction of velocity v2. So, the product of tension and velocity $$(\vec v_2)$$ will be negative.

$$\vec{T.} \vec{v_2} = T \,v_2\; cos 180° = -Tv_2$$  • Applying virtual work method,

$$\sum T v = 0$$

$$\Rightarrow Tv_1 – Tv_2 = 0$$

$$v_1 = v_2$$

Similarly, $$\dfrac{dv_1}{dt}$$ = $$\dfrac{dv_2}{dt}$$

$$\Rightarrow a_1 =a_2$$

#### In the given arrangement, find the acceleration of block B.

A $$\dfrac{a}{2}$$

B $$\dfrac{a}{3}$$

C $$2{a}$$

D $$a$$

×

Let acceleration of block B is $$a'$$ downwards. Applying virtual work method

For block A

= (+ Ta)  Applying virtual work method

For block B

= (– Ta')  By virtual work method,

$$\sum \;Ta$$ = 0

Ta – Ta' = 0

a = a' ### In the given arrangement, find the acceleration of block B. A

$$\dfrac{a}{2}$$

.

B

$$\dfrac{a}{3}$$

C

$$2{a}$$

D

$$a$$

Option D is Correct

# Finding Acceleration of Block by Virtual Work Method

• To find the constraint relation between the motion of different bodies, virtual work method is used.
• According to virtual work method,

$$\sum \;\vec{T.}\vec{v} = 0$$

Where $$\vec{T}$$ is the tension in the string

$$\vec{v}$$ is the velocity of block

• Generally virtual work method is used in string block problems.
• Consider a situation in which two blocks are attached with a common string, as shown in figure.  • If force F is applied on block B, then block A moves with velocity v1 and block B moves with velocity v2.  • For block A, the direction of tension in string T and direction of velocity $$\vec{v}_1$$ is same. So, product of T and v1 is positive.

T.v1 = T v1 cos 0° = T v1  • For block B, the direction of tension in string is opposite to direction of velocity v2. So, product of tension and velocity will be negative.

$$\vec{T.} \vec{v_2} = T \;v_2\; cos 180° = -Tv_2$$  • Applying virtual work method,

$$\sum T v = 0$$

$$\Rightarrow Tv_1 – Tv_2 = 0$$

$$v_1 = v_2$$

Similarly, $$\dfrac{dv_1}{dt}$$ = $$\dfrac{dv_2}{dt}$$

$$\Rightarrow a_1 =a_2$$

#### In the given arrangement, if block A is at rest, block B has acceleration aB = 2 m/s2 upwards, then find the acceleration (a) of block C.

A 6 m/sec2

B 2 m/sec2

C 8 m/sec2

D 3 m/sec2

×

Let T1 and T2 are the respective tension forces in upper string and lower string, as shown in figure. Relation between T1 and T

At pulley 2

T1 – 2T2 = 0

T1 = 2T2 Since, block A is at rest so, for pulley 2, let acceleration of block C is 'a'.  By virtual work method, for block B

= aT

= 2T  For block C

= – aT  According to virtual work method,

$$\sum\, Ta = 0$$

2T – aT = 0

a = 2 m/sec2 ### In the given arrangement, if block A is at rest, block B has acceleration aB = 2 m/s2 upwards, then find the acceleration (a) of block C. A

6 m/sec2

.

B

2 m/sec2

C

8 m/sec2

D

3 m/sec2

Option B is Correct

# Constraint Acceleration between Two Blocks and an Accelerating Pulley

• Consider a pulley system fixed to the ceiling of an elevator, as shown in figure.  • The elevator is moving upwards with an acceleration 'a'.
• Due to acceleration of the elevator, both masses will move with different accelerations with respect to the ground.
• Let acceleration of both masses m1 and m2 are a1 and a2 respectively, as shown in figure.  • Due to acceleration of the lift, the net acceleration of block and tension force are shown in figure.  • Using Virtual Work Method

→ For pulley

=  +2Ta  → For block m

= + Ta1  → For block m

= +Ta2  • According to virtual work method,

$$\sum\;Ta = 0$$

⇒ 2Ta + Ta1 + Ta2 = 0

⇒ 2a + a1 + a2 = 0

#### Consider the given arrangement which is placed in an accelerating elevator, as shown in figure. The acceleration of block A and elevator are aA = 2 m/sec2 and ae = 4 m/sec2 respectively. Find the acceleration of block B.

A 8 m/sec2

B 10 m/sec2

C 4 m/sec2

D 2 m/sec2

×

Due to acceleration of lift, block A and block B will move with different accelerations.

Let 'a' be the acceleration of block B. Applying virtual work method,

For pulley,

Tension × acceleration = 2T × ae

= 2T × 4

= + 8T  For block A,

Tension × acceleration = T × aA

= T × 2

= +2T  For block B

Tension × acceleration = – T × a

= – Ta  According to virtual work method,

$$\sum\; Ta = 0$$

8T + 2T – Ta = 0

10T = Ta

a = 10 m/sec2 ### Consider the given arrangement which is placed in an accelerating elevator, as shown in figure. The acceleration of block A and elevator are aA = 2 m/sec2 and ae = 4 m/sec2 respectively. Find the acceleration of block B. A

8 m/sec2

.

B

10 m/sec2

C

4 m/sec2

D

2 m/sec2

Option B is Correct