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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'.

Illustration Questions

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?

image
A

\(a\)

.

B

\(2a\)

C

\(\dfrac{a}{2}\)

D

zero

Option A 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, \(\dfrac{dv_1}{dt}\) = \(\dfrac{dv_2}{dt}\)

\(\Rightarrow a_1 =a_2\)

Illustration Questions

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.

image

Relation between T1 and T

At pulley 2

T1 – 2T2 = 0

T1 = 2T2

image

Since, block A is at rest so, for pulley 2, let acceleration of block C is 'a'.

image image

By virtual work method, for block B 

= aT

= 2T

image image

For block C 

= – aT

image image

According to virtual work method,

\(\sum\, Ta = 0\)

2T – aT = 0

a = 2 m/sec2

image

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. 

image
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

Illustration Questions

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.

image

Applying virtual work method, 

For pulley, 

Tension × acceleration = 2T × ae 

= 2T × 4 

= + 8T

image image

For block A,

Tension × acceleration = T × aA

 = T × 2  

= +2T

image image

For block B

Tension × acceleration = – T × a 

= – Ta

image image

According to virtual work method, 

\(\sum\; Ta = 0\)

8T + 2T – Ta = 0

10T = Ta

a = 10 m/sec2 

image

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.

image
A

8 m/sec2

.

B

10 m/sec2

C

4 m/sec2

D

2 m/sec2

Option B 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, \(\dfrac{dv_1}{dt}\) = \(\dfrac{dv_2}{dt}\)

\(\Rightarrow a_1 =a_2\)

Illustration Questions

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.

image

Relation between T1 and T2,

At pulley 2

 T1 – 2T2 = 0

T1 = 2T2

 

image

Applying virtual work method, 

For block A

⇒ 2Ta1

image image

For Block B

⇒ – Ta2

image image

For block C

⇒ Ta3

image image

According to virtual work method,

\(\sum\;Ta = 0\)

2Ta1 – Ta2 + Ta3 = 0

2a1 – a2 + a3 = 0 

image

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. 

image
A

2a1 + a2 + a3 = 0

.

B

2a+ a2 – a3 = 0

C

2a– a2 – a3 = 0

D

2a– a2 + a3 = 0

Option D is Correct

Illustration Questions

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.

image

Relation between T1 and T2,

At pulley 2

T1 – 2 T2 = 0

T1 = 2 T

image image

Applying virtual work method,

For block A,

Tension × acceleration = + (2T) a1 

image image

For block B,

T × a

= – Ta2

image image

For block C

T × a 

= – Ta3 

image image

According virtual work method,

\(\sum \;Ta = 0\)

2Ta1 – Ta2 – Ta3 = 0

2a1 – a2 – a3 = 0 

image

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. 

image
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, \(\dfrac{dv_1}{dt}\) = \(\dfrac{dv_2}{dt}\)

\(\Rightarrow a_1 =a_2\)

Illustration Questions

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.

image

Applying virtual work method

For block A

= (+ Ta)

image image

Applying virtual work method

For block B

= (– Ta')

image image

By virtual work method, 

\(\sum \;Ta\) = 0

Ta – Ta' = 0

a = a'

image

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

image
A

\(\dfrac{a}{2}\)

.

B

\(\dfrac{a}{3}\)

C

\(2{a}\)

D

\(a\)

Option D is Correct

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