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Advanced Problems Of Newtons Laws

Practice examples of Newton’s laws and Atwood machine problems, calculation of acceleration using constraint.

Visual Idea of Constraint Motion

  • Consider the given arrangement, as shown in figure.

  • In this arrangement, both blocks will have same acceleration because by visual inspection it can be seen that both blocks are connected with same string.

  • Acceleration of block B = Acceleration of block A

Calculation of Acceleration

  • Consider that mass of block A is \(m_1\) and mass of block B is \(m_2\), as shown in given arrangement. \((m_1>m_2)\)
  • Let tension in the string be T.

Free body diagrams

  • Assume that downward direction is positive.
  • Let the acceleration of blocks be 'a'.
  • Applying Newton's second law

\(m_1g-T=m_1a\)         ....(i)

(Since, no external force is applied, motion will take place only due to \(m_1g\) downward)

\(T=m_2a\)  ...(ii)

(As only one force \(T\) is acting on block \(B\) in the direction of acceleration)

\(m_1g-m_2a=m_1a\)

\(m_2a+m_1a=m_1g\)

\(a=\dfrac{m_1g}{m_2+m_1}\)

Tension in the string 

\(T=\dfrac{m_2.m_1g}{m_1+m_2}\)

Illustration Questions

In the given arrangement, the masses of block A and B are \(m_1=4\,kg\) and \(m_2=2\,kg\) respectively. Determine the acceleration of both the blocks.\((g=10\,m/sec^2)\)

A \(10\,m/s^2\)

B \(6.66\,m/s^2\)

C \(12\,m/s^2\)

D \(8\,m/s^2\)

×

By visual inspection, both the blocks are connected with same string across the pulley. So, the acceleration of both the blocks will be same.

Let tension in the string be T.

image

Free body diagrams

image

Assume that downward direction is positive.

Let the acceleration of blocks be 'a'.

Applying Newton's second law

\(m_1g-T=m_1a\)      ....(i)

(Since, no external force is applied, motion will take place only due to \(m_1g\) downward)

\(T=m_2a\)        ...(ii)

(As only one force \(T\) is acting on block \(B\) in the direction of acceleration)

image

\(m_1g-m_2a=m_1a\)

\(m_2a+m_1a=m_1g\)

\(a=\dfrac{m_1g}{m_2+m_1}\)

Given :  \(m_1= 4\,kg,\,\,\,\,\,\,m_2=2\,kg ,\,\,\,\,\,\,\,\,g=10\,m/sec^2\)

\(a=\dfrac{4×10}{4+2}\)

\(a=6.66\,m/s^2\)

In the given arrangement, the masses of block A and B are \(m_1=4\,kg\) and \(m_2=2\,kg\) respectively. Determine the acceleration of both the blocks.\((g=10\,m/sec^2)\)

image
A

\(10\,m/s^2\)

.

B

\(6.66\,m/s^2\)

C

\(12\,m/s^2\)

D

\(8\,m/s^2\)

Option B is Correct

Atwood Machine 

  • When two objects are hanged by an ideal string passing over an ideal pulley, the arrangement is called Atwood machine.
  • By visual inspection, it is clear that when one block moves upwards, and the other block moves downwards, then their accelerations are of equal magnitude.

Calculation of Acceleration 

  • Consider the given arrangement in which mass of block A is \(m_1\) and mass of block B is \(m_2\) (\(m_2>m_1\)).
  • Let tension in the string be T.

Free body diagrams

  • Assuming upward as positive direction and the acceleration of block be 'a'.
  • Applying Newton's second law

For block A  

\(T-m_1g=m_1a\)   ... (i)

For block B   
\(m_2g-T=m_2a\)   ... (ii)

\(T=(m_1a+m_1g)\)  {By solving equation (i)}

Putting value of T in equation (ii)

\(m_2g-m_1g-m_1a=m_2a\)

\((m_1+m_2)a=(m_2-m_1)g\)

\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\)

 

Illustration Questions

In an Atwood machine, the masses of block A and block B are  \(m_1=2\,kg\) and \(m_2=4\,kg\) respectively. Find the acceleration of each block. \((g=10\,m/s^2)\)

A \(6\,m/s^2\)

B \(4\,m/s^2\)

C \(8\,m/s^2\)

D \(3.33\,m/s^2\)

×

By visual inspection, it is clear that when one block moves upwards, and the other block moves downwards, then their accelerations are of equal magnitude.

image

Let tension in the string be T.

image

Free body diagrams

image image

Assuming upward as positive direction and the acceleration of block be 'a'.

Applying Newton's second law

For block A

 \(T-m_1g=m_1a\)   ... (i)

For block B   

\(m_2g-T=m_2a\)   ... (ii)

image

\(T=(m_1a+m_1g)\)  {By solving equation (i)}

Putting value of T in equation (ii)

\(m_2g-m_1g-m_1a=m_2a\)

\((m_1+m_2)a=(m_2-m_1)g\)

\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\)

image

\(a=\left(\dfrac{4-2}{2+4}\right)10\)

\(a=3.33\,m/s^2\)

image

In an Atwood machine, the masses of block A and block B are  \(m_1=2\,kg\) and \(m_2=4\,kg\) respectively. Find the acceleration of each block. \((g=10\,m/s^2)\)

image
A

\(6\,m/s^2\)

.

B

\(4\,m/s^2\)

C

\(8\,m/s^2\)

D

\(3.33\,m/s^2\)

Option D is Correct

Constraint Relation for Two Blocks hanging on a Pulley on Inclined

  • Consider the given arrangement, as shown in the figure.
  • By visual inspection, if block A moves downward then B moves upward with same magnitude of acceleration.

Calculation of Acceleration

  • In the given arrangement, masses of block of A and block B are  \(m_1\) and  \(m_2\) respectively.  \((m_1>m_2)\)
  • Let \(T\) be the tension in the string and the acceleration of block be \(a\).

Free body diagrams

  • Assuming upward as positive direction.
  • Applying Newton's second law 

\(m_1g\;sin\theta\;-T=m_1a\)  ...(i)

\(T-m_2g=m_2a\)    ...(ii)

\(\Rightarrow T=m_2(g+a)\)

Putting value of T in equation (i)

\(m_1g\;sin\theta\;-m_2g-m_2a=m_1a\)

\((m_1\;sin\theta\;-m_2)g=(m_1+m_2)a\)

\(a=\left(\dfrac{m_1sin\theta-m_2}{m_1+m_2}\right)g\)

Illustration Questions

For the given arrangement, masses of block A and block B are \(m_1=5\,kg\) and \(m_2=1\,kg\) respectively and angle of inclination is \(\theta=37º\). Determine the acceleration of both the blocks. (Take sin37º = 3/5)

A \(3\,m/s^2\)

B \(\dfrac{1}{7}\,m/s^2\)

C \(3.33\,m/s^2\)

D \(\dfrac{5}{3}\,m/s^2\)

×

By visual inspection, if block A moves downward then B moves upward with same magnitude of acceleration.

Let \(T\) be the tension in the string and the acceleration of block be \(a\).

Free body diagrams

image

Assuming upward as positive direction.

Applying Newton's second law 

\(m_1g\;sin\theta\;-T=m_1a\)  ...(i)

\(T-m_2g=m_2a\)    ...(ii)

\(\Rightarrow T=m_2(g+a)\)

Putting value of T in equation (i)

\(m_1g\;sin\theta\;-m_2g-m_2a=m_1a\)

\((m_1g\;sin\theta\;-m_2)g=(m_1+m_2)a\)

\(a=\left(\dfrac{m_1sin\theta-m_2}{m_1+m_2}\right)g\)

\(a=\Bigg(\dfrac{5×sin37º-1}{6}\Bigg)10\)

\(=\dfrac{10}{3}=3.33\,m/s^2\)

For the given arrangement, masses of block A and block B are \(m_1=5\,kg\) and \(m_2=1\,kg\) respectively and angle of inclination is \(\theta=37º\). Determine the acceleration of both the blocks. (Take sin37º = 3/5)

image
A

\(3\,m/s^2\)

.

B

\(\dfrac{1}{7}\,m/s^2\)

C

\(3.33\,m/s^2\)

D

\(\dfrac{5}{3}\,m/s^2\)

Option C is Correct

Calculation of Acceleration of Blocks Hanging on Triangular Shaped Wedge

  • Consider the given arrangement, as shown in the figure.
  • By visual inspection, if block A moves downward then B moves upward with same magnitude of acceleration.

Calculation of Acceleration

  • Consider two blocks A and B of masses  \(m_1\) and  \(m_2\) respectively, and angle of inclination is \(\theta\) with horizontal, as shown in figure.
  • Let \(T\) be the tension in the string and the acceleration of masses be \(a\).

Free body diagrams

  • Taking upward as positive direction.
  • Applying Newton's second law 

\(T-m_1g\;sin\theta=m_1a\)  ...(i)

\(m_2g\;sin\theta-T=m_2a\)   

\(\implies T=m_2g\;sin\theta-m_2a\)

Putting value of T in equation (i)

\(m_2g\;sin\theta\;-m_2a-m_1g\;sin\theta=m_1a\)

\(m_2g\;sin\theta\;-m_1g\;sin\theta=m_1a+m_2a\)

\(\dfrac{(m_2-m_1)g\;sin\theta}{m_1+m_2}=a\)

Illustration Questions

In the given arrangement, masses of two blocks A and B are \(m_1=2\,kg\) and \(m_2=4\,kg\) and angle of inclination is \(\theta=37º\) each, as shown in figure. Find the acceleration of each block.      (Take sin37º = 3/5)

A \(2\,m/sec^2\)

B \(3\,m/sec^2\)

C \(4\,m/sec^2\)

D \(5\,m/sec^2\)

×

By visual inspection, if block A moves downward then B moves upward with same magnitude of acceleration.

Let \(T\) be the tension in the string and the acceleration of masses be \(a\).

Free body diagrams

image

Taking upward as positive direction.

Applying Newton's second law 

\(T-m_1g\;sin\theta=m_1a\)  ...(i)

\(m_2gsin\theta-T=m_2a\)   

\(\implies T=m_2g\;sin\theta-m_2a\)

Putting value of T in equation (i)

\(m_2g\;sin\theta\;-m_2a-m_1g\;sin\theta=m_1a\)

\(m_2g\;sin\theta\;-m_1g\;sin\theta=m_1a+m_2a\)

\(\dfrac{(m_2-m_1)g\;sin\theta}{m_1+m_2}=a\)

\(a=\left(\dfrac{4-2}{4+2}\right)10×sin\;37º\)

\(=\dfrac{1}{3}×10×\dfrac{3}{5}\)

\(a=2\,m/sec^2\)

In the given arrangement, masses of two blocks A and B are \(m_1=2\,kg\) and \(m_2=4\,kg\) and angle of inclination is \(\theta=37º\) each, as shown in figure. Find the acceleration of each block.      (Take sin37º = 3/5)

image
A

\(2\,m/sec^2\)

.

B

\(3\,m/sec^2\)

C

\(4\,m/sec^2\)

D

\(5\,m/sec^2\)

Option A is Correct

Calculation of Acceleration using Constraint

  • Consider the given arrangement, as shown in the figure.
  • By visual inspection, if block A moves downward then B moves upward with same magnitude of acceleration.

Calculation of Acceleration

  • Consider that the masses of two blocks A and B are \(m_1\) and  \(m_2\) respectively.  \((m_1>m_2)\)
  • Let \(T\) be the tension in the string and the acceleration of block be \(a\).

 

Free body diagrams

  • Taking upward direction as positive.
  • Applying Newton's second law 

\(T-m_1g=m_1a\)  ...(i)

\(m_2g-T=m_2a\)    ...(ii)

\(\implies T =m_1g+m_1a\)

Putting value of T in equation (ii)

\(m_2g-m_1g-m_1a=m_2a\)

\((m_2-m_1)g=(m_1+m_2)a\)

\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\) 

Illustration Questions

In the given arrangement, the masses of blocks A and B are \(m_1=2\,kg\) and \(m_2=4\,kg\) respectively. Find the acceleration of each block.

A \(\dfrac{5}{3}\,m/s^2,\dfrac{10}{3}\,m/s^2\)

B \(\dfrac{10}{3}m/s^2\;each\)

C \(\dfrac{5}{3}\,m/s^2\;each\)

D \(\dfrac{10}{3}\,m/s^2,\dfrac{5}{3}\,m/s^2\)

×

By visual inspection, if block A moves downward then B moves upward with same magnitude of acceleration.

Let \(T\) be the tension in the string and the acceleration of block be \(a\).

 

Free body diagrams

image

Taking upward direction as positive.

Applying Newton's second law 

\(T-m_1g=m_1a\)  ...(i)

\(m_2g-T=m_2a\)    ...(ii)

\(\implies T =m_1g+m_1a\)

Putting value of T in equation (ii)

\(m_2g-m_1g-m_1a=m_2a\)

\((m_2-m_1)g=(m_1+m_2)a\)

\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\)

\(a=\dfrac{4-2}{4+2}×10\)

\(=\dfrac{10}{3}m/s^2\;each\)

In the given arrangement, the masses of blocks A and B are \(m_1=2\,kg\) and \(m_2=4\,kg\) respectively. Find the acceleration of each block.

image
A

\(\dfrac{5}{3}\,m/s^2,\dfrac{10}{3}\,m/s^2\)

.

B

\(\dfrac{10}{3}m/s^2\;each\)

C

\(\dfrac{5}{3}\,m/s^2\;each\)

D

\(\dfrac{10}{3}\,m/s^2,\dfrac{5}{3}\,m/s^2\)

Option B is Correct

Illustration Questions

In the given arrangement, the masses of three blocks \(A,\,B,\,C\) are \(m_1=2\,kg\), \(m_2=1\,kg\) and  \(m_3=2\,kg\) respectively. Find the acceleration of three bodies.  

A \(6m/s^2\;up,\,\,\, 6m/s^2\;up,\,\,\,2m/s^2\;down\)

B \(\dfrac{40}{7}m/s^2\;up,\,\,\, \dfrac{60}{7}m/s^2\;down,\,\,\,\dfrac{30}{7}m/s^2\;up\)

C \(5m/s^2\;up, \,\,\,3m/s^2\;up,\,\,\,2m/s^2\;down\)

D \(\dfrac{10}{7}m/s^2\;up, \,\,\,\dfrac{10}{7}m/s^2\;up,\,\,\,\dfrac{30}{7}m/s^2\,down \)

×

By visual inspection, the constraint relation cannot be determined.

Let the acceleration of blocks A, B and C are \(a_1,a_2\) and \(a_3\) respectively.

Let \(T_1\) and \(T_2\) are the tension forces in upper string and lower string respectively, as shown in figure.

image

Relation between  \(T_1\) and \(T_2\)

At Pulley 2

\(T_1-2T_2=0\)

\(T_1=2T_2\)

image

Representation of forces

image

F.B.D for block A

By Newton's 2nd law,

\(2T-m_1g=m_1a_1\)

\(2T-2×10=2×a_1\)....(i)

\(2T-20=2a_1\)

image

F.B.D for block B

By Newton's 2nd law,

\(m_2g-T=m_2a_2\)

\(10-T=a_2\)   ....(ii)

image

F.B.D for block C

By Newton's 2nd law,

\(T-m_3g=m_3a_3\)

\(T-20=2a_3\) ....(iii)

image

To calculate \(T,a_1,a_2\) and \(a_3\), three equations are insufficient. So, the fourth equation will come from constraint relation.

Applying virtual work method, 

for block A

\(\Rightarrow 2\; T a_1\)

image

For block B 

\(\Rightarrow - Ta_2\)

image

For block C

\(\Rightarrow Ta_3\)

image

According to virtual work method,

\(\Sigma Ta = 0\)

\(2Ta_1 - Ta_2 + Ta_3 = 0\)

\(2a_1 - a_2 +a_3 = 0\).............(iv)

Put values of  \(a_1,a_2\) and \(a_3\) from equation (i), (ii) and (iii) in equation (iv)

\(2a_1-a_2+a_3=0\)

\((2T-20)-(10-T)+\left(\dfrac{T-20}{2}\right)=0\)

\(\Rightarrow T=\dfrac{80}{7}N=11.42\,N\)

Put values of \((T)\) in equation (i), (ii) and (iii) and then by solving, we get

\(a_1=\dfrac {10}{7}\,m/s^2\)

\(a_2=-\dfrac{10}{7}\,m/s^2\)

\(a_3=-\dfrac{30}{7}\,m/s^2\)

In the given arrangement, the masses of three blocks \(A,\,B,\,C\) are \(m_1=2\,kg\), \(m_2=1\,kg\) and  \(m_3=2\,kg\) respectively. Find the acceleration of three bodies.  

image
A

\(6m/s^2\;up,\,\,\, 6m/s^2\;up,\,\,\,2m/s^2\;down\)

.

B

\(\dfrac{40}{7}m/s^2\;up,\,\,\, \dfrac{60}{7}m/s^2\;down,\,\,\,\dfrac{30}{7}m/s^2\;up\)

C

\(5m/s^2\;up, \,\,\,3m/s^2\;up,\,\,\,2m/s^2\;down\)

D

\(\dfrac{10}{7}m/s^2\;up, \,\,\,\dfrac{10}{7}m/s^2\;up,\,\,\,\dfrac{30}{7}m/s^2\,down \)

Option D is Correct

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