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Simple Applications Of 2nd Law

Learn newton’s second law of motion examples, tension in a rope when a man climbs on it with some acceleration and apparent weight in an elevator. Practice to find the tension in the string connecting the two blocks.

Newton's Second Law

  • According to Newton's second law a body of mass m, subjected to force \(\vec F_1,\vec F_2,\vec F_3\) ......will under go an acceleration \(\vec a\) which is given by 

          \(\vec a = \dfrac{\vec F_{net}}{m}\)

  •  Net force : \(\vec F_{net} = \vec F_1 + \vec F_2+\vec F_3+......\)

      \(\vec a =\) acceleration of body 

       \(m\) = mass of the object in kg

  • Since, \(m\) is always positive thus, the direction of acceleration is same as that of the direction of net force (\(\vec F_{net}\)).
  • It concludes \(\vec a \propto\vec F_{net}\).

  • Consider a body of mass m on which force F is acting as shown in figure.
  • Assume body is placed on a smooth surface. 
  • According to Newton's second law

\(\Sigma F = ma\)

\(\vec F = ma\)

Acceleration, \(a = \dfrac{\vec F}{m}\)

Illustration Questions

A block of mass m = 2 kg is pulled by a force F = 20 N, as shown in figure. Assuming that the block is placed on a smooth surface, calculate the acceleration of the block.

A 20 m/s2

B 10 m/s2

C 1 m/s2

D 12 m/s2

×

Net force on the body 

\(\Sigma F = 20 \,N\)

image

Applying Newton's second law

\(\Sigma F = ma\)

20 = 2 × a

a = 10 m/s2

image

A block of mass m = 2 kg is pulled by a force F = 20 N, as shown in figure. Assuming that the block is placed on a smooth surface, calculate the acceleration of the block.

image
A

20 m/s2

.

B

10 m/s2

C

1 m/s2

D

12 m/s2

Option B is Correct

Calculation of Acceleration when Force is Acting on the Body at some Angle 

  • Consider a body of mass \(m\), which is placed on a smooth surface.
  • A force is acting on the body at an angle \(\theta\) with the horizontal, as shown in figure.

By visual inspection, acceleration is in horizontal direction.

  • There is no motion in vertical direction,

         \(\therefore \, N + F \,sin\theta = mg\)

          \(\Sigma F_y=0\)

\(\Sigma F_x =F \,cos\theta\)

Acceleration,  \(a= \dfrac{\Sigma F_x}{m}\)

 \(a= \dfrac{Fcos\theta}{m}\)

Illustration Questions

A block of mass \(m = 100 \,kg\), is pulled by a force \( F = 100 \,N\), which is applied at an angle \(\theta=\) \(60^\circ\). Calculate the acceleration of the block for the system shown in figure. 

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

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

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

D \(\dfrac{3}{4}\) \(m/s^2\)

×

By visual inspection, we can see that acceleration is in horizontal direction.

FBD of block A

image

Net force in horizontal direction 

\(\Sigma F _x = T cos \theta \)

Net force in vertical direction 

\(\Sigma F_y = 0\)

[Since, there is no motion in vertical direction ]

Applying Newton's second law 

\(\Sigma F _x = ma\)

\(a = \dfrac{T cos\theta }{m}\)

100 × cos 60º = 100 × a 

\(a= \dfrac{1}{2} m/s^2\)

A block of mass \(m = 100 \,kg\), is pulled by a force \( F = 100 \,N\), which is applied at an angle \(\theta=\) \(60^\circ\). Calculate the acceleration of the block for the system shown in figure. 

image
A

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

.

B

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

C

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

D

\(\dfrac{3}{4}\) \(m/s^2\)

Option B is Correct

Calculation of Normal Force between Two Blocks

  • Consider two blocks A and B which are pushed by a force F, as shown in figure.
  • The hand H pushes block A, A pushes back with same force i.e. \(\vec F_{H \, on\, A} = \vec F_{A \, on \,H}\), thus it is action reaction pair.
  • \(\vec F_H\) is not directly applied on B.
  • Only A interacts with B.

\( F_{A \, on\, B} = F_{B \, on \,A}\)      (Action - reaction pair)

  • As blocks are accelerating to the right, there must be net force towards right side.

  • Free body diagrams

\(\vec F_{A \,on \, H} = \vec F_{H \,on \, A} > \vec F_{B \,on \, A} = \vec F_{A \,on \, B}\)

Illustration Questions

Two blocks A and B of masses m1 and m2 respectively are placed on a frictionless surface and a force F is applied on them, as shown in figure. Calculate the normal force between the blocks. 

A \(N = \dfrac{m_2 F}{m_1 + m_2}\)

B \(N = \dfrac{m_1 F}{m_1 + m_2}\)

C \(N = \dfrac{ F}{m_1 + m_2}\)

D \(N = \dfrac{m_1 \,m_2}{F}\)

×

Free body diagrams 

image

Applying Newton's second law 

\(F - N = m_1 a \)     .......(1)

\(N= m_2 a\)       ........(2)

\(F - m_2a = m_1a\)

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

The normal force between the blocks is,

\(N = \dfrac{m_2 F}{m_1 + m_2}\)

Two blocks A and B of masses m1 and m2 respectively are placed on a frictionless surface and a force F is applied on them, as shown in figure. Calculate the normal force between the blocks. 

image
A

\(N = \dfrac{m_2 F}{m_1 + m_2}\)

.

B

\(N = \dfrac{m_1 F}{m_1 + m_2}\)

C

\(N = \dfrac{ F}{m_1 + m_2}\)

D

\(N = \dfrac{m_1 \,m_2}{F}\)

Option A is Correct

Tension in a String Between Two Blocks

  • Consider two blocks A and B of masses m1 and m2 respectively, which are attached with a string as shown in figure.
  • Block B is now pulled with some force F.
  • As both the blocks are attached to a string, the acceleration of both the blocks will be the same.

Free body diagrams

  • Net force along horizontal direction for block A 

\(\Sigma F _x = T\)

Applying Newton's second law

\(\Sigma F _x = m_1a\)

\(T = m_1a\)      ........(1)

  • Net force along horizontal direction for block B

\(\Sigma F _x = F-T\)

Applying Newton's second law

\(\Sigma F _x = m_2a\)

\(F-T = m_2a\)      ........(2)

\(F- m_1a = m_2a\)

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

Thus, tension is given by,

\(T = \dfrac{m_1F}{m_1+m_2}\)

Illustration Questions

Consider two blocks A and B which are attached with a string and placed over a smooth surface. This system is pulled by a force F = 150 N. If masses of block A and block B are m1 = 5 kg and m2 = 10 kg, respectively, then find the tension in the string.

A 80 N 

B 40 N 

C 15 N

D 50 N 

×

Net force along horizontal direction for block A 

\(\Sigma F _x = T\)

Applying Newton's second law

\(\Sigma F _x = m_1a\)

\(T = m_1a\)      ........(1)

image

Net force along horizontal direction for block B

\(\Sigma F _x = F-T\)

Applying Newton's second law

\(\Sigma F _x = m_2a\)

\(F-T = m_2a\)      ........(2)

image

\(F- m_1a = m_2a\)

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

Thus, tension is given by

\(T = \dfrac{m_1F}{m_1+m_2}\)

\(T = \dfrac{5× 150}{15} = 50 \, N\)

image

Consider two blocks A and B which are attached with a string and placed over a smooth surface. This system is pulled by a force F = 150 N. If masses of block A and block B are m1 = 5 kg and m2 = 10 kg, respectively, then find the tension in the string.

image
A

80 N 

.

B

40 N 

C

15 N

D

50 N 

Option D is Correct

Concept of Reading of Weighing Machine in an Elevator 

Weight and apparent weight 

  • The weighing machine reads the normal force exerted on the surface on top of the machine. 
  • If mass of the man is m.
  • Then using Newton's second law

            \(N = mg \) 

  • Consider a man is standing on a weighing machine inside an accelerating elevator, as shown in figure.

  • Free body diagram of man 

  • Net force in vertical direction 

           \(\Sigma F_y = N -mg\)

  • Applying Newton's second law

         \(N - mg = ma\)  

        \(N= mg + ma\)

      \(N= m(a+ g)\)

  • Here, N >mg, the weight shown by the weighing machine will be greater than actual weight.
  • This is called apparent weight which is greater or lesser than actual weight.

Illustration Questions

Sara is standing on weighing machine which is kept inside a deaccelerating lift having acceleration a, as shown in figure. If mass of Sara is m, then calculate the reading of weighing machine. (Given : g= 10 m/s2)  

A ma

B m(g+a)

C mg

D m(g – a)

×

Free body diagram of Sara

image

Net force in vertical downward direction 

\(\Sigma F _y = mg - N\)

Applying Newton's second law 

mg – N = ma 

N =  m(g – a)

Sara is standing on weighing machine which is kept inside a deaccelerating lift having acceleration a, as shown in figure. If mass of Sara is m, then calculate the reading of weighing machine. (Given : g= 10 m/s2)  

image
A

ma

.

B

m(g+a)

C

mg

D

m(g – a)

Option D is Correct

Reading of Spring Balance in an Accelerating Elevator

  • Sometimes, light spring balance is used to measure weight.
  • This spring balance is assumed to be massless.
  • To calculate the reading of a spring balance, replace spring balance with a string.
  • The value of tension in the string will be the reading of spring balance.
  • Consider a spring balance which is placed in an accelerating elevator, as shown in figure.
  • This spring balance is used to measure mass m.

  • First, replace spring balance with a string.

  • Free body diagram of the block

  • Net force in vertical direction 

\(\Sigma F_y = T - mg\)

T – mg = ma       [Apply Newton's second law]

T = ma + mg

T = m (a + g)

T > mg

Illustration Questions

What will be the reading of a spring balance when the mass m is measured in a de-accelerating elevator having acceleration a, as shown in figure? (given: g=10 m/s2)

A ma

B m (g + a )

C m (g – a )

D mg

×

Replace the spring balance with a string, as shown in figure.

image

Free body diagram of block 

image

 Applying Newton's second law, net force in vertical direction 

mg – T = ma

mg – ma = T 

T = m (g – a )

Reading of spring balance will be

T = m (g – a )

What will be the reading of a spring balance when the mass m is measured in a de-accelerating elevator having acceleration a, as shown in figure? (given: g=10 m/s2)

image
A

ma

.

B

m (g + a )

C

m (g – a )

D

mg

Option C is Correct

Tension in a Heavy Rope Pulled by some Force Horizontally 

  • A heavy rope of length \(\ell\) and mass \(m\) is pulled by a force \(F\) horizontally, as shown in figure.
  • Consider the surface as frictionless.
  • The tension in a rope is to be calculated at distance \(x\) from the end where the force is applied.
  • Let \(\lambda \) be the mass per unit length.

           \(\lambda = \dfrac{m}{\ell}\)

  • The acceleration of rope will be  

F = ma 

\(a = \dfrac{F}{m}\)

  • Each part of rope will have same acceleration. 

Applying Newton's second law 

\(\Sigma F = ma\)

 \(F - T = (\lambda x) \,a\)

(The value of acceleration \(a = \dfrac{F}{m}\) will remain same at this distance)

 \(F - T = \dfrac{m}{\ell} × \dfrac{F}{m} ×x\)

 \(F - \dfrac{F\,x}{\ell} = T \)

 \(F \left( 1-\dfrac{x}{\ell}\right) = T \)

 \(F \left( \dfrac{\ell - x}{\ell}\right) = T \)

Illustration Questions

A heavy rope of mass \(m = 5 \,kg\) is pulled by force \(F = 100 \,N\) on a frictionless surface. What will be the value of tension force at a distance \(x =\)\(2\,m\) from the end where force is applied, if total length of rope is \(\ell =\) \(4\,m\)?

A 25 N 

B 75 N 

C 50 N 

D 100 N

×

Let \(\lambda \) be the mass per unit length.

\(\lambda = \dfrac{m}{\ell}\)

image

The acceleration of rope will be  

F = ma 

\(a = \dfrac{F}{m}\)

image

Each part of rope will have same acceleration.

image

Applying Newton's second law 

\(\Sigma F = ma\)

image

 \(F - T = (\lambda x ) \,a\)

(The value of acceleration \(a = \dfrac{F}{m}\) will remain same at this distance)

 \(F - T = \dfrac{m}{\ell} × \dfrac{F}{m} x\)

 \(F - \dfrac{Fx}{\ell} = T \)

 \(F \left( 1-\dfrac{x}{\ell}\right) = T \)

 \(F \left( \dfrac{\ell - x}{\ell}\right) = T \)

image

\( T = F \left( \dfrac{\ell - x}{\ell}\right) \)

\(T = 100 \left( \dfrac{4 - 2}{4}\right) \)

\(T= 50 \,N\)

image

A heavy rope of mass \(m = 5 \,kg\) is pulled by force \(F = 100 \,N\) on a frictionless surface. What will be the value of tension force at a distance \(x =\)\(2\,m\) from the end where force is applied, if total length of rope is \(\ell =\) \(4\,m\)?

A

25 N 

.

B

75 N 

C

50 N 

D

100 N

Option C is Correct

Tension in a Rope when a Man Climbs on it with some Acceleration 

  • Consider a man of mass m, climbing up the rope with acceleration a, as shown in figure.
  • Let T be tension in the rope.

  • Free body diagram

Applying Newton's second law

T – mg = ma

T  = mg + ma

T = m (g + a)

Illustration Questions

A man is climbing on a vertical fixed rope with an acceleration a = 2 m/s2 as shown in figure. Find the tension in the string if mass of the man is m = 30 kg.          (Given : g = 10 m/s2 ) 

A 250 N

B 360 N

C 300 N

D 400 N

×

Free body diagram 

image

Applying Newton's second law

T – mg  = ma

T = mg + ma

T = m(g + a) 

image

T = 30 (10 + 2)

T = 360 N

image

A man is climbing on a vertical fixed rope with an acceleration a = 2 m/s2 as shown in figure. Find the tension in the string if mass of the man is m = 30 kg.          (Given : g = 10 m/s2 ) 

image
A

250 N

.

B

360 N

C

300 N

D

400 N

Option B is Correct

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