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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, $$a = \dfrac{\vec F}{m}$$

#### 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

$$\Sigma F = 20 \,N$$

Applying Newton's second law

$$\Sigma F = ma$$

20 = 2 × a

a = 10 m/s2

### 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

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,  $$a= \dfrac{\Sigma F_x}{m}$$

$$a= \dfrac{Fcos\theta}{m}$$

#### 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

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.

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

# 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.

#### 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

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)

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

#### 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.

Free body diagram of block

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)

A

ma

.

B

m (g + a )

C

m (g – a )

D

mg

Option C 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

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

#### 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

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.

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,

$$T = \dfrac{m_1F}{m_1+m_2}$$

#### 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)

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

$$T = \dfrac{m_1F}{m_1+m_2}$$

$$T = \dfrac{5× 150}{15} = 50 \, N$$

### 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

Option D 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)

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

$$F - \dfrac{F\,x}{\ell} = T$$

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

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

#### 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.

$$\lambda = \dfrac{m}{\ell}$$

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)

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

$$F - \dfrac{Fx}{\ell} = T$$

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

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

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

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

$$T= 50 \,N$$

### 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)

#### 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

Applying Newton's second law

T – mg  = ma

T = mg + ma

T = m(g + a)

T = 30 (10 + 2)

T = 360 N

### 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

Option B is Correct