← Back to Laws of Motion

Constrained Motion

Question 1:
For the arrangement of pulleys shown in figure the effort (P) required to raise the given load (W) is

200 N

250 N

800 N

1000 N

View Solution

Solution:

Net Force acting on the block in upward direction is 4T which will balance weight of the object. So,

⇒ 4T= 800 N

⇒T = 200 N

Discuss this question

Question 2:
What are the acceleration of the blocks A and B, shown in figure (in m/sec²)

0,0

4 m/s², 4 m/s²

6 m/s², 3 m/s²

4 m/s², 2 m/s²

View Solution

Solution:

Writing equations of motion for the objects as,

40 - T = 4 × 2a ---(i)

and 2T -40 = 4a -----(ii)

Solving (i) and (ii)

a= 2 m/s²

So, acceleration of object A is 4 m/s² and for object B is 2 m/s²

Discuss this question

Question 3:
The masses of the bodies A and B in figure are 20 kg and 10 kg, respectively. They are initially at rest on the floor and are connected by a weightless string passing over a weightless and frictionless pulley. An upward force F is applied to the pulley. Find the acceleration a1 of body A and a2 of body B when F is 340 N :

a1 = 7 m/sec², a2 = 24 m/sec²

a1 = 0, a2 = 0

a1 = 1.5 m/sec², a2 = 7 m/sec²

a1 = 0 m/sec², a2 = 7 m/sec²

View Solution

Solution:

As F= 340 N. Force acting on object A in upward direction is F/2 i.e 170 N. This force is insufficient to pull the object upwards. (weight of A is greater than 170 N)

So, a1= 0 m/s²

Force acting on B in upward direction is 170 N and weight is 10g or 100 N. Net force in upward direction is 70 N. so,

a2= 7 m/s²

Discuss this question

Question 4:
A painter is raising himself and the crate on which he stands with an acceleration of 5m/s² by a massless rope–and–pulley arrangement. Mass of painter is 100 kg and that of the crate is 50 kg. If g = 10 m/s², then :

Tension in the rope 2250 N

Tension in the rope is 1125 N

Force of contact between painter and the floor is 750 N

Force of contact between the painter and the floor is 500 N.

View Solution

Solution:

Let us assume tension force in the wire is T. Then net force acting on man and crate in upward direction is 2T. Gravitational force acting in downward direction is (50+100) g = 1500 N. Writing equation of motion for the system.

2T- 1500 = 150 × a

⇒ 2T- 1500 = 150 × 5

⇒ 2T = 750 +1500 = 2250

⇒ T = 1125 N

Discuss this question

Question 5:
The block A is moving downward with constant velocity v0. Find the velocity of the block B, when the string makes an angle θ with the horizontal.

v0 cos θ

v0 / cos θ

v0 sin θ

View Solution

Solution:

When two objects are connected by a string component of velocity along the string remains equal.

v₁ cosθ = v ₀

⇒v₁= v₀/ cosθ

Discuss this question

Question 6:
The force F needed to keep the block at equilibrium in given figure is (pulley and string are massless)

Mg/5

Mg/4

Mg/2

Mg/3

View Solution

Solution:

For Equilibrium,

2F = Mg ⇒ F = Mg/2

Discuss this question

Question 7:
In the arrangement, shown in figure, pulleys A and B are massless and frictionless and threads are ideal. Block of mass m1 will remain at rest if:

\[ \frac{1}{m_{3}}=\frac{2}{m_{2}}+\frac{3}{m_{1}} \]

\[ m_{1}= m_{2}= m_{3} \]

\[ \frac{4}{m_{1}}=\frac{1}{m_{2}}+\frac{1}{m_{3}}\]

\[ \frac{1}{m_{1}}=\frac{1}{m_{2}}+\frac{1}{m_{3}}\]

View Solution

Solution:

nlm constrained motion

In the movable pulley system, tension in the string connecting m2 and m3 is:

$T = \frac{2 m_2 m_3 g}{m_2 + m_3}$

Since this tension acts twice to balance

$m_1$

, we equate:

$2T = m_1 g \Rightarrow \frac{4 m_2 m_3 g}{m_2 + m_3} = m_1 g$

Cancelling

$g$

and rearranging gives:

$\boxed{ \frac{4}{m_{1}} = \frac{1}{m_{2}} + \frac{1}{m_{3}} }$

Discuss this question

Question 8:
If acceleration of block m1 is a downward then acceleration of block m2 will be:

2a upward

a upward

a/2 upward

2a downward

View Solution

Solution:

For ideal pulleys, the product of tension and acceleration remains constant:

$T_1 a = T_2 a_2$

Since $T_2 = 2T_1$ , substitute to get:

$T_1 a = 2T_1 a_2 \Rightarrow a_2 = \frac{a}{2}$

$\boxed{\text{So, } M_2 \text{ accelerates upward with } \frac{a}{2}}$

Discuss this question

Question 9:
What is the minimum value of F needed so that block begins to move upward on frictionless incline plane as shown?

Mg tan(θ/2)

Mg cot(θ/2)

Mg.sinθ/(1+sinθ)

Mgsin(θ/2)

View Solution

Solution:

🔹 Step-by-step:

Block of mass M is on a frictionless incline of angle $\theta$
Force $F$ is applied via a pulley system, split into two components:
- One acts up along the incline: F
- One acts horizontally, which when resolved along the incline becomes: F $F \cos \theta$

Total upward force along incline = $F + F \cos \theta$

Downward component of weight = $Mg \sin \theta$

🔹 For the block to just start moving upward:

$F + F \cos θ = M g \sin θ$

$F (1 + \cos θ) = M g \sin θ$

$F = \frac{Mg \sin \theta}{1 + \cos \theta}$

Now use the identity:

$\frac{\sin \theta}{1 + \cos \theta} = \cot\left(\frac{\theta}{2}\right)$

Final Answer:

$\boxed{F = Mg \cot\left(\frac{\theta}{2}\right)}$

Discuss this question