Name: Pulley-Block System
Uploaded: 2019-07-19
Duration: 10 min 30 s
Description: We must notice the pulley block system to lift, move anything in real life. Do you know how we can estimate the forces and mechanism? Let us analyse this in...

Question

A chain consisting of five links each with mass $100 gm$ is lifted vertically with constant acceleration of $2 m s^{- 2}$ $(↑)$ as shown. Find $(g = 10 m s^{- 2})$

(a) the forces acting between adjacent links.

(b) the force $F$ exerted on the top link by the agent lifting the chain.

(c) the net force on each link.

Accepted Answer

Given,

Mass of each link in the chain, $m = 100 gm = 0.1 kg$ ,

Acceleration of the whole chain or all the links, $a = 2 m s^{- 2}$ and

Acceleration due to gravity, $g = 10 m s^{- 2}$ .

Solution for (a),

From the free-body diagram shown above,

the force between the last two links $(T_{4})$ is

$T_{4} - m g = m a$

$\Rightarrow T_{4} = m g + m a$

$\Rightarrow T_{4} = 0.1 (10 + 2) = 1.2 N$

The force between the third and the fourth link $(T_{3})$ is

$T_{3} - (T_{4} + m g) = m a$

$\Rightarrow T_{3} = T_{4} + m g + m a$

$\Rightarrow T_{3} = 1.2 + 0.1 (10 + 2)$

$\Rightarrow T_{3} = 2.4 N$

The force between the second and the third link $(T_{2})$ is

$T_{2} - (T_{3} + m g) = m a$

$\Rightarrow T_{2} = T_{3} + m g + m a$

$\Rightarrow T_{2} = 2.4 + 0.1 (10 + 2)$

$\Rightarrow T_{2} = 3.6 N$

the force between the first and the second link $(T_{1})$ is

$T_{1} - (T_{2} + m g) = m a$

$\Rightarrow T_{1} = T_{2} + m g + m a$

$\Rightarrow T_{1} = 3.6 + 0.1 (10 + 2)$

$\Rightarrow T_{1} = 4.8 N$

Solution for (b),

From the free-body diagram of the top(first) link,

$F - (T_{1} + m g) = m a$

$\Rightarrow F = T_{1} + m g + m a$

$\Rightarrow F = 4.8 + 0.1 (10 + 2)$

$\Rightarrow F = 6 N$

Solution for (c),

All the links accelerate at the rate, $a = 2 m s^{- 2}$ .

Therefore, the net force on each link is

$F_{net} = m a$

$\Rightarrow F_{net} = (0.1) (2)$

$\Rightarrow F_{net} = 0.2 N$ .

Important Questions on Laws of Motion

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A chain consisting of five links each with mass 100 gm is lifted vertically with constant acceleration of 2 m s-2 ↑ as shown. Find (g=10 m s-2) (a) the forces acting between adjacent links. (b) the force F exerted on the top link by the agent lifting the chain. (c) the net force on each link.

Important Questions on Laws of Motion

Learn and Prepare for any exam you want !