# Pulleys  (Page 5/5)

 Page 5 / 5

## Exercises

In the arrangement shown, the acceleration of block “A” is 1 $m/{s}^{2}$ . Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

(a) The acceleration of block “B” is 1 $m/{s}^{2}$ .

(b) The acceleration of block “B” is 2 $m/{s}^{2}$ .

(c) The acceleration of block “B” is 0.5 $m/{s}^{2}$ .

(d) The acceleration of block “B” is 0.75 $m/{s}^{2}$ .

We are required to determine acceleration of block "B" to answer this question. By inspection, we observe that if block “A” moves “x” distance towards right, then the string length is distributed in two equal halves about hanging pulley. Clearly, if “A” moves by “x”, then “B” moves by x/2. As such, acceleration of “A” is twice that of “B”. As it is given that acceleration of “A” is 1 $m/{s}^{2}$ , the acceleration of “B” is 2 $m/{s}^{2}$ .

Hence, option (b) is correct.

Given : mass of block “A” = 0.5 kg and acceleration of pulley = 1 $m/{s}^{2}$ . Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

(a) The force “F” is 1 N.

(b) The force “F” is 2 N.

(c) The tension in the string is 1 N.

(d) The tension in the string is 0.5 N.

Let the acceleration of block “A” is “a”. We know that the acceleration of block “A” is two times that of pulley. Hence,

$a=1X2=2\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Let the tension in the string be “T”. Considering forces on block “A”, we have :

$T=ma=0.5X2=1\phantom{\rule{1em}{0ex}}N$

Considering forces on the mass-less pulley, we have :

$⇒F=2T=2X1=2\phantom{\rule{1em}{0ex}}N$

Hence, options (b) and (c) are correct.

In the arrangement shown in the figure, accelerations of blocks A,B and C are ${a}_{A}$ , ${a}_{B}$ and ${a}_{C}$ respectively. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then

$\mathrm{\left(a\right)}\phantom{\rule{2pt}{0ex}}2{a}_{A}+2{a}_{B}+{a}_{C}=0$

$\mathrm{\left(b\right)}\phantom{\rule{2pt}{0ex}}{a}_{A}+{a}_{B}+2{a}_{C}=0$

$\mathrm{\left(c\right)}\phantom{\rule{2pt}{0ex}}{a}_{A}+{a}_{B}+{a}_{C}=0$

$\mathrm{\left(d\right)}\phantom{\rule{2pt}{0ex}}2{a}_{A}+{a}_{B}+{a}_{C}=0$

Here, we need to determine relation of accelerations of blocks using constraint of string length.

${x}_{A}+{x}_{A}-{x}_{0}+{x}_{B}-{x}_{0}+{x}_{B}-{x}_{0}+{x}_{C}-{x}_{0}=L$ $⇒2{x}_{A}+2{x}_{B}+{x}_{C}=L+3{x}_{0}=\text{constant}$

Differentiating two times, we have :

$⇒2{a}_{A}+2{a}_{B}+{a}_{C}=0$

Hence, option (a) is correct.

Given : mass of block “A” = 10 kg, spring constant = 200 N/m, extension in the string = 0.1 m. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

(a) The force “F” is 40 N.

(b) The acceleration of block A is 4 $m/{s}^{2}$ .

(c) The acceleration of pulley is 1 $m/{s}^{2}$

(d)The tension in the string is 10 N.

Let “T” be the tension in the string. Here, spring force is equal to tension in the string.

$T=kx=200X0.1=20\phantom{\rule{1em}{0ex}}N$

Let acceleration of block “A” is “a”. Considering forces on block “A”, we have :

$⇒T=ma=10Xa$

$⇒a=\frac{T}{10}=\frac{20}{10}=2\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Considering forces on the mass-less pulley, we have :

$⇒F=2T=2X20=40\phantom{\rule{1em}{0ex}}N$

Now, acceleration of pulley is half of the acceleration of block. Hence, acceleration of pulley is 1 $m/{s}^{2}$ .

Hence, options (a) and (c) are correct.

In the arrangement shown, the mass of “A” and “B” are 1 kg and 2 kg respectively. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

(a) The tension in the string connected to “A” is 1 N.

(b) The tension in the string connected to “A” is 20/3 N.

(c) The tension in the string connected to “B” is 2 N.

(d) The tension in the string connected to “B” is 20/3 N.

Let the tensions in the strings connected to blocks “A” and “B” be “ ${T}_{1}$ ” and “ ${T}_{2}$ ” respectively. We see here that only force making block “A” to move on the table is tension in the string connected to it. Let acceleration of block “A” is “a”. From constraint relation, we know that the acceleration of “A” is twice the acceleration of “B”. Thus, acceleration of block “B” is “a/2”. Applying law of motion for block “A”, we have :

$⇒{T}_{1}={m}_{1}Xa=1Xa=a$

Consideration of mass-less pulley,

$⇒{T}_{2}=2{T}_{1}=2Xa=2a$

Applying law of motion for block “A”, we have :

$⇒{m}_{2}g-{T}_{2}={m}_{2}X\frac{a}{2}$ $⇒2X10-2a=2X\frac{a}{2}=a$ $⇒a=\frac{20}{3}\phantom{\rule{2pt}{0ex}}m/{s}^{2}$

Putting this value in the expressions of tensions, we have :

$⇒{T}_{1}=a=\frac{20}{3}\phantom{\rule{2pt}{0ex}}N$ $⇒{T}_{2}=2a=\frac{40}{3}\phantom{\rule{2pt}{0ex}}N$

Hence, options (b) is correct.

Two blocks of mass “m” and “nm” are hanging over a pulley as shown in the figure. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. If n>1 and acceleration of the blocks is g/3, then value of “n” is :

$\left(a\right)\frac{3}{2}\phantom{\rule{1em}{0ex}}\left(b\right)\frac{5}{3}\phantom{\rule{1em}{0ex}}\left(c\right)3\phantom{\rule{1em}{0ex}}\left(d\right)2$

We consider blocks and string as one system. The net external force on the system is

$F=nmg-mg=\left(n-1\right)mg$

Total mass of the system, M, is :

$M=nm+m=\left(n+1\right)m$

The acceleration of the system i.e. the acceleration of either block is :

$⇒a=\frac{\left(n-1\right)mg}{\left(n+1\right)m}=\frac{\left(n-1\right)g}{\left(n+1\right)}$

According to question,

$⇒a=\frac{\left(n-1\right)g}{\left(n+1\right)}=\frac{g}{3}$ $⇒3n-3=n+1$ $⇒2n=4$ $⇒n=2$

Hence, option (d) is correct.

Given : mass of block “A” = 0.5 kg, mass of block “B”=1 kg and F = 2.5 N. Consider string and pulley to be mass-less and friction absent everywhere in the arrangement. Then,

(a) The acceleration of block “B” is 1 $m/{s}^{2}$ .

(b) The acceleration of block “A” is 1 $m/{s}^{2}$ .

(c) The tension in the string connecting block “B” and pulley is 1 N.

(d) The tension in the string connected to block “A” is 2 N.

Let the acceleration of block “B” is “a”. Considering forces on block of mass “B”, we have :

$F-{T}_{1}=Ma$ $⇒2.5-{T}_{1}=1Xa=a$

Considering free body diagram of mass-less pulley,

$⇒{T}_{1}=2{T}_{2}$

The pulley and the block are connected by inextensible string. As such, their accelerations are same. Further, we know by constraint analysis that acceleration of block is twice that of pulley. Hence, acceleration of block “B” is “2a”. Considering forces on block “A”, we have :

$⇒{T}_{2}=mX2a=0.5X2a=a$

Thus,

$⇒{T}_{1}=2Xa=2a$

Putting this value in the force equation of block “A”,

$2.5-2a=0.5a$ $⇒a=1\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Clearly,

$⇒{T}_{1}=2a=2X1=2\phantom{\rule{1em}{0ex}}N$ $⇒{T}_{2}={T}_{1}/2=1\phantom{\rule{1em}{0ex}}N$

Hence, option (a) is correct.

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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