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Expert-verified Found in: Page 117 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Figure 5.33 shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 98 N on the wall to which it is attached. The tensions in the three shorter cords are${{\mathbit{T}}}_{{\mathbf{1}}}$${\mathbf{=}}{\mathbf{58}}{\mathbf{.}}{\mathbf{8}}{\mathbf{N}}{\mathbf{,}}{{\mathbit{T}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{49}}{\mathbf{.}}{\mathbf{0}}{\mathbf{N}}$and ${{\mathbit{T}}}_{\mathbf{3}\mathbf{}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{8}}{\mathbf{N}}$. (a) What is the mass of disk A (b) What is the mass of disk B, (c) What is the mass of disk C, and (d) What is the mass of disk D? a) Mass of Disk A is $4.0\mathrm{kg}$.

b) Mass of Disk B is $1.0\mathrm{kg}$.

c) Mass of Disk C is $4.0\mathrm{kg}$.

d) Mass of Disk D is $1.0\mathrm{kg}$.

See the step by step solution

## Step 1: The given data

1. The magnitude of force is, ${F}_{\mathrm{Pull}}=98\mathrm{N}$.
2. The tensions in three strings is,${T}_{1}$$=58.8\mathrm{N}$,${T}_{2}=49.0\mathrm{N}$,${T}_{3}=9.8\mathrm{N}$ .
3. The acceleration due to gravity is,$g=9.8\mathrm{m}/{\mathrm{s}}^{2}$ .

## Step 2: Understanding the concept of tension and weight

The tension will be equal to the weight of the mass attached to that string. With this, we can find the masses of all disks. Since the system is in equilibrium, the net force on all the disks is zero.

Formula:

Weight of the block, $W=mg$ (i)

Where m is the mass of disk, g is the acceleration due to gravity.

Tension in the string balances the weight of the block, $T-w=0$ (ii)

## Step 3: (d) Calculation for mass of disk D

As T3 has attached to only disk D, so from equation (ii), we get

${T}_{3}={m}_{D}g$

Substitute the values in the above equation.

$9.8\mathrm{N}={m}_{D}×9.8\mathrm{m}/{\mathrm{s}}^{2}$

${m}_{D}=1.0\mathrm{kg}$${m}_{D}=1.0\mathrm{kg}$

Hence, the value of mass of disk D is $1.0\mathrm{kg}$

## Step 4: (c) Calculation for mass of disk C

As, T2 has attached to disk C and D, so from equation (ii), we get

${T}_{2}=\left({m}_{c}+{m}_{D}\right)g\phantom{\rule{0ex}{0ex}}49\mathrm{N}=\left({\mathrm{m}}_{\mathrm{c}}+1\mathrm{kg}\right)×9.8\mathrm{m}/{\mathrm{s}}^{2}\left(\because {m}_{\mathit{D}}=1.0\mathrm{kg}\right)\phantom{\rule{0ex}{0ex}}\frac{49}{9.8}={m}_{c}+1\phantom{\rule{0ex}{0ex}}{m}_{c}+1=5\phantom{\rule{0ex}{0ex}}{m}_{c}=4.0\mathrm{kg}$

## Step 5: (b) Calculation for mass of disk B

As T1 has attached to disk B, C and D, so from equation (ii), we get

${T}_{1}=\left({m}_{B}+{m}_{C}+{m}_{D}\right)g\phantom{\rule{0ex}{0ex}}58.8\mathrm{N}=\left({\mathrm{m}}_{\mathrm{B}}+4\mathrm{kg}+1\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^{2}\left(\because {m}_{D}=1.0\mathrm{kg}&{m}_{c}=4.0\mathrm{kg}\right)\phantom{\rule{0ex}{0ex}}\left({m}_{B}\mathit{+}\mathit{5}\right)=6\phantom{\rule{0ex}{0ex}}{\mathrm{m}}_{\mathrm{B}}=1.0\mathrm{kg}$

Hence, the value of mass of disk B is $1.0\mathrm{kg}$

## Step 6: (a) Calculation of mass of disk A

As the pulling force given to us is 98 N. So that would be equal to the sum of weights of all disks.

${F}_{pull}=\left({m}_{A}+{m}_{B}+{m}_{C}+{m}_{D}\right)g\phantom{\rule{0ex}{0ex}}98\mathrm{N}=\left({m}_{A}+1\mathrm{kg}+4\mathrm{kg}+1\mathrm{kg}\right)×9.8\mathrm{m}/{\mathrm{s}}^{2}\left(\because {m}_{B}=1.0\mathrm{kg},{m}_{D}=1.0\mathrm{kg}&{\mathrm{m}}_{\mathrm{C}}=4.0\mathrm{kg}\right)\phantom{\rule{0ex}{0ex}}\left({m}_{A}+6\right)=10\phantom{\rule{0ex}{0ex}}{m}_{A}=4.0\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{disk}\mathrm{A}\mathrm{is}4.0\mathrm{kg}$ ### Want to see more solutions like these? 