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Found in: Page 147

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68 , and the coefficient of kinetic friction is 0.56 . In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) 222 N , (b) 334 N , (c) 445 N , and (d) 556 N . For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

(a) The magnitude of the frictional force is f = 222 N .

(b) The magnitude of the frictional force is f = 334 N .

(c) The magnitude of the frictional force is f = 311 N .

(d) The magnitude of the frictional force is f = 311 N .

See the step by step solution

Step 1: Given data:

Weight of the cabinet, W = 556 N

Coefficient of static friction between the cabinet and the floor, ${\mu }_{s}=0.68$

Coefficient of kinetic friction, ${\mu }_{k}=0.56$

Step 2: Understanding the concept:

In order to move a filing cabinet, the force applied must be able to overcome

the frictional force.

Apply Newton’s second law

${{\mathbf{F}}}_{{\mathbf{push}}}{\mathbf{-}}{\mathbf{f}}{\mathbf{=}}{{\mathbf{F}}}_{{\mathbf{net}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{=}}{\mathbf{ma}}$

If you find the applied force ${{\mathbit{F}}}_{\mathbf{p}\mathbf{u}\mathbf{s}\mathbf{h}}$ to be less than ${{\mathbit{f}}}_{\mathbf{s}\mathbf{,}\mathbf{max}}$ , the maximum static frictional force, our conclusion would then be “no, the cabinet does not move” (which means is actually zero and the frictional force is simply ${\mathbit{f}}{\mathbf{=}}{{\mathbit{F}}}_{\mathbf{p}\mathbf{u}\mathbf{s}\mathbf{h}}$ ). On the other hand, if you obtain a > 0 then the cabinet moves (so ${\mathbit{f}}{\mathbf{=}}{{\mathbit{f}}}_{{\mathbf{k}}}$ ).

For ${{\mathbit{f}}}_{\mathbf{s}\mathbf{,}\mathbf{m}\mathbf{a}\mathbf{x}}$ and ${{\mathbit{f}}}_{{\mathbf{k}}}$ use Eq. 6-1 and Eq. 6-2 (respectively), and in those formulas set the magnitude of the normal force to the weight of the cabinet.

Step 3: Calculate the maximum static friction and kinetic friction:

The maximum static frictional force is,

${\mathrm{f}}_{\mathrm{s},\mathrm{max}}={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}$

Here, the weight of the cabinet will be balanced by the normal force. Therefore,

${\mathrm{f}}_{\mathrm{s},\mathrm{max}}={\mathrm{\mu }}_{\mathrm{s}}\mathrm{W}$

And the kinetic frictional force is,

${\mathrm{f}}_{\mathrm{k}}={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{f}}_{\mathrm{k}}\phantom{\rule{0ex}{0ex}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{W}\phantom{\rule{0ex}{0ex}}=\left(0.56\right)\left(556\mathrm{N}\right)\phantom{\rule{0ex}{0ex}}=311\mathrm{N}$

Step 4: (a) Define the magnitude of the frictional force with horizontal force of magnitude 222 N :

Calculate the magnitude of the frictional force on the cabinet from the floor when it is pushed with horizontal force of magnitudes, which is,

.${F}_{push}=222N$

Here, the magnitude of the horizontal pushing force is less than the maximum static force. Therefore,

${F}_{push}<{f}_{s,max}\phantom{\rule{0ex}{0ex}}222N<378N$

Hence, the cabinet does not move. So, acceleration of the cabinet will be zero.

$a=0m/{s}^{2}$

The frictional force is,

${F}_{push}-f=ma\phantom{\rule{0ex}{0ex}}=m\left(0m/{s}^{2}\right)\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}f={F}_{push}\phantom{\rule{0ex}{0ex}}=222N$

Here, the magnitude of the frictional force between the carbonate and the floor is f = 222N .

Step 5: (b) Determine the magnitude of the frictional force with horizontal force of magnitude 334 N :

Calculate the magnitude of the frictional force on the cabinet from the floor when it is pushed with horizontal force of magnitudes, which is

${F}_{push}=334N$

Here, the magnitude of the horizontal pushing force is less than the maximum static force. Thus,

${F}_{push}<{f}_{s,max}\phantom{\rule{0ex}{0ex}}334N<378N$

Hence, the cabinet does not move. So, acceleration of the cabinet will be zero.

$a=0m/{s}^{2}$

The frictional force is,

${F}_{push}-f=ma\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}f={F}_{push}\phantom{\rule{0ex}{0ex}}=334N$

Step 6: (c) Define the magnitude of the frictional force with horizontal force 445 N :

Calculate the magnitude of the frictional force on the cabinet from the floor when it is pushed with horizontal force of magnitudes, which is,

${F}_{push}=445N$.

Here, the magnitude of the horizontal pushing force is less than the maximum static force. Thus,

${F}_{push}>{f}_{s,max}\phantom{\rule{0ex}{0ex}}445N>378N$

Hence, the cabinet will move. So, the frictional force will be he kinetic friction. Hence, the friction force in this case between the cabinet and the floor is,

$f={F}_{k}\phantom{\rule{0ex}{0ex}}=311N$

Step 7: (d) The magnitude of the frictional force with horizontal force 556 N :

Calculate the magnitude of the frictional force on it from the floor when it is pushed with horizontal force of magnitudes, which is,

${F}_{push}=556N$.

Again, you have

${F}_{push}>{f}_{s,max}\phantom{\rule{0ex}{0ex}}556N>378N$

Which means the cabinet moves.

Hence, the cabinet will move. So, the frictional force will be he kinetic friction. Hence, the friction force in this case between the cabinet and the floor is,

$f={f}_{k}\phantom{\rule{0ex}{0ex}}=311N$

Step 8: (e) Find out in which of the attempts does the cabinet move:

As in part (c) and (d) you have ${F}_{push}>{f}_{s,max}$ which means the cabinet moves.

Hence, the cabinet moves in (c) and (d).