StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q11P

Expert-verifiedFound in: Page 140

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

** A ${\mathbf{68}}{\mathbf{}}{\mathbf{kg}}$ crate is dragged across a floor by pulling on a rope attached to the crate and inclined ${\mathbf{15}}{\mathbf{\xb0}}$ above the horizontal. (a) If the coefficient of static friction is ${\mathbf{0}}{\mathbf{.}}{\mathbf{50}}$ , what minimum force magnitude is required from the rope to start the crate moving? (b) If ${{\mathbf{\mu}}}_{{\mathbf{k}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{35}}$ , what is the magnitude of the initial acceleration of the crate?**

(a) The minimum force magnitude is required from the rope to start the crate moving is 304 N.

(b) The magnitude of the initial acceleration of the crate is 1.3 m/s2.

Mass of the block, $\mathrm{m}=68\mathrm{kg}$

Coefficient of static friction, ${\mathrm{\mu}}_{\mathrm{a}}=0.50$

Coefficient of kinetic friction, ${\mathrm{\mu}}_{\mathrm{k}}=0.35$

Angle, $\mathrm{\theta}=15\xb0$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\sum _{}\mathrm{ma}$

where, F is the net force, m is mass and a is an acceleration.

Free body diagram of the block:

By using Newton’s 2nd law along y direction, block is not moving along y.So, the acceleration is 0, it gives,

$\sum _{}{\mathrm{F}}_{\mathrm{x}}=0\phantom{\rule{0ex}{0ex}}\mathrm{N}-\mathrm{mg}+\mathrm{F}\mathrm{sin}\left(15\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{N}=\mathrm{mg}-\mathrm{F}\mathrm{sin}\left(15\right)\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\mathrm{along}\mathrm{x}\mathrm{direction},\phantom{\rule{0ex}{0ex}}\sum _{}{\mathrm{F}}_{\mathrm{x}}=0\phantom{\rule{0ex}{0ex}}\mathrm{since}\mathrm{block}\mathrm{is}\mathrm{at}\mathrm{rest},\mathrm{a}=0\phantom{\rule{0ex}{0ex}}\mathrm{Fcos}\left(15\right)-{\mathrm{f}}_{\mathrm{s}}=0\phantom{\rule{0ex}{0ex}}{\mathrm{f}}_{\mathrm{s}}=\mathrm{Fcos}\left(15\right)\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{frictional}\mathrm{force}\mathrm{is},\phantom{\rule{0ex}{0ex}}{\mathrm{f}}_{\mathrm{s}}={\mathrm{\mu}}_{\mathrm{s}}\mathrm{N}={\mathrm{\mu}}_{\mathrm{s}}(\mathrm{mg}-\mathrm{Fsin}(15\left)\right)\phantom{\rule{0ex}{0ex}}\mathrm{Equating}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\phantom{\rule{0ex}{0ex}}\mathrm{Fcos}\left(15\right)={\mathrm{\mu}}_{\mathrm{s}}(\mathrm{mg}-\mathrm{Fsin}(15\left)\right)\phantom{\rule{0ex}{0ex}}\mathrm{Fcos}\left(15\right)+{\mathrm{\mu}}_{\mathrm{s}}\mathrm{Fsin}\left(15\right)={\mathrm{\mu}}_{\mathrm{s}}\mathrm{mg}\phantom{\rule{0ex}{0ex}}\mathrm{F}=\frac{{\mathrm{\mu}}_{\mathrm{s}\mathrm{mg}}}{\mathrm{cos}\left(15\right)+{\mathrm{\mu}}_{\mathrm{s}}\mathrm{sin}\left(15\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(0.50\right)\left(68\right)\left(9.81\right)}{\mathrm{cos}\left(15\right)+\left(0.50\right)\mathrm{sin}\left(15\right)}\phantom{\rule{0ex}{0ex}}=304\mathrm{N}\phantom{\rule{0ex}{0ex}}$

Therefore, the minimum force magnitude is required from the rope to start the crate moving is 304N.

By using Newton’s 2nd law along y direction,block is not moving along y.So, the acceleration is 0, it gives,

$\underset{}{\sum {\mathrm{F}}_{\mathrm{y}}=0}\phantom{\rule{0ex}{0ex}}\mathrm{N}=\mathrm{mg}+\mathrm{F}\mathrm{sin}\left(15\right)+0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{N}=\mathrm{mg}-\mathrm{F}$

The kinetic frictional force is,

${\mathrm{f}}_{\mathrm{k}}={\mathrm{\mu}}_{\mathrm{k}}\mathrm{N}\phantom{\rule{0ex}{0ex}}={\mathrm{\mu}}_{\mathrm{k}}\left(\mathrm{mg}-\mathrm{Fsin}\left(15\right)\right)$

According to the Newton’s 2nd law along horizontal direction i.e. x is as the block movingalong x.Let, the acceleration of the block is a,

$\underset{}{\sum {\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}}}\phantom{\rule{0ex}{0ex}}\mathrm{Fcos}\left(15\right)-{\mathrm{f}}_{\mathrm{x}}=\mathrm{ma}\phantom{\rule{0ex}{0ex}}\mathrm{a}=\frac{\mathrm{Fcos}\left(15\right)-{\mathrm{\mu}}_{\mathrm{x}}\left(\mathrm{mg}-\mathrm{Fsin}\left(15\right)\right)}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}=\frac{\left(304\right)\mathrm{cos}\left(15\right)-{\mathrm{\mu}}_{\mathrm{k}}\left(\mathrm{mg}-\mathrm{Fsin}\left(15\right)\right)}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}=1.3\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}$

Therefore, the magnitude of the initial acceleration of the crate is 1.3 m/s2.

94% of StudySmarter users get better grades.

Sign up for free