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Expert-verified Found in: Page 140 ### Fundamentals Of Physics

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 ${\mathbf{15}}{\mathbf{°}}$ 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.

See the step by step solution

## Step 1: Given

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, $\mathrm{\theta }=15°$

## Step 2: Determining the concept

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.

## Step 3: Determining the free body diagram of the block

Free body diagram of the block: ## Step 4: (a) Determining theminimum force required from the rope to start the crate moving

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

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

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

## Step 5: (b) Determining themagnitude of the initial acceleration of the crate

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. ### Want to see more solutions like these? 