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Q16P

Expert-verifiedFound in: Page 141

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A loaded penguin sled weighing ${}^{\mathbf{80}\mathbf{}\mathbf{N}}$ rests on a plane inclined at angle ${}^{\mathbf{\theta}\mathbf{=}{\mathbf{20}}^{\mathbf{0}}}$to the horizontal (Fig. 6-23). Between the sled and the plane, the coefficient of static friction is ${}^{\mathbf{0}\mathbf{.}\mathbf{25}}$, and the coefficient of kinetic friction is ${}^{\mathbf{0}\mathbf{.}\mathbf{15}}$. (a) What is the least magnitude of the force parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude ${}^{{\mathbf{F}}}$ that will start the sled moving up the plane? (c) What value of ${}^{{\mathbf{F}}}$is required to move the sled up the plane at constant velocity?**

(a) The least magnitude of force ${}^{F}$ that prevents the sled moving down is ${}^{8.6\mathrm{N}}$

(b) The minimum magnitude of F that will start the sled moving up the plane is ${}^{46\mathrm{N}}$

(c) The value of F, that is required to move the sled up the plane at constant velocity is ${}^{39\mathrm{N}}$

Weight, ${}^{\mathrm{W}=80\mathrm{N}}$

Coefficient of static friction, ${}^{{\mathrm{\mu}}_{\mathrm{s}=0.25}}$

Coefficient of kinetic friction, ${}^{{\mathrm{\mu}}_{\mathrm{k}}=0.15}$

Inclined angle, ${}^{\mathrm{\theta}={20}^{0}}$

**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:**

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

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

Free body Diagram of sled over the inclined plane:

First, from the weight, calculate the mass of the system,

$W=mg\phantom{\rule{0ex}{0ex}}m=\frac{W}{g}\phantom{\rule{0ex}{0ex}}=\frac{80\mathrm{N}}{9.81}\phantom{\rule{0ex}{0ex}}=8.2\mathrm{kg}$

By using Newton’s 2nd law along vertical direction (along y),

$\sum {F}_{y}=m{a}_{y}$

since, block is not moving upward, ${a}_{y}=0$

$N-{F}_{g}\mathrm{cos}\left(\theta \right)=0\phantom{\rule{0ex}{0ex}}N={F}_{g}\mathrm{cos}\left(\theta \right)$

Relation between static frictional force and normal force is,

${f}_{s}={\mu}_{s}N={N}_{s}{F}_{g}\mathrm{cos}\left(\theta \right)$ (i)

When sled is moving down frictional force is upward,

Now, applying Newton’s 2nd along the x direction, for fig. a,

$\sum {F}_{s}=m{a}_{x}$

since, block is not moving, localid="1654149316961" ${a}_{x}=0$

$F+{f}_{s}-{F}_{g}\mathrm{sin}\left(\theta \right)=0\phantom{\rule{0ex}{0ex}}F={f}_{g}\mathrm{sin}\left(\theta \right)-{f}_{s}$

By using equation (i) in above equation,

$F={F}_{g}\mathrm{sin}\left(\theta \right)-{\mu}_{s}{F}_{g}\mathrm{cos}\left(\theta \right)\phantom{\rule{0ex}{0ex}}F={F}_{g}(\mathrm{sin}\left(\theta \right)-{\mu}_{s}\mathrm{cos}\left(\theta \right))\phantom{\rule{0ex}{0ex}}=mg(\mathrm{sin}\left(20\right)-(0.25\left)\mathrm{cos}\left(20\right)\right)\phantom{\rule{0ex}{0ex}}=8.6N\phantom{\rule{0ex}{0ex}}$

Thus, the least magnitude of force ${}^{F}$ that prevents the sled moving down is ${}^{8.6\mathrm{N}}$

To find the F to start the sled moving up is, again we have to use Newton’s 2nd law along y and x direction, but in this case as the sled moving up frictional force is downward.So, refer to fig. b.

Along vertical direction (y),

$\sum {F}_{y}=m{a}_{y}$

since block is not moving upward,${a}_{y}=0$

$N-{F}_{g}\mathrm{cos}\left(\theta \right)=0\phantom{\rule{0ex}{0ex}}N={F}_{g}\mathrm{cos}\left(\theta \right)$

Then static frictional force,

${f}_{s}={\mu}_{s}N={\mu}_{s}{F}_{g}\mathrm{cos}\left(\theta \right)$

Along horizontal direction (x),

$\sum {F}_{x}=m{a}_{x}$

since, block is not moving, ${a}_{x}=0$

$F-{f}_{s}-{F}_{g}\mathrm{sin}\left(\theta \right)=0\phantom{\rule{0ex}{0ex}}F={F}_{g}\mathrm{sin}\left(\theta \right)+{f}_{s}\phantom{\rule{0ex}{0ex}}={F}_{g}(\mathrm{sin}\theta +{\mu}_{s}\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}=46N$

Hence, the minimum magnitude of F that will start the sled moving up the plane is ${}^{46\mathrm{N}}$.

To find ${}^{F}$ to move the sled up the plane at constant velocity, use kinetic frictional force,

${f}_{k}={\mu}_{k}N\phantom{\rule{0ex}{0ex}}={\mu}_{k}{F}_{g}\mathrm{cos}\left(\theta \right)$

Along horizontal direction (x),

$\sum {F}_{x}=m{a}_{x}$

since, block is moving with constant velocity, ${a}_{x}=0$

$F-{f}_{k}-{F}_{g}\mathrm{sin}\left(\theta \right)=0\phantom{\rule{0ex}{0ex}}F={F}_{g}\mathrm{sin}\left(\theta \right)+{f}_{k}\phantom{\rule{0ex}{0ex}}={F}_{g}(\mathrm{sin}\theta +{\mu}_{k}\mathrm{cos}\theta )\phantom{\rule{0ex}{0ex}}=39N$

Hence, the value of F, that is required to move the sled up the plane at constant velocity is ${}^{39\mathrm{N}}$

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