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## A block weighing #14 kg# is on a plane with an incline of #pi/12# and friction coefficient of #1/5#. How much force, if any, is necessary to keep the block from sliding down?

Morgan
Featured 4 months ago

I get $\approx 9.0 N$

#### Explanation:

We want to know if the forces currently acting on the block will cause it to accelerate (i.e. move down the ramp), and if so, how much force would need to be applied to the block to prevent this. We can test both of these things simultaneously.

A force diagram of the situation:

Where ${\vec{f}}_{s}$ is the force of static friction, $\vec{n}$ is the normal force, ${\vec{F}}_{g}$ is the force of gravity, and ${\vec{F}}_{g x}$ and ${\vec{F}}_{g y}$ are the parallel and perpendicular components of the force of gravity, respectively.

Note that I used the force of static friction, which we want to keep the object from moving. If this is not enough, and the object is accelerating under the given conditions, this is the force of kinetic friction. Whether you call this static or kinetic friction initially will not affect the calculations.

Next, we can put together statements for the parallel and perpendicular forces:

$\sum {F}_{x} = {\vec{F}}_{g x} - {\vec{f}}_{s} = m {\vec{a}}_{x}$

$\sum {F}_{y} = \vec{n} - {\vec{F}}_{g y} = m \cancel{{\vec{a}}_{y}} = 0$

Note that I have defined down the ramp as positive , and so the parallel component of gravity is positive, while the perpendicular component is negative. This can be seen from the force diagram.

Note there is no movement in the vertical direction either way, and so the vertical acceleration is zero, setting the sum of the perpendicular forces equal to zero.

For the object to remain at rest and not slide down the ramp, the net horizontal acceleration must also be zero. We want to know what force, if any, is necessary to make the horizontal acceleration equal zero. We can call this other force $\vec{F}$ and rewrite our net parallel force statement:

$\sum {F}_{x} = {\vec{F}}_{g x} - {\vec{f}}_{s} - \vec{F} = 0$

Note that $\vec{F}$ is negative. We want it to oppose the motion, which would occur down the ramp (positive).

We can solve for this unknown force:

$\vec{F} = {\vec{F}}_{g x} - {\vec{f}}_{s}$

If no external force is necessary, we should get a value of at most $0$. If not, it indicates that the object would accelerate under the given conditions (only force of gravity, normal force, and force of friction).

• We know the equation for friction: $\vec{f} = \mu \vec{n}$

• From the sum of forces statement for perpendicular forces, we know that $\vec{n} = {\vec{F}}_{g y}$

From the force diagram, we can produce equations for the parallel and perpendicular components of the force of gravity. By the magic of geometry, the angle between the vector for the force of gravity (total) and the perpendicular component is the same as the angle of incline of the plane. We see that the parallel component of gravity is opposite the angle, so we will use $\sin \left(\theta\right)$ to find it. Similarly, we see that the perpendicular component of gravity is adjacent to the angle, so we will use $\cos \left(\theta\right)$ to find it.

We know that the force of gravity is given by ${\vec{F}}_{g} = m g$

${\vec{F}}_{g x} = m g \sin \left(\theta\right)$

${\vec{F}}_{g y} = m g \cos \left(\theta\right)$

Therefore, $\vec{n} = m g \cos \left(\theta\right)$.

Substituting these equations back into the equation we derived for the unknown force:

$\vec{F} = m g \sin \left(\theta\right) - \mu m g \cos \left(\theta\right)$

$\implies \vec{F} = m g \left(\sin \left(\theta\right) - \mu \cos \left(\theta\right)\right)$

We can now plug in our known values to solve for $\vec{F}$:

$\vec{F} = \left(14 k g\right) \left(9.8 \frac{m}{s} ^ 2\right) \left(\sin \left(\frac{\pi}{12}\right) - \left(\frac{1}{5}\right) \cos \left(\frac{\pi}{12}\right)\right)$

There is no pleasant exact value (not a great angle), but we get $\vec{F} = 9.00496831 N$.

$\vec{F} \approx 9.0 N$

We can confirm that the object would be accelerating without the added force:

${\vec{F}}_{x} = {\vec{F}}_{g x} - \vec{f} = m {\vec{a}}_{x}$

$\implies {\vec{a}}_{x} = \frac{\cancel{m} g \sin \left(\theta\right) - \mu \cancel{m} g \cos \left(\theta\right)}{\cancel{m}}$

$\implies {\vec{a}}_{x} = g \left(\sin \left(\theta\right) - \mu \cos \left(\theta\right)\right)$

$\implies {\vec{a}}_{x} = \left(9.8 \frac{m}{s} ^ 2\right) \left(\sin \left(\frac{\pi}{12}\right) - \left(\frac{1}{5}\right) \cos \left(\frac{\pi}{12}\right)\right)$

$\implies {\vec{a}}_{x} \approx 0.64 \frac{m}{s} ^ 2$

$\therefore$ the object would accelerate down the ramp at $0.64 \frac{m}{s} ^ 2$ under the original conditions.

We can also check our answer:

$\sum {F}_{x} = {\vec{F}}_{g x} - \vec{f} - 9.00496831 N = m {\vec{a}}_{x}$

$\implies {\vec{a}}_{x} = \frac{m g \sin \left(\theta\right) - \mu m g \cos \left(\theta\right) - 9.00496831 N}{m}$

$\implies {\vec{a}}_{x} = g \sin \left(\theta\right) - \mu g \cos \left(\theta\right) - 9.00496831 N$

$\implies {\vec{a}}_{x} = \frac{\left(14 k g\right) \left(9.8 \frac{m}{s} ^ 2\right) \sin \left(\frac{\pi}{12}\right) - \left(\frac{1}{5}\right) \left(14\right) \left(9.8 \frac{m}{s} ^ 2\right) \cos \left(\frac{\pi}{12}\right) - 9.00496831 N}{14 k g}$

$\implies {\vec{a}}_{x} = 0$

## A bulb of resistance R=16 ohms is attached in series with an infinite resistor network with identical resistances r ohms. A 10 V battery drives current in the circuit. What should be the value of 'r' such that the bulb dissipates about 1 W of power?

A08
Featured 4 months ago

$14.8 \Omega$, rounded to one decimal place.

#### Explanation:

Let ${R}_{e}$ be equivalent resistance of an infinite resistor network with identical resistances of $r$ ohms. Since bulb is connected in series with network,

Total resistance seen by battery $= R + {R}_{e}$
Current in the bulb is given by the expression
$I = \frac{V}{R + {R}_{e}}$ ......(1)
Power dissipated in the bulb is
$P = {I}^{2} R$ .....(2)
Inserting given values we get from (1)
$I = \frac{10}{16 + {R}_{e}}$ and inserting this value and other given numbers we get from (2)
$1 = {\left(\frac{10}{16 + {R}_{e}}\right)}^{2} \times 16$
Taking square root of both sides we get
$1 = \frac{10}{16 + {R}_{e}} \times 4$
$\implies 16 + {R}_{e} = 40$
$\implies {R}_{e} = 40 - 16 = 24 \Omega$ ......(3)

To calculate ${R}_{e}$
As there are infinite many resistors, there will still be infinite many resistors if we detach the first two resistors from the front of nodes $X \mathmr{and} Y$ as shown in figure. Resistance seen looking to the right will be the same as the resistance seen between nodes $A \mathmr{and} B$

As such the network reduces to sum of two resistances 1. resistance $r$ between nodes $A \mathmr{and} B$ and 2. resistance equivalent to two parallel resistors $r$ and ${R}_{e}$ between nodes $X \mathmr{and} Y$.

For infinite resistor network we have an equation

${R}_{e} = r + \frac{r \times {R}_{e}}{r + {R}_{e}}$
$\implies \left({R}_{e} - r\right) = \frac{r \times {R}_{e}}{r + {R}_{e}}$
$\implies \left({R}_{e} - r\right) \times \left(r + {R}_{e}\right) = \left(r \times {R}_{e}\right)$
$\implies \left({R}_{e}^{2} - {r}^{2}\right) - \left(r \times {R}_{e}\right) = 0$

Using (3) we get

$\left({24}^{2} - {r}^{2}\right) - 24 r = 0$
$\implies {r}^{2} + 24 r - {24}^{2} = 0$

Solving the quadratic and choosing positive root as resistance can not be negative
$r = \frac{- 24 \pm \sqrt{{24}^{2} + 4 \times {24}^{2}}}{2}$
$r = 14.8 \Omega$, rounded to one decimal place.

-.-.-.-.-.-.-.

Quadratic equation can also be solved using inbuilt graphic utility

## What are values of ΔH, ΔU, ΔS for 1 mole of H20?

Truong-Son N.
Featured 4 months ago

The way the question is worded is quite confusing, but if that's the way it's worded, here's how I'm interpreting this.

Since the definitions of the three thermodynamic functions given are:

$\Delta {H}_{{H}_{2} O \left(l\right)} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \left(l\right) \mathrm{dT}$
$\Delta {S}_{{H}_{2} O \left(l\right)} = {\int}_{{T}_{1}}^{{T}_{2}} \frac{{C}_{P} \left(l\right)}{T} \mathrm{dT}$
$\Delta {U}_{{H}_{2} O \left(l\right)} = \Delta {H}_{{H}_{2} O \left(l\right)} - \Delta \left(P V\right)$

we can work from there.

It seems reasonable (though tedious) that the question wants us to calculate for $\text{293 K" -> "373 K}$, through the vaporization process, and on to the heating process of $\text{373 K" -> "520 K}$, actually...

This would illustrate the definitions given above.

CHANGE IN ENTHALPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR

$\Delta {H}_{{H}_{2} O \left(l\right)} = {\int}_{\text{293 K")^("373 K}} {C}_{P} \left(l\right) \mathrm{dT}$

In this temperature range, we may assume that ${C}_{P} \left(l\right)$ is approximately constant, so that:

#color(blue)(DeltaH_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") dT#

$= \left(\text{75.36 J/mol"cdot"K")("373 - 293 K}\right)$

$=$ $\textcolor{b l u e}{\text{6028.8 J/mol}}$

Include the $\Delta {H}_{\text{vap}}$ of $\textcolor{b l u e}{\text{40695.7 J/mol}}$ for the vaporization process.

Next:

#color(blue)(DeltaH_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") dT#

$= \left(\text{36 J/mol"cdot"K")("520 - 373 K}\right)$
(not $36000$; the kilo was a typo.)

$=$ $\textcolor{b l u e}{\text{5292 J/mol}}$

This gives a total $\Delta H$ for the heating, vaporization, and further heating process as:

$\boldsymbol{\textcolor{b l u e}{n \Delta H}} = n \left(\Delta {H}_{{H}_{2} O \left(l\right)}^{293 \to \text{373 K") + DeltaH_"vap" + DeltaH_(H_2O(g))^(373 -> "520 K}}\right)$

#= ("1 mol")(6028.8 + 40695.7 + 5292) = "52016.5 J"#

$=$ $\boldsymbol{\textcolor{b l u e}{\text{52.017 kJ}}}$

CHANGE IN ENTROPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR

Similarly, the change in entropy over a temperature range can be calculated by the same kind of approximation about ${C}_{P} \left(l\right)$ and ${C}_{P} \left(g\right)$:

#color(blue)(DeltaS_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") 1/TdT#

$= \left(\text{75.36 J/mol"cdot"K}\right) \ln \left(\frac{373}{293}\right)$

$=$ $\textcolor{b l u e}{\text{18.19 J/mol"cdot"K}}$

At constant temperature and pressure, which is how it is at the phase equilibrium we establish for vaporization, $\Delta G = 0$, so that $\Delta H - T \Delta S = 0$. Therefore:

$\textcolor{b l u e}{\Delta {S}_{\text{vap") = (DeltaH_"vap")/T_b = ("40695.7 J/mol")/("373 K}}}$

$=$ #color(blue)("109.10 J/mol"cdot"K"#

Finally, a similar process for $\Delta S$ going from $\text{373 K}$ to $\text{520 K}$ would be:

#color(blue)(DeltaS_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") 1/TdT#

$= \left(\text{36 J/mol"cdot"K}\right) \ln \left(\frac{520}{373}\right)$

$=$ $\textcolor{b l u e}{\text{11.96 J/mol"cdot"K}}$

So, the total $\Delta S$ for this heating process on $\text{1 mol}$ of water would be:

$\boldsymbol{\textcolor{b l u e}{n \Delta S}} = n \left(\Delta {S}_{{H}_{2} O \left(l\right)}^{293 \to \text{373 K") + DeltaS_"vap" + DeltaS_(H_2O(g))^(373 -> "520 K}}\right)$

$= \left(\text{1 mol}\right) \left(18.19 + 109.10 + 11.96\right)$ $\text{J/mol"cdot"K}$

$=$ $\boldsymbol{\textcolor{b l u e}{\text{139.25 J/K}}}$

CHANGE IN INTERNAL ENERGY FOR HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR

Recall that $\Delta H = \Delta U + \Delta \left(P V\right) = \Delta U + P \Delta V + V \Delta P$. Well, we've been at constant pressure, so we don't have to worry about the $V \Delta P$ term.

For the heating process, we'd need the densities of water at both temperatures to get accurate molar volumes (which are necessary because liquids are fairly incompressible). I think we can assume that $\Delta U \approx \Delta H$ for this first part because $P \Delta V$ is going to be small.

But just to see...

#rho_("293 K") = "998.21 g/L"#
#rho_("373 K") = "958.37 g/L"#

#barV_("293 K") = M_m/rho = "g"/"mol" xx "L"/"g" = 18.015/998.21 = "0.018047 L/mol"#

#barV_("373 K") = M_m/rho = 18.015/958.37 = "0.018798 L/mol"#

So for $\text{1 mol}$ of liquid water, $\Delta V = 0.018798 - 0.018047 = \text{0.000751 L}$, and so, from:

$\Delta H = \Delta U + \Delta \left(P V\right) = \Delta U + P \Delta V$,
(at constant pressure)

#color(blue)(nDeltaU_(H_2O(l))^(293 -> "373 K")) = nDeltaH_(H_2O(l))^(293 -> "373 K") - PDeltaV_(H_2O(l))^(293 -> "373 K")#

$=$ $\left(\text{1 mol")("6028.8 J/mol") - ("1.01325 bar")("0.000751 L")xx("8.314472 J")/("0.0831345 L"cdot"bar}\right)$

$=$ $\textcolor{b l u e}{\text{6028.7 J") ~~ DeltaH_(H_2O(l))^(293 -> "373 K}}$

Indeed, the $P \Delta V$ term hardly mattered for liquid water.

For the vaporization, a similar idea follows in that $\Delta {V}_{\left(l\right) \to \left(g\right)} \approx {V}_{\left(g\right)}$. At high temperatures, we can treat water approximately as an ideal gas, and so:

$P \Delta {V}_{\left(l\right) \to \left(g\right)} \approx P {V}_{\left(g\right)} = n R T$

#= ("1 mol")("8.314472 J/mol"cdot"K")("373 K") = "3101.30 J"#

Therefore:

#color(blue)(nDeltaU_"vap") = nDeltaH_"vap" - PDeltaV_((l)->(g))#

$= \left(\text{1 mol")("40695.7 J/mol") - ("3101.30 J}\right)$

$=$ $\textcolor{b l u e}{\text{37594.4 J}}$

Finally, for the heating of the gas, again, we can assume ideality to get the change in volume of the water vapor. At constant $P$ and $n$:

$P \Delta V = n R \Delta T$

$= \left(\text{1 mol")("8.314472 J/mol"cdot"K")("520 - 373 K}\right)$

$=$ $\text{1222.23 J}$

So:

#color(blue)(nDeltaU_(H_2O(g))^(373 -> "520 K")) = nDeltaH_(H_2O(g))^(373 -> "520 K") - PDeltaV#

$= \left(\text{1 mol")("5292 J/mol") - ("1222.23 J}\right)$

$=$ $\textcolor{b l u e}{\text{4069.77 J}}$

So, the overall $\Delta U$ is:

$\boldsymbol{\textcolor{b l u e}{n \Delta U}} = n \left(\Delta {U}_{{H}_{2} O \left(l\right)}^{293 \to \text{373 K") + DeltaU_"vap" + DeltaU_(H_2O(g))^(373 -> "520 K}}\right)$

$= \left(\text{1 mol}\right) \left(6028.7 + 37594.4 + 4069.77\right)$ $\text{J/mol}$

$=$ $\boldsymbol{\textcolor{b l u e}{\text{47.693 kJ}}}$

## Two mirrors are inclined at an angle theta. light ray is incident parallel to one of the mirrors. the ray will start retracing its path after third reflection if?

A08
Featured 2 months ago

${30}^{\circ}$

#### Explanation:

The two mirrors are inclined at an angle $\theta$. The ray diagram has been drawn for an arbitrary angle. Actual angle need to be found out. As such actual ray diagram will be different.

It is clear that for third reflection the ray must hit the mirror normally or make angle of reflection $= {0}^{\circ}$

As dotted line at $B$ is normal to the mirror, retracing the path from $C ,$ we see that four angles at point $B$, are

$90 - \theta$, (sum of three angles of a $\Delta = {180}^{\circ}$)
$\theta \mathmr{and} \theta$, (angles of incidence and of reflection)
$90 - \theta$, (straight line has angle $= {180}^{\circ}$)

Incident ray at $A$ is parallel to the other mirror and dotted line is normal to mirror. This indicates angles $\theta$ at $A$

Now in triangle $A B C$ sum of its three angles is equal to ${180}^{\circ}$. We have the equation
$90 + \left(\theta + \theta\right) + \theta = 180$
$\implies 3 \theta = 180 - 90$
$\implies \theta = {30}^{\circ}$

## Why is work a scalar but torque is a vector? thanks for answers.

dk_ch
Featured 1 month ago

Following the definition of work written above we see that work done (W) by a constant force ($\vec{F}$) is the product of the magnitude displacement ($\vec{d}$) caused by it and the component of the force in the direction of displacement.

So mathematically

$W = \left\mid \vec{d} \right\mid \cdot \left\mid \vec{F} \right\mid \cos \theta$, where $\theta$ represents the angle between $\vec{F} \mathmr{and} \vec{d}$

So vectorially

$W = \vec{F} \cdot \vec{d}$

This means work is scalar product of two vectors quantities. So work is a scalar quantity.

TORQUE

When force is applied to rotate a body around an axis the magnitude of rotational effect caused by the force depends on three quantities (1) the distance of point of application of force from axis of rotation,the magnitude of radius vector ($\left\mid \vec{r} \right\mid$), (2) the magnitude of force $\left\mid \vec{F} \right\mid$ and (3) sine of the the angle $\theta$ between $\vec{F}$ and $\vec{r}$ and the product of these three quantities is the measure of the rotational effect caused and is known as torque

Mathematically

Magnitude of Torque

$\tau = \left\mid \vec{r} \right\mid \left\mid \vec{F} \right\mid \sin \theta$

So vectorially it is the cross product of two vectors $\vec{r} \mathmr{and} \vec{F}$ i.e.

$\vec{\tau} = \vec{r} \times \vec{F}$

So torque is a vector quantity. The direction of torque is determined by the thumb rule as shown in above figure.

## Q8. A circular iron ring has a mean circumference of #1.5m# and cross sectional area of #0.01m^2#. A saw cut of# 4mm# wide is made in the ring. Calculate the magnetizing current required to produce a flux of # 0.8mWb# in the air gap if the ring is wound ?

A08
Featured 1 week ago

$3.43 \text{ A}$, rounded to two decimal places.

#### Explanation:

Refer to the figure above, we need to treat iron ring and gap in the ring separately.

Magnetic ring.

Let the current to produce required magnetic flux in ring be$= I$
Magnetizing force produced in the iron ring $H = \frac{N I}{l}$
where $N$ is the number of turns and
$l$ is the mean circumference.
$\therefore$ $H = \frac{175 I}{1.5} = 116. \overline{6} I$ AT${\text{m}}^{-} 1$

We know that for all media except vacuum Flux density $B$ is given as
$\frac{B}{H} = {\mu}_{0} {\mu}_{r}$
$\implies B = H {\mu}_{0} {\mu}_{r}$
Inserting given values we get
$\implies B = 116. \overline{6} I \times \left(4 \pi \times {10}^{-} 7\right) \times 400 \text{ T}$
$\implies B = 0.0586 I \text{ T}$
Magnetic Flux density $B = \text{Magnetic flux"/"Area}$
Inserting vales we get
Magnetic flux $\Phi = B \times \text{Area"=0.0586Ixx0.01=5.86xx10^-4 I" Wb}$

The ratio of the total flux produced $\Phi$ to the useful flux ${\Phi}_{u}$ created in the air gap is called leakage factor or leakage coefficient. After ignoring fringing we have
$\frac{\Phi}{\Phi} _ u = 1.45 = \frac{5.86 \times {10}^{-} 4 I}{\Phi} _ u$

$\implies {\Phi}_{u} = \frac{5.86 \times {10}^{-} 4 I}{1.45}$
$\implies {\Phi}_{u} = 4.041 \times {10}^{-} 4 I$

This equals flux in the air gap.
$0.8 \times {10}^{-} 3 = 4.041 \times {10}^{-} 4 I$
$\implies I = \frac{0.8 \times {10}^{-} 3}{4.041 \times {10}^{-} 4}$
$\implies I = 1.979 \text{ A}$ ......(1)

Air gap.

Magnetic flux for air gap ${\Phi}_{g} = 0.8 \times {10}^{-} 3 W b$
#B_g=Phi_g/a=(0.8×10^-3)/0.01=0.08Wbm^-2#

#H_g=B_g/μ_0=0.08/(4π×10^-7)=63662ATm#

To produce flux in the gap required Magnetizing force ${H}_{g} = \frac{N {I}_{g}}{l} _ g$
$\implies {I}_{g} = \frac{{H}_{g} {l}_{g}}{N}$

Inserting various values we get
${I}_{g} = \frac{63662 \times 0.004}{175}$

$\implies {I}_{g} = 1.455 \text{ A}$ .......(2)

from (1) and (2)

Total current $= I + {I}_{g} = 1.979 + 1.455 = 3.43 \text{ A}$, rounded to two decimal places.

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