Dynamics - Inclined plane

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Dynamics - Inclined plane

Question 1

A particle of mass $\var{mass} \mathrm{kg}$ is projected with speed $\var{u}\ \mathrm{ms^{-1}}$ up a rough plane. The coefficient of friction between the particle and the plane is $\var{mu}$. The plane is inclined at an angle $\theta = \var{theta}^{\circ}$ to the horizontal.

The acceleration due to gravity is $g = 9.8 \mathrm{ms^{-2}}$.

a)

What is the normal reaction, $R$, between the particle and the plane?

Give your answer in Newtons to 3 decimal places.

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b)

Using the value for $R$ from part a) find the deceleration of the particle as it moves up the plane.

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c)

Find the distance that the particle will move before it comes to instantaneous rest.

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Advice

A diagram can be drawn to show the forces acting on the particle.

Here the particle is drawn in three positions, showing its original speed when it is projected from the bottom of the slope, its position at some point up the slope and its position when it instantaneously comes to rest further up the slope.

The slope is rough so friction ($\muR \mathrm{N}$) acts down the slope, against the direction of motion.

a)

The normal reaction $R$, is found by resolving the forces perpendicular to the plane.

\begin{align}
R - mg \cos \theta & = 0, \\
R & = mg \cos \theta, \\
& = (\var{mass} \times 9.8) \cos (\var{theta}^{\circ}), \\
&= \var{precround(R,3)} \mathrm{N}.
\end{align}

The normal reaction force between the particle and the plane is $\var{precround(R,3)} \mathrm{N}$.

b)

To find the acceleration of the particle we resolve the forces parrallel to the plane.

\begin{align}
- mg \sin \theta - \mu R & = ma, \\
- \var{mass} \times 9.8 \sin (\var{theta}^{\circ}) - (\var{precround(R,3)} \times \var{mu}) & = \var{mass}a, \\
a & = \frac{ - \var{mass} \times 9.8 \sin (\var{theta}^{\circ}) - (\var{precround(R,3)} \times \var{mu}) }{\var{mass}}, \\
& = \var{precround(a,3)} \mathrm{ms^{-2}}.
\end{align}

Therefore the deceleration of the particle is $\var{precround(-a,3)} \mathrm{ms^{-2}}$ as it is the negative of the acceleration.

c)

The particle will travel up the slope until it comes to instantaneous rest; at this point its velocity will be $0$. We can use the SUVAT equation $v^2 = u^2 + 2as$ in order to find the distance the particle will travel.

We know that $u = \var{u}, v= 0$ and $a = \var{precround(a,3)}$.

\begin{align}
v^2 & = u^2 + 2as, \\
0 & = \simplify{{u}^2+{precround(2a,3)}s}, \\
s & = \frac{\var{u}^2}{\var{precround(-2a,3)}}, \\[0.5em]
& = \var{precround(u^2/(-2*a),3)} \mathrm{m}.
\end{align}

The distance the particle will travel up the plane before it comes to instantaneous rest is $\var{precround(u^2/(-2*a),3)} \mathrm{m}$.

Question 2

A block of mass $\var{mass}\mathrm{kg}$ slides down a rough plane which is inclined at an angle $\theta = \var{theta}^{\circ}$ to the horizontal. The mass begins at rest and accelerates at $\var{a}\mathrm{ms^{-2}}$.

The acceleration due to gravity is $g=9.8\mathrm{ms^{-2}}.$

a)

Find the normal reaction, $R \ \mathrm{N}$ between the block and the plane, to 3 decimal places.

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b)

Using $R$ find the coefficient of friction, $\mu$, between the block and the plane, to 3 decimal places.

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c)

Suppose that the block hits the bottom of the slope at a speed of $\var{v}\mathrm{ms^{-1}}$. To 3 decimal places, how far in metres ($\mathrm{m}$) has the block slid from its initial position?

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Advice

a)

To find the normal reaction force $R$, we resolve the forces perpendicular to the plane.

\begin{align}
R - mg \cos \theta & = 0, \\
R & = mg \cos \theta, \\
& = (\var{mass} \times 9.8) \cos (\var{theta}^{\circ}), \\
& = \var{R} \ \mathrm{N}.
\end{align}

b)

To find the coefficient of friction, $\mu$, we resolve parallel to the plane and use our $R$ value from part a).

\begin{align}
mg \cos (90^{\circ} - \theta) - \mu R & = ma, \\
\mu & = \frac{mg \cos(90^{\circ} - \theta) - ma}{R}, \\
& = \frac{ (\var{mass} \times 9.8) \cos (\var{90 - theta}^{\circ}) - (\var{mass} \times \var{a})}{\var{R}}, \\
& = \var{precround(mu,3)}. \end{align}

The coefficient of friction between the block and the plane is $\var{precround(mu,3)}$.

c)

This question can be solved using the SUVAT equations.

We know that the block was initially at rest, so $u = 0$.

The acceleration $a = \var{a}$.

The final velocity $v = \var{v}$.

We can use the equation $v^2 = u^2 + 2as$, and solve for the distance, $s$.

\begin{align}
v^2 & = u^2 + 2 as, \\
\var{v}^2 & = 0 + (2 \times \var{a}s), \\
s & = \simplify[]{{v}/(2*{a})}, \\
& = \var{precround(slide_distance,3)}\mathrm{m}. \end{align}

The block slid $\var{precround(v^2/2*a,3)}\mathrm{m}$ down the slope before reaching the given speed.

Question 3

A box of mass $\var{mass}\mathrm{kg}$ is being pushed up an inclined plane by a horizontal force $P = \var{P}\mathrm{N}$. The plane is inclined at an angle of $\theta=\var{theta}^{\circ}$. above the horizontal.

In the following the acceleration due to gravity is taken as $g=9.8\mathrm{ms^{-2}}$.

a)

Find the normal reaction between the box and the plane, in $\mathrm{N}$ to 3 decimal places.

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b)

Assuming the plane is smooth, find the acceleration of the box up the plane, in $\mathrm{ms^{-2}}$ to 3 decimal places.

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Advice

a)

We can draw a diagram to show the forces acting on the box. The plane is smooth so there is no friction.

To find the normal reaction between the box and the plane we solve $F=ma$ perpendicular to the plane.

\begin{align}
F & = ma, \\
R - mg \cos \theta - P \cos(90^{\circ} - \theta) & = m \times 0, \\
R & = mg \cos \theta - P \cos(90^{\circ} - \theta), \\
& = (\var{mass} \times 9.8 \cos(\var{theta}^{\circ})) - (\var{P} \cos(\var{90-theta}^{\circ})), \\
& = \var{R}\mathrm{N}.
\end{align}

The normal reaction, $R$, is $\var{R}\mathrm{N}$ to 3d.p.

b)

To find the acceleration of the box up the plane we resolve parallel to the plane. The plane is smooth so there is no friction.

\begin{align}
F& = ma,\\
P \cos \theta - mg \cos (90^{\circ} - \theta) & = ma, \\
a & = \frac{ P \cos \theta - mg \cos (90^{\circ} - \theta)}{m}, \\
&= \frac{ \var{P} \cos(\var{theta}^{\circ}) - (\var{mass} \times 9.8 \cos(\var{90-theta}^{\circ}))}{\var{mass}}, \\
&= \var{precround((P*cos(radians(theta))-mass*9.8*cos(radians(90-theta)))/mass,3)}\mathrm{ms^{-2}}.
\end{align}

The acceleration of the box up the plane is $\var{precround((P*cos(radians(theta))-mass*9.8*cos(radians(90-theta)))/mass,3)}\mathrm{ms^{-2}}.$

Question 4

Suppose that there is a plane which is inclined to the horizontal at an angle of $\var{theta}^{\circ}$. A particle of mass $\var{mass}\mathrm{kg}$ is placed upon the plane.

All parts should be answered to 3d.p and the acceleration due to gravity should be taken as $g = 9.8 \mathrm{ms^{-2}}$.

a)

What is the acceleration in $\mathrm{ms^{-2}}$ of the particle as it slides down the plane if the plane is smooth?

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b)

What is the acceleration in $\mathrm{ms^{-2}}$ of the particle as it slides down the plane if the coefficient of friction is $\var{mu}$?

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Advice

You can draw a diagram of the forces acting upon the particle.

a) When the plane is smooth we have that the coefficient of friction $\mu = 0$.

We then resolve $F=ma$ in the direction down the slope.

\begin{align} F &= ma, \\
mg \cos ((90 - \theta)^{\circ}) & = ma, \\
\var{mass} \times 9.8 \times \cos (\var{90 - theta}^{\circ}) & = \var{mass} \times a, \\
9.8 \cos (\var{90 - theta}^{\circ}) & = a, \\
\var{precround(9.8*cos(radians(90-theta)),3)} \mathrm{ms^{-2}} & = a. \end{align}

So the acceleration of the particle when the slope is smooth is $ \var{precround(9.8*cos(radians(90-theta)),3)} \mathrm{ms^{-2}}$.

b) If the plane is rough and the coefficient of friction is $\mu = \var{mu}$ then we first resolve the forces perpendicular to the plane to find $R$.

\begin{align} F & = ma, \\
R - mg \cos \theta & = 0, \\
R &= mg \cos \theta, \\
& = \var{mass} \times 9.8 \cos( \var{theta}^{\circ})\mathrm{N}.
\end{align}

We then resolve $F=ma$ in the direction down the slope, using the value we have for $R$.

\begin{align} F & = ma, \\
mg \sin \theta - \mu R & = ma, \\
mg \sin \theta - \mu mg \cos \theta & = ma, \\
g \sin \theta - \mu g \cos \theta & = a, \\
\left(9.8\sin(\var{theta}^{\circ})\right) - \left(\var{mu}\times 9.8 \cos(\var{theta}^{\circ})\right)& = a, \\
\var{precround(9.8*sin(radians(theta))-mu*9.8*cos(radians(theta)),3)} \mathrm{ms^{-2}} & = a. \end{align}

So the acceleration of the acceleration of the particle is $\var{precround(9.8*sin(radians(theta))-mu*9.8*cos(radians(theta)),3)} \mathrm{ms^{-2}}$ when the coefficient of friction is $mu=\var{mu}$.

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Dynamics - Inclined plane

Dynamics - Inclined plane

Question 1

Question 1

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Question 2

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Question 3

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Question 4

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