Dynamics - maximum frictional force

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Dynamics - maximum frictional force

Question 1

A $\var{mass} \, \mathrm{kg}$ mass is resting on a rough horizontal plane. The acceleration due to gravity is $\var{g} \mathrm{ms^{-2}}$.

a)

Find, in $\mathrm{N}$ to 3 decimal places, the maximum frictional force which can act on the mass when the coefficient of friction is $\var{friction1}$.

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b)

Find, in $\mathrm{N}$ to 3 decimal places, the maximum frictional force which can act on the mass when the coefficient of friction is $\var{friction2}$.

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c)

Suppose a horizontal force of $\var{P1}N$ is applied to the mass. The coefficient of friction is still $\var{friction2}$.

What is the magnitude of the frictional force acting on the mass, in $\mathrm{N}$ to 3 decimal places?

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d)

What is the acceleration of the mass, in $\mathrm{ms^{-2}}$ to 3 decimal places?

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e)

Suppose a horizontal force of $\var{Fmax}N$ is applied to the mass. The coefficient of friction is still $\var{friction2}$.

What is the magnitude of the frictional force acting on the mass, in $\mathrm{N}$ to 3 decimal places?

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f)

What is the acceleration of the mass, in $\mathrm{ms^{-2}}$ to 3 decimal places?

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g)

Suppose a horizontal force of $\var{Fmax+2*a}N$ is applied to the mass. The coefficient of friction is still $\var{friction2}$.

What is the magnitude of the frictional force acting on the mass, in $\mathrm{N}$ to 3 decimal places?

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h)

What is the acceleration of the mass, in $\mathrm{ms^{-2}}$ to 3 decimal places?

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Advice

a)

The maximum frictional force is $F_{\text{MAX}} = \mu R$, where $R$ is found by resolving the forces in the upwards direction. Here, acceleration is zero.

\begin{align}
F &= ma \\
R - mg & = ma \\
R - \left(\var{mass} \times 9.8 \right) & = \var{mass} \times 0 \\
R & = \var{mass} \times 9.8 \\
&= \var{mass*9.8}
\end{align}

So $R = \var{mass*9.8}$.

Therefore when the coefficient of friction $\mu = \var{friction1}$ we have

\begin{align}
F_{\text{MAX}} & = \mu R \\
& = \var{friction1} \times \var{mass*9.8} \\
& = \var{precround(friction1*mass*9.8,3)}
\end{align}

So the maximum frictional force is $F_{\text{MAX}} = \var{precround(friction1*mass*9.8,3)}N.$

b)

We have from part a) that $R = \var{mass*9.8}$.

Therefore when the coefficient of friction $\mu = \var{friction2}$ we have

\begin{align}
F_{\text{MAX}} & = \mu R \\
& = \var{friction2} \times \var{mass*9.8} \\
& = \var{precround(friction2*mass*9.8,3)}
\end{align}

So the maximum frictional force is $F_{\text{MAX}} = \var{precround(friction2*mass*9.8,3)}N.$

c)

Now we have that $F_{\text{MAX}} = \var{precround(friction2*mass*9.8,3)}N$ so any horizontal force less than or equal to $F_{\text{MAX}}$ will not move the mass.

The frictional force acting on the mass will be equal to the horizontal force in order to prevent the mass from moving.

d)

The force applied to the mass and the frictional force cancel each other out, so acceleration is zero.

e)

Now we have that $F_{\text{MAX}} = \var{precround(friction2*mass*9.8,3)}N$ so any horizontal force less than or equal to $F_{\text{MAX}}$ will not move the mass.

When the horizontal force is equal to the maximum frictional force the friction will need to be at its maximum value to stop the mass from moving.

f)

The mass is in limiting equilibrium but still isn't moving so acceleration is zero.

g)

Now that a horizontal force greater than $F_{\text{MAX}}$ is applied the mass will move.

The friction will be unable to stop the mass moving but it will remain at its maximum value of $F_{\text{MAX}}$.

h)

The acceleration of the mass can be found by resolving $F = ma$.

\begin{align}
F & = ma \\
\var{2*a + Fmax} - \var{Fmax} & = \var{mass} \times a \\
a & = \frac{\var{2*a + Fmax} - \var{Fmax}}{\var{mass}} \\
& = \var{precround(2*a/mass,3)}
\end{align}

So the acceleration of the mass is $\var{precround(2*a/mass,3)} \, \mathrm{ms^{-2}}$.

Question 2

A block of mass $\var{mass} \ \mathrm{kg}$ lies at rest on a rough horizontal floor. The coefficient of friction between the mass and the floor is $\var{mu}$. A force $P \ \mathrm{N}$ is applied to the mass which can push or pull the mass horizontally along.

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

a)

Find the magnitude in Newtons (to 3 d.p.) that $P$ must exceed to achieve movement if $P$ is applied horizontally.

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b)

Find the magnitude in Newtons (to 3 d.p.) that $P$ must exceed to achieve movement if $P$ is applied at an angle $\alpha$ above the horizontal, where $ \tan \alpha = \frac{\var{O}}{\var{A}}$.

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c)

Find the magnitude in Newtons (to 3 d.p.) that $P$ must exceed to achieve movement if $P$ is applied at an angle $\alpha$ below the horizontal, where $\tan \alpha = \frac{\var{O2}}{\var{A2}}$.

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Advice

a)

If $P$ is applied horizontally to the mass which lies at rest, we can resolve $F = ma$ with acceleration $a=0$ to find the normal reaction $R$, and from this find the maximum frictional force $F_{\text{MASS}}=\mu R$.

In the upwards direction we have

\begin{align}
F & = ma \\
R - mg & = 0 \\
R & = mg \\
& = \var{mass} \times 9.8 \\
& = \var{mass*9.8}
\end{align}

The maximum possible frictional force is $F_{\text{MASS}} = \mu R = \var{mu} \times \var{R} = \var{precround(mu*R,3)} \ \mathrm{N}$. Therefore to achieve movement $P$ must exceed $\var{precround(mu*R,3)} \ \mathrm{N}$.

b)

Note that if $ \tan \alpha = \frac{\var{O}}{\var{A}}$ then using SOHCAHTOA where $O = \var{O}$, $A = \var{A}$ and $H = \sqrt{\var{O}^2 + \var{A}^2}$ we have that $\sin \alpha = \frac{\var{O}}{\sqrt{\var{O}^2 + \var{A}^2}}$ and $\cos \alpha = \frac{\var{A}}{\sqrt{\var{O}^2 + \var{A}^2}}$.

$P$ is applied at an angle above the horizontal as shown in the diagram. We can resolve $F=ma$ in the upward direction where acceleration $a=0$.

\begin{align} R + P \sin \alpha - mg & = ma \\
R + \frac{\var{O}}{\sqrt{\var{O}^2 + \var{A}^2}}P - (\var{mass} \times 9.8)& = 0 \\
R & = \var{mass*9.8} - \frac{\var{O}}{\sqrt{\var{O}^2 + \var{A}^2}}P
\end{align}

We can now resolve $F=ma$ in the horizontal direction, where acceleration is also zero and $F_{\text{MAX}} = \mu R = \var{mu} \times \left( \var{mass*9.8} - \frac{\var{O}}{\sqrt{\var{O}^2 + \var{A}^2}}P \right)$.

\begin{align} P \cos \alpha - F_{\text{MAX}} &= ma \\
P \cos \alpha & = F_{\text{MAX}} \\
\frac{\var{A}}{\sqrt{\var{O}^2 + \var{A}^2}} P & = \var{mu} \times \left( \var{mass*9.8} - \frac{\var{O}}{\sqrt{\var{O}^2 + \var{A}^2}}P \right) \\
& = \var{mu*mass*9.8} - \frac{\var{mu*O}}{\sqrt{\var{O}^2 + \var{A}^2}} P \\
\left( \frac{\var{A}}{\var{precround(H,3)}} + \var{precround(mu*O/H,3)} \right) P & = \var{mu*mass*9.8} \\
P & = \frac{ \var{mu*mass*9.8}}{ \left( \frac{\var{A}}{\var{precround(H,3)}} + \var{precround(mu*O/H,3)} \right)} \\
& = \var{precround((mu*mass*9.8)/(A/H + mu*O/H),3)} \end{align}

Therefore $P$ must exceed $\var{precround((mu*mass*9.8)/(A/H + mu*O/H),3)} \ \mathrm{N}$ in order for the mass to start moving.

c)

Note that if $\tan \alpha = \frac{\var{O2}}{\var{A2}}$ then using SOHCAHTOA where $O = \var{O2}$, $A = \var{A2}$ and $H = \sqrt{\var{O2}^2 + \var{A2}^2}$ we have that $\sin \alpha = \frac{\var{O2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}$ and $\cos \alpha = \frac{\var{A2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}$.

We can resolve $F = ma$ in the upward direction where acceleration is zero.

\begin{align}
R - mg - P \sin \alpha & = m \times 0 \\
R & = mg + P \sin\alpha \\
& = \var{mass*9.8} + \frac{\var{O2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}P
\end{align}

From this we can calculate $F_{\text{MAX}} = \mu \times R = \var{mu} \times \left(\var{mass*9.8} + \frac{\var{O2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}P\right)$.

We then resolve $F = ma$ in the horizontal direction, where once again acceleration is zero.

\begin{align} P \cos \alpha - F_{\text{MAX}} & = m \times 0 \\
\frac{\var{A2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}P & = \var{mu}\left(\var{mass*9.8} + \frac{\var{O2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}P\right) \\
\left(\frac{\var{A2}}{\sqrt{\var{O2}^2 + \var{A2}^2}} - \frac{\var{mu} \times \var{O2}}{\sqrt{\var{O2}^2 + \var{A2}^2}}\right)P & = \var{mu} \times \var{mass*9.8} \\
\var{precround((A2/H2) - (mu*O2/H),3)} P & = \var{mu*9.8*mass} \\
P & = \var{precround((mu*9.8*mass)/(A2/H2 - mu*O2/H),3)} \end{align}

Therefore $P$ must exceed $\var{precround((mu*9.8*mass)/(A2/H2 - mu*O2/H),3)}N$ in order for the mass to start moving.

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Dynamics - maximum frictional force

Dynamics - maximum frictional force

Question 1

Question 1

a)

b)

c)

d)

e)

f)

g)

h)

Advice

a)

b)

c)

d)

e)

f)

g)

h)

Question 2

Question 2

a)

b)

c)

Advice

a)

b)

c)