Kvant Physics Problem 197

Verified: yes
Verdicts: PASS + PASS
Solve time: 14m37s
Source on kvant.digital

Problem

On a body of mass $m$ resting on a horizontal rough surface with coefficient of friction $k$, at time $t=0$ a force proportional to time began to act at an angle $\alpha$ to the horizontal. Determine the velocity $v$ of the body after $\tau$ seconds.

A. V. Ustinova

Setup and Assumptions

A body of mass $m$ lies on a fixed horizontal rough surface. The coefficient of friction between the body and the surface is $k$. At time $t=0$ a force $\mathbf{F}(t)$ begins to act on the body, with magnitude proportional to time,

$F(t)=\beta t,$

where $\beta$ has units $\mathrm{N/s}$. The force is directed at a constant angle $\alpha$ above the horizontal.

The body is initially at rest at $t=0$. Motion is described in an inertial reference frame attached to the ground. Air resistance is neglected. The normal force adjusts according to vertical force balance while contact with the surface is maintained. The time interval of interest is $0\le t\le \tau$, during which the body is assumed to remain in contact with the surface.

Since the body is initially at rest, the onset of motion must be checked before applying the equation with friction force $kN$. The horizontal component of the applied force is

$F_x(t)=\beta t\cos\alpha.$

While the body remains at rest, static friction balances this component. Motion begins when the required static friction reaches its limiting value. Using the friction coefficient denoted in the problem by $k$, the threshold condition is

$\beta t_0\cos\alpha=kN(t_0).$

With

$N(t)=mg-\beta t\sin\alpha,$

the instant $t_0$ at which sliding starts satisfies

$\beta t_0\cos\alpha=k\bigl(mg-\beta t_0\sin\alpha\bigr).$

Solving for $t_0$,

$\beta t_0(\cos\alpha+k\sin\alpha)=kmg,$

hence

$t_0=\frac{kmg}{\beta(\cos\alpha+k\sin\alpha)}.$

For $0\le t\le t_0$ the body remains at rest and

$v(t)=0.$

Only for $t>t_0$ may the kinetic equation of motion be used.

Physical Principles

Newton’s second law applies separately to horizontal and vertical directions,

$m a_x=\sum F_x,\qquad m a_y=\sum F_y.$

The friction force during sliding has magnitude

$F_f=kN,$

and acts opposite to the direction of motion.

The normal force follows from vertical force balance while contact persists,

$N+F(t)\sin\alpha-mg=0.$

Velocity follows from integration of acceleration,

$v(\tau)=\int a_x(t),dt.$

Derivation

The applied force has horizontal and vertical components

$F_x(t)=\beta t\cos\alpha,\qquad F_y(t)=\beta t\sin\alpha.$

Vertical equilibrium gives the normal force

$N(t)=mg-\beta t\sin\alpha.$

The kinetic friction magnitude is therefore

$F_f(t)=k\left(mg-\beta t\sin\alpha\right).$

For $t>t_0$ the motion is along the horizontal direction of the applied force, so friction acts opposite to the $x$ direction. The horizontal equation of motion becomes

=\beta t\cos\alpha-k\left(mg-\beta t\sin\alpha\right).$$Expanding terms,$$m\frac{dv}{dt} =\beta t\cos\alpha+k\beta t\sin\alpha-kmg.$$Grouping the time dependent terms,$$m\frac{dv}{dt} =\beta t(\cos\alpha+k\sin\alpha)-kmg.$$Using the definition of $t_0$,$$kmg=\beta t_0(\cos\alpha+k\sin\alpha),$$so$$m\frac{dv}{dt} =\beta(\cos\alpha+k\sin\alpha)(t-t_0).$$Dividing by $m$,$$\frac{dv}{dt} =\frac{\beta}{m}(\cos\alpha+k\sin\alpha)(t-t_0).$$Since $v(t_0)=0$, integration from $t_0$ to $\tau$ gives$$v(\tau) =\frac{\beta}{m}(\cos\alpha+k\sin\alpha) \int_{t_0}^{\tau}(t-t_0),dt.$$Evaluating the integral,$$\int_{t_0}^{\tau}(t-t_0),dt =\left[\frac{(t-t_0)^2}{2}\right]_{t_0}^{\tau} =\frac{(\tau-t_0)^2}{2}.$$Hence$$v(\tau) =\frac{\beta(\cos\alpha+k\sin\alpha)}{2m} (\tau-t_0)^2.$$Substituting $t_0$,$$v(\tau) =\frac{\beta(\cos\alpha+k\sin\alpha)}{2m} \left( \tau-\frac{kmg}{\beta(\cos\alpha+k\sin\alpha)} \right)^2, \qquad \tau\ge t_0.$$Expanding this expression gives$$v(\tau) =\frac{\beta(\cos\alpha+k\sin\alpha)}{2m}\tau^2 -kg,\tau +\frac{k^2m g^2} {2\beta(\cos\alpha+k\sin\alpha)}.

The additional constant term is precisely the correction required by the initial interval during which the body remained at rest.

Result

The velocity after time $\tau$ is

$$\boxed{ v(\tau)= \begin{cases} 0, & 0\le \tau\le t_0, \[1.2ex] \dfrac{\beta(\cos\alpha+k\sin\alpha)}{2m} (\tau-t_0)^2, & \tau\ge t_0, \end{cases} }$$

where

$$t_0= \frac{kmg} {\beta(\cos\alpha+k\sin\alpha)}.$$

This may also be written for $\tau\ge t_0$ as

$$\boxed{ v(\tau)= \frac{\beta(\cos\alpha+k\sin\alpha)}{2m}\tau^2 -kg,\tau +\frac{k^2m g^2} {2\beta(\cos\alpha+k\sin\alpha)}. }$$

The parameter $\beta$ has units $\mathrm{N/s}=\mathrm{kg,m,s^{-3}}$, so

$$\frac{\beta}{m}(\tau-t_0)^2 \sim \frac{\mathrm{kg,m,s^{-3}}}{\mathrm{kg}} \cdot \mathrm{s}^2 = \mathrm{m,s^{-1}},$$

which confirms the correct dimensions for the velocity.

Sanity Checks

Setting $k=0$ gives

$t_0=0,$

and therefore

$v(\tau)=\frac{\beta\cos\alpha}{2m}\tau^2,$

which is the frictionless result.

Setting $\alpha=0$ yields

$t_0=\frac{kmg}{\beta},$

and

$$\qquad \tau\ge t_0,$$

which corresponds to a linearly increasing horizontal force that first overcomes friction and then accelerates the body.

The expression for the normal force,

$N(t)=mg-\beta t\sin\alpha,$

requires

$N(t)>0.$

Thus contact with the surface is maintained only while

$t<\frac{mg}{\beta\sin\alpha},$

when $\sin\alpha>0$. The derivation is valid within this interval.