Friction — Question 8

Question 8

A changing pull produces a changing static friction force

A 4 kg crate starts at rest on a fixed horizontal floor. Static friction can supply up to 12 N. A horizontal pull is $P(t)=(4\ \mathrm{N/s})t$ rightward for $0\le t\le4$ s. Once sliding begins, friction has magnitude 8 N. Equality at the static limit is treated as impending slip.

1. Give friction and acceleration before the threshold.

2. Find the threshold time and the right-hand acceleration just after slipping starts.

3. Find velocity at 4 s using the time-dependent acceleration.

Question 8 - Solution

A changing pull produces a changing static friction force

(a) Match a subthreshold pull.

Before sliding, force balance requires $$\boxed{f_s(t)=-(4\ \mathrm{N/s})t,\qquad a(t)=0}\quad(0\leq t\leq3\ \mathrm s).$$ The endpoint is the stipulated impending-slip state, where $|f_s|=12$ N.

(b) Locate the change of model.

The static limit is reached at $$4t_*=12\Longrightarrow\boxed{t_*=3\ \mathrm s}.$$ Immediately after slipping, the resistance drops to 8 N: $$\boxed{a(3^+)=\frac{12-8}{4}=1\ \mathrm{m/s^2}}.$$

(c) Integrate over the sliding interval only.

During sliding, $a(t)=t-2$ in SI numerical units. With $v(3)=0$, $$v(4)=\int_3^4(t-2)\,dt=\left[\frac{t^2}{2}-2t\right]_3^4.$$ $$\boxed{v(4)=1.5\ \mathrm{m/s}\text{ rightward}}.$$ The first three seconds contribute no velocity change; the finite force step does not produce an instantaneous velocity jump.