A small ball of mass m = 20 g is held against the front of a…
A small ball of mass $m = 20\ \mathrm{g}$ is held against the front of a vertical plate that moves horizontally with constant acceleration $a_1 = 4g$. (a) What force does the ball exert on the plate? (b) If the coefficient of friction between the ball and the plate is $\mu = 0.5$, at what acceleration $a_2$ does the ball start to slide down? (c) If the plate's acceleration is $a_3 = g$, how long does it take the ball to slide down a height $h = 10\ \mathrm{cm}$?
Explanation of Method
A ball held against an accelerating plate.
Step 1: (a) Force on the plate
The plate pushes the ball forward with $N = ma_1 = 4mg$. Friction must hold the ball up, $f = mg$; it can, since $\mu N = 2mg \gt mg$. The ball presses on the plate with the reaction to both:
\[ F = \sqrt{N^2 + f^2} = mg\sqrt{16 + 1} = 0.02 \cdot 9.81 \cdot 4.123. \]
\[F = mg\sqrt{17} \approx 0.81\ \mathrm{N}\]
(Perpendicular to the plate 0.785 N, and 0.196 N downward along it.)
Step 2: (b) When it starts to slide
The ball stays put as long as the largest friction can carry its weight: $\mu m a \ge mg \;\Longleftrightarrow\; a \ge \frac{g}{\mu}$.
\[a_2 = \frac{g}{\mu} = 2g \approx 19.6\ \mathrm{m/s^2}\]
For any smaller acceleration the ball slides down.
Step 3: (c) Sliding down at $a_3 = g$
Now $N = mg$ and sliding friction is $\mu mg = 0.5\,mg$ upward, so the ball slides down with
\[ a_{\downarrow} = g - \mu a_3 = 4.905\ \mathrm{m/s^2}, \]
\[ h = \tfrac12 a_{\downarrow} t^2. \]
\[ t = \sqrt{\frac{2h}{a_{\downarrow}}} = \sqrt{\frac{0.2}{4.905}}. \]
\[t \approx 0.20\ \mathrm{s}\]
Result
\[\boxed{\,F = mg\sqrt{17} \approx 0.81\ \mathrm{N}\,}\]
\[\boxed{\,a_2 = \frac{g}{\mu} = 2g \approx 19.6\ \mathrm{m/s^2}\,}\]
\[\boxed{\,t \approx 0.20\ \mathrm{s}\,}\]
Remark
About the original solution. Only part (a) was solved, and its result $mg\sqrt{17}$ is correct (≈ 0.81 N), though it mixed the inertial force with the normal force in the reasoning. Parts (b) and (c) were missing; they are worked out above.