A man of mass m = 85 kg stands on the floor of a tram moving…
A man of mass $m = 85\ \mathrm{kg}$ stands on the floor of a tram moving with constant acceleration $a = 3\ \mathrm{m/s^2}$. The coefficient of friction between the floor and his shoes is $\mu = 0.1$. Find: (a) the inertial force on the man; (b) the force he must supply by holding a handle so as not to slide; (c) how hard he would have to press down on the floor (without a handle) so as not to slide.
Explanation of Method
Standing in an accelerating tram.
Step 1: (a) Inertial force
In the tram's frame the man feels a backward inertial force equal to the force needed to accelerate him with the tram: \[F_i = ma = 85 \cdot 3 = 255\ \mathrm{N}\]
Step 2: (b) Handle force
Friction from the floor can give at most
\[ F_{f,\max} = \mu mg = 0.1 \cdot 85 \cdot 9.81 \approx 83.4\ \mathrm{N}. \]
The rest must come from the handle: $F_h = ma - \mu mg \approx 255 - 83.4 = 171.6\ \mathrm{N}$.
Step 3: (c) Pressing on the floor
Without a handle, friction alone must give 255 N. Friction is $\mu$ times the force pressing on the floor, so that force $N$ must be
\[ \mu N \ge ma \;\Longrightarrow\; N \ge \frac{ma}{\mu} = \frac{255}{0.1}. \]
\[\begin{gathered} N \ge 2550\ \mathrm{N} \\ \text{(about 1716 N more than his weight)} \end{gathered}\]
He could only do that by bracing himself against something above him, e.g. pushing up against the ceiling.
Result
\[\boxed{\,F_i = ma = 85 \cdot 3 = 255\ \mathrm{N}\,}\]
\[\boxed{\,F_h = ma - \mu mg \approx 255 - 83.4 = 171.6\ \mathrm{N}\,}\]
\[\boxed{\,\begin{gathered} N \ge 2550\ \mathrm{N} \\ \text{(about 1716 N more than his weight)} \end{gathered}\,}\]
Remark
About the original solution. Parts (a) and (b) are correct. In (c) it gave just his weight (833.85 N), but pressing with only his weight gives the 83.4 N of friction from (b), which is not enough. The question asks what pressing force prevents sliding: 2550 N in total, three times his weight.