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202209271004 Exercise 4.5 (Q31)

An elevator accelerates downward at 2.4\,\mathrm{m\, s^{-2}}. What force does the elevator’s floor exert on a 52\,\mathrm{kg} passenger?

Extracted from R. Wolfson. (2016). Essential University Physics.

Roughwork.

Take downward positive, \downarrow\textrm{+ve}. By Newton’s 2nd Law,

\mathbf{F}_{\textrm{net}}=m\mathbf{a}.

Given m=52\,\mathrm{kg}. Let F be the unknown magnitude of force:

\begin{aligned} \mathbf{a} & = 2.4\,\hat{\mathbf{k}} \\ \mathbf{F}_\textrm{net} & = -F\,\hat{\mathbf{k}}+mg\,\hat{\mathbf{k}} \\ & = (-F+mg)\,\hat{\mathbf{k}} \\ \end{aligned}

Then,

\begin{aligned} -F+mg & = 2.4m \\ F & = (g-2.4)m \\ & = (9.81-2.4)(52) \\ & = 385\,\mathrm{N}\qquad \textrm{(3 s.f.)} \end{aligned}

\therefore The force the elevator’s floor exerts on the passenger is 385\,\mathrm{N} upward.