Equilibrium — Question 7

Question 7

A laboratory equilibrium can disappear in an accelerating frame

A $3\,\mathrm{kg}$ block remains at rest on a level laboratory table with no horizontal real force. An observer accelerates right at $2\,\mathrm{m/s^2}$ relative to the laboratory.

1. Find the block acceleration in the observer frame.

2. What fictitious force restores the form of Newton's equation in that frame, and is the block static there?

Question 7 - Solution

A laboratory equilibrium can disappear in an accelerating frame

(a) Derivation and calculation

For a nonrotating translating frame with origin position $X(t)$, $$x'=x-X(t),\qquad a'=a-\ddot X=a-A.$$ With $a=0$ and $A=+2\ \mathrm{m/s^2}$, $$\boxed{a'=-2\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

The frame equation is $$ma'=F_{\mathrm{real}}-mA.$$ Therefore the required inertial term is $$\boxed{F_{\mathrm{fict}}=-mA=-(3)(2)=-6\ \mathrm N}.$$ The block is not static in the accelerating frame. If the observer is initially at rest relative to the laboratory, $$x'(t)=x'_0-\tfrac12At^2.$$

Physical check

The fictitious force is a coordinate correction, not another interaction acting on the block.