Some Particular Forces — Question 3

Question 3

A contact force cannot pull a passenger down

A 40 kg passenger is initially touching an elevator floor. The floor is prescribed to accelerate downward at 12 m/s$^2$. Take $g=10$ m/s$^2$, neglect air and assume no straps or ceiling contact.

1. What normal force would shared acceleration require?

2. Is that value physically possible for an ordinary floor?

3. Find the passenger’s actual acceleration and separation from the floor after 0.5 s, starting with zero relative velocity.

Question 3 - Solution

A contact force cannot pull a passenger down

(a) Try the contact constraint.

$$N-mg=ma_f\ \Rightarrow\ N=40(10-12)=-80\,\mathrm N$$ Upward positive gives $N-400=40(-12)$, hence $N=-80$ N.

(b) Reject an inadmissible contact solution.

$$N\ge0\quad\text{(unilateral contact)};\qquad\boxed{N_{\mathrm{actual}}=0}$$ A floor normal can push upward but cannot pull downward. Negative N means the assumed continued contact fails. The correct detached state has N=0.

(c) Evolve the detached bodies.

$$a_p=-10,\quad a_{p/f}=-10-(-12)=2,\quad\Delta y_{p/f}=\tfrac12(2)(0.5)^2=0.25\,\mathrm m$$ The passenger falls at $-10$ m/s$^2$ while the floor accelerates at $-12$ m/s$^2$. Relative acceleration is $+2$ m/s$^2$, so the passenger is $\tfrac12(2)(0.5)^2=0.25$ m above the floor after 0.5 s.