The Requirements of Equilibrium — Question 8

Question 8

The floor resultant must lie within the contact footprint

A rectangular block of weight $200\,\mathrm N$ has base width $0.80\,\mathrm m$. A horizontal $40\,\mathrm N$ force is applied $1.0\,\mathrm m$ above the floor. Its center of gravity is vertically above the base midpoint. Assume static equilibrium is possible.

1. Find the normal resultant and its horizontal offset from the base midpoint.

2. Find the required static-friction coefficient and test the normal-resultant location.

Question 8 - Solution

The floor resultant must lie within the contact footprint

(a) Derivation and calculation

Vertical balance gives $$\boxed{N=W=200\ \mathrm N}.$$ About the base midpoint, the horizontal push has clockwise torque $-Fh$. Thus $$Nx-Fh=0\quad\Longrightarrow\quad \boxed{x=\frac{(40)(1.0)}{200}=.20\ \mathrm m}.$$ The normal resultant shifts toward the pushed side.

(b) Result and interpretation

Horizontal balance requires friction $f=-40$ N. Its admissibility condition is $$|f|\le\mu_sN\quad\Longrightarrow\quad\boxed{\mu_s\ge.20}.$$ The footprint condition is $$|x|=.20\ \mathrm m<\frac{.80}{2}=.40\ \mathrm m.$$ Thus the normal resultant is inside the base; tipping is not yet limiting, provided sufficient friction exists.

Physical check

The normal-pressure moment is $200(.20)=40$ N m, exactly opposing the push moment.