Question

The 75-kg man pushes on the 150 kg crate with a horizontal force F The coefficients of static and kinetic friction between the crate and the surface are us=0.3 and uk = 0.18, and the coefficient of static friction between the man's shoes and the surface is u's=0.8. (Figure 1) What is the greatest acceleration the man can give the crate if the man's acceleration is zero at this instant? (Ctri)

Answer 1

Applying conditions of equillibrium on man I Fy = 0 > Nm - mg -O i Nm > mg = 75 X 9.81 = 735.75 N E Fx -0 > F-.f = 0 f = F = Us Nm = 0.8 x 735.75 . f = F = 5886 N is i max. force applied by the man F = 588.6 N Now for bona crate: Ify so Nc- Mg = 0 = Ne= Mg i No 150 x 9.81 -14715 Ify song EF- F - f = 0 F = fc = ls Nc = 0.3 x 1471-5 .. F = 441.45 N < 588.6 N . F Ms Ne - crate will move with maximum acceleration

I Fa= Ma FMR Ne = Ma 588.6 - 0.18 X 150x 9.81 = 150 x al 588.6- 264.87= 150 a sa = 323.73 150 i, a = 2.16 m/s2