Question

In the figure, a slab of mass m₁ = 40 kg rests on a frictionless floor, and a block of mass m₂ = 8 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force F of magnitude 105 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?

Answer 1

If you liked the answer then give a thumb up.

m2 alls - 0.6 Figure is not given as said in the question But according to statement, I suppose the figure will be like this: criven, mayotg mg = 81 slab,m, elk= 0.4 q=1050 (a) free-body diagram of block ma: frictional force pe med Gtatic friction Kinetic friction (fw) for deg. Na fx lloc. Na - 0.6x Mag -0.4x mag Co.6x8x9.8 -0.4x8x9.8 -- 47.04N = 31.36N Cince q=10507fs=47.04N so body will accelerate and fix will act. É-fix = mga -) a=9.205 m2 22 in unit-vector notation, ar - 9. 205 m/s2 i 1 22 (M = n2a 105 - 31.36=84 (b) freebody diagram of slab mit for any for = mia fk ama a= fk - - ma 31.36 40 -0.784 on 1 22 in vector form, a 0.784 m 2