Question

Question 5 P 0.8 m A rigid uniform square block of mass m = 60 kg is resting on the ground as shown above. A force P is applied to the block at 0.8 m above the ground. The coefficient of kinetic friction between the ground and the block is given by uk = 0.2. It can be shown that the free-body diagram for the situation above is as given below, where F is the friction, N is the normal force and d locates the line of action of N. Vila (a) Show from first principles that, for a square block with uniformly distributed mass, the moment of inertia about the centre of gravity is given by I=ml where L is the length of the sides of the square block and m is the total mass (6 marks). (b) Write the equations, valid for arbitrary P, for the acceleration of centre of mass (ām, āy) and the angular acceleration a anticlockwise is positive) for the instant shown above (8 marks). (c) If the block slides on the ground without tipping over, determine the maximum P in this case (6 marks)

Answer 1

Given that,

P is the applied force,

mg is the weight of the block acting downwards,

N is the normal force acting at a distance d from the axis,

F is the friction force acting on the block

1 Answen O8m m ma Foee bady diagoom may Im mat a) Detemining the cenderof goauty moment of inetiaabut the I mat) =m (212

mL2 TRe momat of inestHa f the bleck oabout the center of gauity fs mL2 The bleck given hoe is sauaae, thoehoa a- b=L Is used n the calculat fen of mass momentot fnertfa bApplying tha dynamic equsltbatum cerdlition along the heeizontal diration EFe mae P-F max PresNmaz Apply fng the dyhamic eauilibatum Condition alena the vestHcas direction. Fmay N-ma may N-ma m

Applytthemement dynamic eauilibadum cordlition about the cendoeid Plo3)Nd - F (o.5)=I PCo.3+ Nd-FCO.5) Tre eauctfons for the accelorattons ce N-mg esnd m -PCo.3Nd F(0-5) 9 eteomining the maximum? if the bleck slides on the qsound without +rpping oven. Foethe above cse Faom eauation and@ N-mg m Nmg

2 PCOS+ Nd-F(0.s) -P Co.3) +Nd- FCOS) 0 The momont at a dislance d fom 4he axis tor hetipping PO.smgcd) o PCo s) mg Oubstituding (o-8) for d ma uKN Por F, ad mfor N in equoion -N5) mg OPles)0sP)-mgo) Subetitus ing 6okg for m, O.2 4or uCnd 9.81mle2for q. PCo.3 J+ Co. 8)P-o-2 X60 x4.81 x9 O P-17.72 N The maximu m P if the Heck slidles on the without f ppiíng ove ts li,42N