hi friends and saw problem on a daily Mertz principle see what is given in problem find P required to acetate the block shown in figure below at two point five meter per second square that means no acceleration of box or block is given which is two point five meter per second square for this acceleration what should be magnitude of force P first step will draw Fb D of this block it will draw Fb D of block always in this D alembert’s principle Freebody diagram is very very important once will represent Freebody diagram applying delivers principle in one step we get solution of problem now what is weight of this block or mass of the block the mass of this block is given here as what 40 kg is mass which is given already in problem that means weight of this box even represent which is 40 into acceleration due to gravity G that is what nine point eight one what will get your normal reaction in upward direction that force P what is acting here is inclined at an angle of 15 degree with horizontal this one is 15 degree due to component of force P block definitely accelerates in right to a direction that means this is a direction of acceleration a if acceleration is a right toward inertia of force leftward mass is given as a 40 kg therefore 40 a as body accelerated right toward frictional force acts in leftward direction where this frictional force if I represent like this μ is given as a point three and this becomes RN this one is complete free body diagram now first we’ll find out here Rn applying equation along Y axis but RN direct is not possible which is in unknown for speak in terms of falsely summation of fy equal to 0 assuming say upward force is positive R an upward next one what will get upward component top P angle with horizontal is given vertical component becomes sine of this angle 15 degree then minus 40 G means 9.81 equal to 0 therefore this RN equal to this 14 to 9.81 is a 390 2.4 – this piece sine of 15° that means this is value of RF say equation number 1 next we’ll apply equation along x-axis summation of fx equal to 0 assuming rightward force is positive now component of P what we’ll get here P cos of 15 degrees frictional force minus 0.3 into RF and inertia force is what 40 into a equal to 0 now a is given in problem that is 2 point 5 meter per second square will substitute in next P cos of 15 minus 0.3 Rn value I’ll substitute from Equation number one three ninety two point four minus P sine of 15 minus 40 into isolation is two point five which is equal to 0 that means equation is in terms of only one unknown that is P it will calculate using calculator p value comes out to be two hundred and eight point six three Newton that means a problem is very easy because a problem is a single body problem therefore simply it will apply D alembert’s principle easily solution is possible thank you