Question

4- i- A block is moving on the frictionless road as shown. It is passing by points A, B and C at velocities of va, VB and vc respectively. The radii of the circular parts are R. Write the centripetal accelerations a at points A, B, C and draw the vectors a (where necessary). (6 pts) fi- a- Two blocks of masses my and m2 are connected by a massless and inextensible rope, passing over a massless pulley, my being on the inclined plane of angle 0 as shown. The coefficient of friction for the inclined part is u but the horizontal part is frictionless. Drawing the free-body diagrammes and writing the equations of motion, find the maximum value of the force F acting horizontally on m2 such that the blocks are still at rest. (9 pts) m2 V u=0 u a- Now a horizontal force of 2F (twice the force in part ii- a) is exerted on m2. What is the acceleration a of the system of blocks? (10 pts) a=? 2F

Answer 1

Ocentibetal acceleration is always direction towards Centre of curvature . aco a Ne mogcoso ifs MN Ni T= migsino tf > Tam, gsino + remigcoso m migsinsk migroso f a Nz F =T mah F - | F = migsinot fm, gcose hence maximum value of force F is (migsino thum, g cose) b assuming acceleration of system of block is a Nz miz →2F T N₂= mg ta mg 2F - T = mza N2- mog cose T f = f No Hm, g-coso хо x T-mig sino-f= my a m gcoso mgsino - T-mig sino-uma gcoso = ma 2 adding eq @ & ego 2F-Tty-mig sino-Himig coso = m, atm, a * 2F - m, gsino-um, gcoso = (mitme) a - 2 (mi gsino + rum, geoso) -mig sino - umi gcoso = mitme) a T mig sino tamigcoso = (m, tm₂) a : migsino tumigcoso a- mit m2 Scanned with CamScanner