Question

Investigation 1 - Connected Particles (Maximum Mark 45) A block mass M kg is resting on the horizontal top of a cube. The coefficient of friction between the block and the cube is u. Two strings are attached to the sides of the block. One string passes over a pulley and is attached to an object of mass m kg which is hanging freely a height h metres above the ground. The other string passes over another pulley on the opposite side of the cube and is attached to an object of mass m2 kg which is also hanging freely a height h metres above the ground. The system is released from rest. You may assume that the cube is large enough that the block never hits one of the pullies. You should make all the usual modelling assumptions needed to simplify the system. Task 1 - Modelling (Maximum Mark 10) Draw a detailed diagram to model the problem and state any modelling assumptions you are making.

Answer 1

M au mig- T= mia, 0 m2 . mig Ti - Tz- ung=ma My az= T2-M2g - Tz Ti H here a= a = a2) +M₂ja nig-T, +72 - M2g = (mitma ja (m, - m2) g - (Hi-T2) from eqn (m.-m.) g- umg - ma = u mg - ma = (mitm) a mi g-m2 g - umg my tm, tmy a

dv dt = -m2-4 mj g mitmyt my mo nama-um) [ +3. t. mi + m₂ +M3 va m, - m2 - um jg (t) (m, +M2+ m₂ is mi = mat um then the system will always rest then the object (mass) if mi> met UM mi start going downward. 07 m₂ > m, tum then down word. mass m2 start going

Task - 3 m, = 20 M=40 kg M=o.l h=50cm kg M₂ = 15kg a = mi M2 - M umjg mi + m2 + M a = 20- 15 - 408 0.1 40 + 20 tis 40 X 0:1.Jx1° a= x 10 5 - 4 75 a = 10 75 = 2 15 mig2 and V = m, -M2-um) gt (M, tm.tm V 3 t using 2 15 v2 = u2+2 as v= ut at IT (2) 2 = (2+ 2 *( ) (7)

()() 2 (2) ر) 2 2 ) = IS t = 2S t 2013 second In this case 0 = ا and V - 22